Matrix Theory and Systems of Equations

This chapter comprehensively covers the foundational aspects of matrix theory and systems of linear equations, essential for advanced mathematical understanding. Mastery of determinants, matrix inversion, and solving linear systems is critical for success in the CUET PG examination, as these topics are consistently featured.

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Chapter Contents

| # | Topic |
|---|-------|
| 1 | Determinants |
| 2 | Solutions of Systems of Linear Equations |
| 3 | Rank and Inverse of a Matrix |

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We begin with Determinants.

Part 1: Determinants

Determinants are scalar values associated with square matrices, providing critical insights into matrix properties such as invertibility and the solvability of linear systems. We observe their fundamental role in various areas of linear algebra and multivariate calculus.

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Core Concepts

1. Definition of a Determinant

The determinant, denoted $\det(A)$ or $|A|$, is a scalar value associated with a square matrix $A$. Its calculation method varies with the matrix order. For a $1 \times 1$ matrix $A = [a]$, we define $\det(A) = a$.

For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is given by $ad - bc$. For higher order matrices, we employ cofactor expansion.

Quick Example: $2 \times 2$ Determinant

Consider the matrix $A = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$. We calculate its determinant.

Step 1: Apply the $2 \times 2$ determinant formula.

>

Step 2: Perform the arithmetic.

>

Answer: $2$

:::question type="MCQ" question="Let $M = \begin{bmatrix} 5 & -2 \\ 3 & 4 \end{bmatrix}$. What is $\det(M)$?" options=["$14$","$26$","$6$","$-26$"] answer="$26$" hint="Apply the standard formula for a $2 \times 2$ determinant." solution="Step 1: Identify the elements $a, b, c, d$.
>

Step 2: Apply the formula $\det(M) = ad - bc$.
>
>
>
The determinant of $M$ is $26$."
:::

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2. Minors and Cofactors

For an $n \times n$ matrix $A$, the minor $M_{ij}$ of an element $a_{ij}$ is the determinant of the $(n-1) \times (n-1)$ matrix obtained by deleting the $i$-th row and $j$-th column of $A$. The cofactor $C_{ij}$ of $a_{ij}$ is defined as $C_{ij} = (-1)^{i+j} M_{ij}$.

The determinant of an $n \times n$ matrix can be computed by cofactor expansion along any row $i$: $\det(A) = \sum_{j=1}^{n} a_{ij} C_{ij}$, or along any column $j$: $\det(A) = \sum_{i=1}^{n} a_{ij} C_{ij}$.

Quick Example: Minors and Cofactors for a $3 \times 3$ Matrix

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$. We compute the minor $M_{11}$ and cofactor $C_{11}$.

Step 1: To find $M_{11}$, delete the first row and first column.

>

Step 2: Calculate the determinant of the resulting $2 \times 2$ matrix.

>

Step 3: Calculate the cofactor $C_{11}$ using $C_{ij} = (-1)^{i+j} M_{ij}$.

>

Answer: $M_{11} = -3$, $C_{11} = -3$.

:::question type="MCQ" question="For the matrix $B = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -1 & 4 \\ 0 & 5 & 6 \end{bmatrix}$, what is the cofactor $C_{23}$?" options=["$5$","$-5$","$1$","$-1$"] answer="$-5$" hint="First find the minor $M_{23}$ by removing row 2 and column 3, then apply the cofactor formula." solution="Step 1: To find $M_{23}$, remove the second row and third column of $B$.
>

Step 2: Calculate $M_{23}$.
>
Step 3: Calculate $C_{23}$ using $C_{ij} = (-1)^{i+j} M_{ij}$.
>
The cofactor $C_{23}$ is $-5$."
:::

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3. Properties of Determinants

Determinants possess several properties crucial for efficient calculation and theoretical understanding. We present these properties, often without formal proof, but with illustrative examples.

3.1. Transpose Property

The determinant of a matrix remains unchanged when its rows and columns are interchanged. That is, $\det(A^T) = \det(A)$.

Quick Example:

Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$.
Then $A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$.

Step 1: Calculate $\det(A)$.

Step 2: Calculate $\det(A^T)$.

Answer: $\det(A) = \det(A^T) = -2$.

:::question type="MCQ" question="If $A$ is a $3 \times 3$ matrix such that $\det(A) = 5$, what is $\det(A^T)$?" options=["$5$","$-5$","$1/5$","$25$"] answer="$5$" hint="Recall the property relating the determinant of a matrix to the determinant of its transpose." solution="According to the property of determinants, $\det(A^T) = \det(A)$.
Given $\det(A) = 5$.
Therefore, $\det(A^T) = 5$."
:::

3.2. Product Property

The determinant of a product of two square matrices is the product of their individual determinants. This implies $\det(AB) = \det(A)\det(B)$.

Quick Example:

Let $A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}$. We verify $\det(AB) = \det(A)\det(B)$.

Step 1: Calculate $\det(A)$ and $\det(B)$.

Step 2: Calculate $AB$.

Step 3: Calculate $\det(AB)$.

Step 4: Compare $\det(AB)$ with $\det(A)\det(B)$.

Answer: $\det(AB) = -3$ and $\det(A)\det(B) = -3$, confirming the property.

:::question type="MCQ" question="If $P$ and $Q$ are $3 \times 3$ matrices with $\det(P) = 2$ and $\det(Q) = 3$, what is $\det(PQ)$?" options=["$5$","$6$","$1$","$0$"] answer="$6$" hint="Recall the property for the determinant of a matrix product." solution="Using the determinant product property, $\det(PQ) = \det(P)\det(Q)$.
Given $\det(P) = 2$ and $\det(Q) = 3$.

The determinant of $PQ$ is $6$."
:::

3.3. Inverse Property

If a matrix $A$ is invertible, its determinant is non-zero, and the determinant of its inverse $A^{-1}$ is the reciprocal of the determinant of $A$. Specifically, $\det(A^{-1}) = \frac{1}{\det(A)}$.

Quick Example:

Let $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$. We find $\det(A^{-1})$.

Step 1: Calculate $\det(A)$.

Step 2: Apply the inverse determinant property.

Answer: $\det(A^{-1}) = 1$.

:::question type="MCQ" question="Given a matrix $A = \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}$, what is $\det(A^{-1})$?" options=["$1$","$-1$","$1/11$","$-1/11$"] answer="$1$" hint="First calculate $\det(A)$, then use the property for the determinant of an inverse matrix." solution="Step 1: Calculate the determinant of $A$.

Step 2: Use the property $\det(A^{-1}) = \frac{1}{\det(A)}$.

The determinant of $A^{-1}$ is $1$."
:::

3.4. Scalar Multiplication Property

If $A$ is an $n \times n$ matrix and $k$ is a scalar, then $\det(kA) = k^n \det(A)$. Each of the $n$ rows is multiplied by $k$, thus multiplying the determinant by $k$ $n$ times.

Quick Example:

Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $k=3$. We find $\det(3A)$.

Step 1: Calculate $\det(A)$.

Step 2: Apply the scalar multiplication property. $A$ is a $2 \times 2$ matrix, so $n=2$.

Answer: $\det(3A) = -18$.

:::question type="MCQ" question="If $A$ is a $4 \times 4$ matrix with $\det(A) = 5$, what is $\det(2A)$?" options=["$10$","$20$","$40$","$80$"] answer="$80$" hint="Remember that for an $n \times n$ matrix $A$ and scalar $k$, $\det(kA) = k^n \det(A)$." solution="Given $A$ is a $4 \times 4$ matrix, so $n=4$.
Given $\det(A) = 5$.
We need to find $\det(2A)$.
Using the property $\det(kA) = k^n \det(A)$:

The determinant of $2A$ is $80$."
:::

3.5. Determinant of Triangular Matrices

The determinant of a triangular matrix (upper triangular, lower triangular, or diagonal) is the product of its diagonal elements.

Quick Example:

Let $A = \begin{bmatrix} 2 & 0 & 0 \\ 3 & 4 & 0 \\ 5 & 6 & 7 \end{bmatrix}$. This is a lower triangular matrix.

Step 1: Identify the diagonal elements.

Step 2: Multiply the diagonal elements.

Answer: $56$.

:::question type="MCQ" question="What is the determinant of the matrix $P = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}$?" options=["$120$","$24$","$10$","$0$"] answer="$24$" hint="Recognize that $P$ is an upper triangular matrix. The determinant of a triangular matrix is the product of its diagonal elements." solution="The matrix $P$ is an upper triangular matrix. For any triangular matrix (upper, lower, or diagonal), its determinant is the product of its diagonal elements.
The diagonal elements are $1, 4, 6$.

The determinant of $P$ is $24$."
:::

3.6. Effect of Elementary Row/Column Operations

Elementary row operations affect the determinant in specific ways:
* Swapping two rows/columns: Multiplies the determinant by $-1$.
* Multiplying a row/column by a scalar $k$: Multiplies the determinant by $k$.
* Adding a multiple of one row/column to another: Does not change the determinant.

Quick Example:

Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, with $\det(A) = -2$.

Step 1: Swap $R_1 \leftrightarrow R_2$ to get $A'$.

Observe $\det(A') = -\det(A)$.

Step 2: Multiply $R_1$ by $k=5$ in $A$ to get $A''$.

Observe $\det(A'') = 5 \det(A)$.

Step 3: Add $2R_1$ to $R_2$ in $A$ to get $A'''$.

Observe $\det(A''') = \det(A)$.

Answer: The determinant changes as described by the properties.

:::question type="MCQ" question="If a $3 \times 3$ matrix $A$ has $\det(A) = 7$, and matrix $B$ is obtained from $A$ by swapping the first and third rows, then what is $\det(B)$?" options=["$7$","$-7$","$1/7$","$49$"] answer="$-7$" hint="Recall how row swaps affect the determinant." solution="Swapping two rows of a matrix multiplies its determinant by $-1$.
Given $\det(A) = 7$.
Since $B$ is obtained by swapping two rows of $A$,

The determinant of $B$ is $-7$."
:::

3.7. Zero Determinant Conditions

A matrix has a determinant of zero if:
* It has a row or column consisting entirely of zeros.
* It has two identical rows or columns.
* It has two proportional rows or columns.
* Its rows (or columns) are linearly dependent. This is equivalent to saying the matrix is singular (non-invertible).

Quick Example:

Consider $A = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$. It has a row of zeros. $\det(A) = (1)(0) - (2)(0) = 0$.

Consider $B = \begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix}$. It has two identical rows. $\det(B) = (1)(2) - (2)(1) = 0$.

Consider $C = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$. Row 2 is $2 \times$ Row 1. $\det(C) = (1)(4) - (2)(2) = 0$.

Answer: All matrices have a determinant of $0$.

:::question type="MCQ" question="Which of the following matrices has a determinant of zero?" options=["$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$","$\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$","$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$","$\begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix}$"] answer="$\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$" hint="Check each matrix for properties that lead to a zero determinant, such as proportional rows/columns." solution="We calculate the determinant for each option:

$\det \left( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \right) = (1)(4) - (2)(3) = 4 - 6 = -2 \neq 0$

$\det \left( \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} \right) = (2)(2) - (4)(1) = 4 - 4 = 0$. (Rows are proportional: $R_1 = 2R_2$)

$\det \left( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = (1)(1) - (0)(0) = 1 \neq 0$

$\det \left( \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix} \right) = (3)(5) - (1)(2) = 15 - 2 = 13 \neq 0$

The matrix $\begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix}$ has a determinant of zero."
:::

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4. Adjoint of a Matrix

The adjoint of a square matrix $A$, denoted $\operatorname{adj}(A)$, is the transpose of the cofactor matrix of $A$. That is, if $C$ is the matrix where $C_{ij}$ is the cofactor of $a_{ij}$, then $\operatorname{adj}(A) = C^T$.

A fundamental property relating a matrix, its adjoint, and its determinant is $A \operatorname{adj}(A) = \operatorname{adj}(A) A = \det(A) I$, where $I$ is the identity matrix.

Another key property is $\det(\operatorname{adj}(A)) = (\det(A))^{n-1}$ for an $n \times n$ matrix $A$.

Quick Example: Adjoint of a $2 \times 2$ Matrix

For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the cofactor matrix $C = \begin{bmatrix} d & -c \\ -b & a \end{bmatrix}$.
Then $\operatorname{adj}(A) = C^T = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

Let $A = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$.

Step 1: Calculate the cofactors.
$C_{11} = (-1)^{1+1} \det([2]) = 2$
$C_{12} = (-1)^{1+2} \det([4]) = -4$
$C_{21} = (-1)^{2+1} \det([1]) = -1$
$C_{22} = (-1)^{2+2} \det([3]) = 3$

Step 2: Form the cofactor matrix $C$.

Step 3: Transpose $C$ to find $\operatorname{adj}(A)$.

Answer: $\operatorname{adj}(A) = \begin{bmatrix} 2 & -1 \\ -4 & 3 \end{bmatrix}$.

:::question type="MCQ" question="Let $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. What is $\operatorname{adj}(A)$?" options=["$\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$","$\begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix}$","$\begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix}$","$\begin{bmatrix} -1 & 2 \\ 3 & -4 \end{bmatrix}$"] answer="$\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$" hint="Find the cofactor matrix and then its transpose. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$." solution="Step 1: Calculate the cofactors of $A$.
$C_{11} = (-1)^{1+1} \det([4]) = 4$
$C_{12} = (-1)^{1+2} \det([3]) = -3$
$C_{21} = (-1)^{2+1} \det([2]) = -2$
$C_{22} = (-1)^{2+2} \det([1]) = 1$
Step 2: Form the cofactor matrix $C$.

Step 3: Transpose $C$ to get $\operatorname{adj}(A)$.

The adjoint of $A$ is $\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$."
:::

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5. Inverse of a Matrix using Determinants

An $n \times n$ matrix $A$ is invertible if and only if $\det(A) \neq 0$. If $A$ is invertible, its inverse $A^{-1}$ can be found using the formula involving the adjoint matrix and its determinant.

Quick Example:

Let $A = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}$. We find $A^{-1}$.
From previous examples, we know $\det(A) = 2$ and $\operatorname{adj}(A) = \begin{bmatrix} 2 & -1 \\ -4 & 3 \end{bmatrix}$.

Step 1: Apply the inverse formula.

Step 2: Perform scalar multiplication.

Answer: $A^{-1} = \begin{bmatrix} 1 & -1/2 \\ -2 & 3/2 \end{bmatrix}$.

:::question type="MCQ" question="Given $A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$, find $A^{-1}$." options=["$\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$","$\begin{bmatrix} -3 & 1 \\ 5 & -2 \end{bmatrix}$","$\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$","$\begin{bmatrix} 1/2 & 1 \\ 1/5 & 1/3 \end{bmatrix}$"] answer="$\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$" hint="First calculate $\det(A)$ and $\operatorname{adj}(A)$, then apply the formula $A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)$." solution="Step 1: Calculate $\det(A)$.

Step 2: Calculate $\operatorname{adj}(A)$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\operatorname{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

Step 3: Apply the inverse formula $A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)$.

The inverse of $A$ is $\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$."
:::

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6. Cramer's Rule

Cramer's Rule is a method for solving systems of linear equations using determinants. For a system of $n$ linear equations in $n$ variables, $AX = B$, where $\det(A) \neq 0$, the solution for each variable $x_i$ is given by $x_i = \frac{\det(A_i)}{\det(A)}$. Here, $A_i$ is the matrix formed by replacing the $i$-th column of $A$ with the column vector $B$.

Quick Example: Cramer's Rule for a $2 \times 2$ System

Solve the system:

Step 1: Write the coefficient matrix $A$ and the constant vector $B$.

Step 2: Calculate $\det(A)$.

Step 3: Form $A_1$ by replacing the first column of $A$ with $B$, and calculate $\det(A_1)$.

Step 4: Form $A_2$ by replacing the second column of $A$ with $B$, and calculate $\det(A_2)$.

Step 5: Apply Cramer's Rule to find $x$ and $y$.

Answer: $x=1, y=2$.

:::question type="MCQ" question="Using Cramer's Rule, what is the value of $x$ in the system:

?" options=["$1$","$2$","$3$","$4$"] answer="$1$" hint="Identify $A$, $B$, and $A_1$. Calculate their determinants and apply $x = \frac{\det(A_1)}{\det(A)}$." solution="Step 1: Define the coefficient matrix $A$ and the constant vector $B$.

Step 2: Calculate $\det(A)$.

Step 3: Form $A_1$ by replacing the first column of $A$ with $B$.

Step 4: Calculate $\det(A_1)$.

Step 5: Apply Cramer's Rule for $x$.

The value of $x$ is $1$."
:::

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Advanced Applications

We examine a complex problem combining multiple determinant properties and matrix operations, similar to those encountered in competitive examinations.

Worked Example: Determinant of $A^2 + A$

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$. Find $\det(A^2 + A)$. (Related to PYQ 1)

Step 1: Calculate $A^2$. Since $A$ is a diagonal matrix, $A^2$ is formed by squaring its diagonal elements.

>

Step 2: Calculate $A^2 + A$. Add the corresponding elements of $A^2$ and $A$.

>

>

Step 3: Calculate $\det(A^2 + A)$. Since $A^2+A$ is a diagonal matrix, its determinant is the product of its diagonal elements.

>

Answer: $\det(A^2 + A) = 144$.

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Worked Example: Determinant Identity for $2 \times 2$ Matrices

Let $A$ and $B$ be $2 \times 2$ matrices. Prove that $\det(A+B) + \det(A-B) = 2 \det(A) + 2 \det(B)$. (Related to PYQ 2)

Step 1: Define generic $2 \times 2$ matrices $A$ and $B$.

>

Step 2: Calculate $\det(A)$ and $\det(B)$.

>

>

Step 3: Calculate $A+B$ and $\det(A+B)$.

>

> \begin{align*}
> \det(A+B) &= (a+e)(d+h) - (b+f)(c+g) \\
> &= (ad+ah+ed+eh) - (bc+bg+fc+fg) \\
> &= ad+ah+ed+eh-bc-bg-fc-fg
> \end{align*}

Step 4: Calculate $A-B$ and $\det(A-B)$.

>

> \begin{align*}
> \det(A-B) &= (a-e)(d-h) - (b-f)(c-g) \\
> &= (ad-ah-ed+eh) - (bc-bg-fc+fg) \\
> &= ad-ah-ed+eh-bc+bg+fc-fg
> \end{align*}

Step 5: Add $\det(A+B)$ and $\det(A-B)$.

> \begin{align*}
> \det(A+B) + \det(A-B) &= (ad+ah+ed+eh-bc-bg-fc-fg) + (ad-ah-ed+eh-bc+bg+fc-fg) \\
> &= 2ad + 2eh - 2bc - 2fg \\
> &= 2(ad-bc) + 2(eh-fg)
> \end{align*}

Step 6: Substitute $\det(A)$ and $\det(B)$ back into the expression.

>

Answer: The identity is proven.

:::question type="NAT" question="Let $A = \begin{bmatrix} x & 3 \\ 2 & y \end{bmatrix}$. If $\det(A) = 4$ and $\det(A) = 2$ when $x=2$, find the value of $y$." answer="4" hint="Use the definition of determinant and the given conditions." solution="Step 1: Write the determinant of $A$.
>

Step 2: Use the first given condition: $\det(A) = 4$.
> \begin{align*}
> xy - 6 &= 4 \\
> xy &= 10 \quad (*)
> \end{align*}
Step 3: Use the second given condition: When $x=2$, the determinant of $A$ is $2$.
Substitute $x=2$ into the determinant expression and set it to $2$.
> \begin{align*}
> (2)y - 6 &= 2 \\
> 2y &= 8 \\
> y &= 4
> \end{align*}
Step 4: Verify the conditions.
If $y=4$, then from $xy=10$ (from Step 2), we have $x(4)=10 \implies x=10/4 = 5/2$.
So, the matrix is $A = \begin{bmatrix} 5/2 & 3 \\ 2 & 4 \end{bmatrix}$.
Its determinant is $\det(A) = (5/2)(4) - (3)(2) = 10 - 6 = 4$. This satisfies the first condition.
Now, check the second condition: When $x=2$, $\det(A)=2$.
If $x=2$, the matrix becomes $A' = \begin{bmatrix} 2 & 3 \\ 2 & 4 \end{bmatrix}$.
Its determinant is $\det(A') = (2)(4) - (3)(2) = 8 - 6 = 2$. This satisfies the second condition.
Both conditions are satisfied with $y=4$.
Answer: \boxed{4}"
:::

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Part 2: Solutions of Systems of Linear Equations

We examine the techniques for determining the existence and nature of solutions for systems of linear equations, a fundamental concept in linear algebra. This analysis is critical for understanding various mathematical models and is a recurrent theme in competitive examinations. We focus on methods involving matrix operations and rank.

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Core Concepts

1. Definition and Matrix Representation

A system of linear equations consists of one or more linear equations involving the same set of variables. We typically represent such a system in a compact matrix form $Ax=b$, where $A$ is the coefficient matrix, $x$ is the column vector of variables, and $b$ is the column vector of constants.

The augmented matrix $[A|b]$ combines the coefficient matrix and the constant vector, providing a consolidated representation for solving the system through row operations.

Quick Example:
Represent the system $2x - y + 3z = 9$, $x + 2y - z = 4$, $3x + y + 2z = 10$ in augmented matrix form.

Step 1: Identify coefficients and constants.
The coefficients are $a_{ij}$ and constants are $b_i$.

Step 2: Construct the augmented matrix.
>

Answer: The augmented matrix is $\left[\begin{array}{ccc|c} 2 & -1 & 3 & 9 \\ 1 & 2 & -1 & 4 \\ 3 & 1 & 2 & 10 \end{array}\right]$.

:::question type="MCQ" question="Which of the following augmented matrices correctly represents the system of equations: $x_1 - 2x_2 = 5$, $3x_2 + x_3 = 7$, $4x_1 + x_3 = 1$?" options=["

","

","

","

"] answer="

" hint="Ensure all variables are accounted for in each equation, using a coefficient of 0 if a variable is missing." solution="The given system is:

Forming the augmented matrix directly from these coefficients and constants yields:

"
:::

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2. Elementary Row Operations and Matrix Forms

Elementary Row Operations (EROs) are fundamental transformations applied to the rows of a matrix that do not alter the solution set of the corresponding linear system. These operations are used to simplify a matrix into a more manageable form, such as Row Echelon Form (REF) or Reduced Row Echelon Form (RREF).

The three types of EROs are:

Swapping two rows ($R_i \leftrightarrow R_j$).

Multiplying a row by a non-zero scalar ($kR_i \to R_i$, $k \ne 0$).

Adding a multiple of one row to another row ($R_i + kR_j \to R_i$).

Quick Example:
Transform the matrix into Row Echelon Form:

Step 1: Make entries below the first pivot zero ($R_2 - 2R_1 \to R_2$, $R_3 + R_1 \to R_3$).
>

Step 2: Make the entry below the second pivot zero ($R_3 + \frac{5}{3}R_2 \to R_3$).
>

Answer: The Row Echelon Form is $\begin{bmatrix} 1 & 2 & -1 \\ 0 & -3 & 5 \\ 0 & 0 & \frac{28}{3} \end{bmatrix}$.

:::question type="MCQ" question="Which of the following matrices is in Reduced Row Echelon Form?" options=["

","

","

","

"] answer="

" hint="Recall the conditions for RREF, especially regarding leading 1s and zeros in their columns." solution="A matrix in RREF must have leading 1s in each non-zero row, and all other entries in the column containing a leading 1 must be zero.
Option 1:

This matrix satisfies all RREF conditions.
Option 2: The entry $a_{12}=2$ is not zero, but the column 2 has a leading 1 in $a_{22}$. This violates RREF condition 2.
Option 3: The leading 1s are not in a staircase pattern, and the first row starts with 0. This violates REF condition 2.
Option 4: The leading 1 in the second row is $a_{23}$, while the leading 1 in the third row is $a_{32}$. This violates REF condition 2. Also, the third row should be above the second row if ordered by leading 1 position."
:::

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3. Rank of a Matrix

The rank of a matrix is a fundamental property reflecting the dimensionality of the vector space spanned by its rows or columns. It is defined as the maximum number of linearly independent row (or column) vectors in the matrix.

We determine the rank by transforming the matrix into Row Echelon Form (REF) using EROs. The rank is then the number of non-zero rows in its REF.

Quick Example:
Find the rank of the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}$.

Step 1: Transform $A$ into REF.
$R_2 - 2R_1 \to R_2$
$R_3 - 3R_1 \to R_3$
>

Step 2: Count the number of non-zero rows.
The REF has one non-zero row.

Answer: The rank of $A$ is 1.

:::question type="NAT" question="What is the rank of the matrix $M = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 7 & 10 \\ 3 & 10 & 14 \end{bmatrix}$?" answer="2" hint="Apply elementary row operations to reduce the matrix to Row Echelon Form and then count the number of non-zero rows." solution="Step 1: Start with the given matrix.
>

Step 2: Perform $R_2 - 2R_1 \to R_2$ and $R_3 - 3R_1 \to R_3$.
>

Step 3: Perform $R_3 - R_2 \to R_3$.
>

The matrix is now in Row Echelon Form.

Step 4: Count the number of non-zero rows.
There are two non-zero rows.

Answer: The rank of $M$ is 2."
:::

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4. Consistency of Linear Systems (Rouché-Capelli Theorem)

The Rouché-Capelli Theorem provides a definitive criterion for determining whether a system of linear equations $Ax=b$ has a solution (is consistent) or not (is inconsistent). This involves comparing the rank of the coefficient matrix $A$ with the rank of the augmented matrix $[A|b]$.

Quick Example:
Determine the consistency of the system: $x + y = 1$, $2x + 2y = 3$.

Step 1: Form the augmented matrix.
>

Step 2: Reduce the augmented matrix to REF.
$R_2 - 2R_1 \to R_2$
>

Step 3: Determine $\operatorname{rank}(A)$ and $\operatorname{rank}([A|b])$.
The coefficient matrix $A = \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}$ reduces to $\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$, so $\operatorname{rank}(A) = 1$.
The augmented matrix $[A|b]$ reduces to $\left[\begin{array}{cc|c} 1 & 1 & 1 \\ 0 & 0 & 1 \end{array}\right]$, so $\operatorname{rank}([A|b]) = 2$.

Step 4: Compare the ranks.
Since $\operatorname{rank}(A) = 1 \ne \operatorname{rank}([A|b]) = 2$, the system is inconsistent.

Answer: The system is inconsistent.

:::question type="MCQ" question="Consider the system of linear equations: $x_1 - x_2 + x_3 = 1$, $2x_1 + x_2 - x_3 = 2$, $3x_1 + 0x_2 + 0x_3 = 4$. Determine its consistency." options=["Consistent with a unique solution","Consistent with infinitely many solutions","Inconsistent","Cannot be determined from the given information"] answer="Inconsistent" hint="Form the augmented matrix and reduce it to Row Echelon Form to compare the ranks of the coefficient and augmented matrices." solution="Step 1: Form the augmented matrix $[A|b]$.
>

Step 2: Reduce the augmented matrix to REF.
$R_2 - 2R_1 \to R_2$
$R_3 - 3R_1 \to R_3$
>

$R_3 - R_2 \to R_3$
>

Step 3: Determine the ranks.
From the REF of $[A|b]$, the coefficient matrix $A$ (first three columns) has 2 non-zero rows, so $\operatorname{rank}(A) = 2$.
The augmented matrix $[A|b]$ has 3 non-zero rows, so $\operatorname{rank}([A|b]) = 3$.

Step 4: Compare the ranks.
Since $\operatorname{rank}(A) = 2 \ne \operatorname{rank}([A|b]) = 3$, the system is inconsistent."
:::

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5. Types of Solutions for Consistent Systems

For a consistent system ($Ax=b$), the nature of its solutions (unique or infinitely many) depends on the relationship between the rank of the coefficient matrix and the number of variables.

Quick Example:
Determine the type of solution for the system: $x + y + z = 6$, $2x - y + z = 3$, $x - 2y + 2z = 3$.

Step 1: Form the augmented matrix.

Step 2: Reduce the augmented matrix to REF.
$R_2 - 2R_1 \to R_2$
$R_3 - R_1 \to R_3$

$R_3 - R_2 \to R_3$

Step 3: Determine the ranks and number of variables.
$\operatorname{rank}(A) = 3$ (3 non-zero rows in the coefficient part).
$\operatorname{rank}([A|b]) = 3$ (3 non-zero rows in the augmented matrix).
Number of variables $n = 3$.

Step 4: Compare ranks and number of variables.
Since $\operatorname{rank}(A) = \operatorname{rank}([A|b]) = n = 3$, the system has a unique solution.

Answer: $\boxed{\text{The system has a unique solution.}}$

:::question type="MCQ" question="For the system of equations: $x - y + z = 0$, $2x - 2y + 2z = 0$, $3x - 3y + 3z = 0$. What type of solution does it have?" options=["No solution","Unique solution","Infinitely many solutions","Dependent on a parameter"] answer="Infinitely many solutions" hint="This is a homogeneous system. Reduce the augmented matrix to REF and compare ranks to the number of variables." solution="Step 1: Form the augmented matrix $[A|b]$.

Step 2: Reduce the augmented matrix to REF.
$R_2 - 2R_1 \to R_2$
$R_3 - 3R_1 \to R_3$

Step 3: Determine the ranks and number of variables.
$\operatorname{rank}(A) = 1$ (1 non-zero row).
$\operatorname{rank}([A|b]) = 1$ (1 non-zero row).
Number of variables $n = 3$.

Step 4: Compare ranks and number of variables.
Since $\operatorname{rank}(A) = \operatorname{rank}([A|b]) = 1 < n = 3$, the system is consistent and has infinitely many solutions. This is also a homogeneous system, so it always has at least the trivial solution. Since its rank is less than the number of variables, it must have non-trivial solutions.
Answer: $\boxed{\text{Infinitely many solutions.}}$"
:::

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6. Homogeneous Systems of Linear Equations

A homogeneous system of linear equations is one where all the constant terms are zero, i.e., $Ax=0$. Such systems always have at least one solution, the trivial solution where all variables are zero ($x=0$). The primary question for homogeneous systems is whether non-trivial solutions exist.

Quick Example:
Determine if the system $x + y - z = 0$, $2x - y + z = 0$, $3x + 0y + 0z = 0$ has non-trivial solutions.

Step 1: Form the coefficient matrix $A$.

Step 2: Reduce $A$ to REF to find its rank.
$R_2 - 2R_1 \to R_2$
$R_3 - 3R_1 \to R_3$

$R_3 - R_2 \to R_3$

Step 3: Determine $\operatorname{rank}(A)$ and compare to $n$.
$\operatorname{rank}(A) = 2$.
Number of variables $n = 3$.

Step 4: Conclude on non-trivial solutions.
Since $\operatorname{rank}(A) = 2 < n = 3$, the system has infinitely many non-trivial solutions.

Answer: $\boxed{\text{The system has infinitely many non-trivial solutions.}}$

:::question type="MCQ" question="For a homogeneous system $Ax=0$ where $A$ is a $4 \times 3$ matrix, if $\operatorname{rank}(A) = 3$, which statement is true?" options=["The system has no solution.","The system has a unique non-trivial solution.","The system has infinitely many non-trivial solutions.","The system has only the trivial solution."] answer="The system has only the trivial solution." hint="Recall the conditions for homogeneous systems: they are always consistent. The nature of the solution (trivial or non-trivial) depends on the rank relative to the number of variables." solution="A homogeneous system $Ax=0$ is always consistent. The number of variables is $n=3$.
Given $\operatorname{rank}(A) = 3$.
Since $\operatorname{rank}(A) = n = 3$, the system has a unique solution. For a homogeneous system, this unique solution is always the trivial solution ($x_1=0, x_2=0, x_3=0$). Therefore, the system has only the trivial solution.
Answer: $\boxed{\text{The system has only the trivial solution.}}$"
:::

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7. Solving Systems using Gaussian/Gauss-Jordan Elimination

Gaussian elimination uses EROs to transform the augmented matrix into Row Echelon Form (REF), from which solutions can be found using back-substitution. Gauss-Jordan elimination extends this by transforming the matrix into Reduced Row Echelon Form (RREF), allowing direct reading of the solutions.

Gaussian Elimination Steps:

Form the augmented matrix $[A|b]$.

Use EROs to transform $[A|b]$ into REF.

Write out the equivalent system of equations from the REF.

Use back-substitution to solve for the variables.

Gauss-Jordan Elimination Steps:

Form the augmented matrix $[A|b]$.

Use EROs to transform $[A|b]$ into RREF.

Read the solutions directly from the RREF.

Worked Example:
Solve the system: $x - y + 2z = 9$, $2x + y + z = 7$, $3x - 2y - 2z = 5$ using Gauss-Jordan elimination.

Step 1: Form the augmented matrix.

Step 2: Transform to REF.
$R_2 - 2R_1 \to R_2$
$R_3 - 3R_1 \to R_3$

$R_2 \leftrightarrow R_3$ (to get a 1 in the pivot position)

$R_3 - 3R_2 \to R_3$

Step 3: Continue to RREF.
$\frac{1}{21}R_3 \to R_3$

$R_2 + 8R_3 \to R_2$
$R_1 - 2R_3 \to R_1$

$R_1 + R_2 \to R_1$

Step 4: Read the solution.
$x = \frac{19}{7}$, $y = -\frac{22}{21}$, $z = \frac{55}{21}$.

Answer: $\boxed{x = \frac{19}{7}, y = -\frac{22}{21}, z = \frac{55}{21}.}$

:::question type="NAT" question="Solve the system $x + 2y = 5$, $3x - y = 1$ for $x$ using Gaussian elimination. Provide the value of $x$." answer="1" hint="Form the augmented matrix, perform EROs to get to REF, then use back-substitution." solution="Step 1: Form the augmented matrix.

Step 2: Reduce to REF.
$R_2 - 3R_1 \to R_2$

Step 3: Convert to equations and back-substitute.
From the second row: $-7y = -14 \implies y = 2$.
Substitute $y=2$ into the first equation: $x + 2(2) = 5 \implies x + 4 = 5 \implies x = 1$.

Answer: $\boxed{1}$"
:::

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8. Cramer's Rule

Cramer's Rule is an alternative method for solving systems of linear equations, applicable only when the system has a unique solution. It involves calculating determinants of matrices derived from the original coefficient matrix.

Worked Example:
Solve the system $x + y = 3$, $2x - 3y = 1$ using Cramer's Rule.

Step 1: Identify $A$, $b$, and calculate $\det(A)$.

Since $\det(A) = -5 \ne 0$, a unique solution exists.

Step 2: Form $A_1$ and $A_2$ and calculate their determinants.

(replace column 1 of $A$ with $b$)

(replace column 2 of $A$ with $b$)

Step 3: Calculate $x$ and $y$.

Answer: $\boxed{x = 2, y = 1.}$

:::question type="MCQ" question="Using Cramer's Rule, find the value of $y$ for the system: $3x - 2y = 1$, $x + 4y = 7$." options=["1","2","3","4"] answer="2" hint="Calculate $\det(A)$ and $\det(A_y)$ (where column 2 of $A$ is replaced by the constant vector)." solution="Step 1: Write the coefficient matrix $A$ and constant vector $b$.

Step 2: Calculate $\det(A)$.

Step 3: Form $A_y$ (replace the second column of $A$ with $b$) and calculate $\det(A_y)$.

Step 4: Apply Cramer's Rule for $y$.

Answer: $\boxed{\frac{10}{7}}$"
:::

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Advanced Applications

Systems with Parameters

Analyzing systems of linear equations containing unknown parameters is crucial for determining conditions under which the system is consistent, has a unique solution, or infinitely many solutions. This often involves reducing the augmented matrix to REF and examining the rows containing the parameters.

Worked Example 1 (Consistency with Parameters - from PYQ 7 pattern):
Consider the system of linear equations:

Determine the conditions on $m$ and $k$ for the system to be consistent.

Step 1: Form the augmented matrix.
>

Step 2: Reduce the augmented matrix to REF.
$R_2 - R_1 \to R_2$
$R_3 - R_1 \to R_3$
>

$R_3 - R_2 \to R_3$
>

>

Step 3: Analyze consistency using Rouché-Capelli Theorem.
The system is consistent if $\operatorname{rank}(A) = \operatorname{rank}([A|b])$.
* If $4-m \ne 0$ (i.e., $m \ne 4$), then $\operatorname{rank}(A) = 3$ and $\operatorname{rank}([A|b]) = 3$. In this case, the system has a unique solution, hence consistent.
* If $4-m = 0$ (i.e., $m = 4$), then the third row of the coefficient matrix becomes all zeros.
* If $m=4$, the augmented matrix becomes:
>

* For consistency when $m=4$, we must have the last row of the augmented matrix also be all zeros, meaning $k-5=0$, so $k=5$.
In this case, $\operatorname{rank}(A) = 2$ and $\operatorname{rank}([A|b]) = 2$. The system has infinitely many solutions.
* If $m=4$ and $k \ne 5$, then the last row is $\left[\begin{array}{ccc|c} 0 & 0 & 0 & k-5 \end{array}\right]$ where $k-5 \ne 0$.
In this case, $\operatorname{rank}(A) = 2$ and $\operatorname{rank}([A|b]) = 3$. The system is inconsistent.

Step 4: Combine the conditions for consistency.
The system is consistent if:

$m \ne 4$ (unique solution)

$m = 4$ AND $k = 5$ (infinitely many solutions)

Answer: The system is consistent if $m \ne 4$ or ($m=4$ and $k=5$).

Worked Example 2 (Unique Solution with Parameters - from PYQ 8 pattern):
The system of linear equations $x+y+z=6$, $x+2y+5z=10$, $2x+3y+\lambda z=\mu$ has a unique solution. Determine the conditions on $\lambda$ and $\mu$.

Step 1: Form the augmented matrix.
>

Step 2: Reduce the augmented matrix to REF.
$R_2 - R_1 \to R_2$
$R_3 - 2R_1 \to R_3$
>

$R_3 - R_2 \to R_3$
>

>

Step 3: Analyze conditions for a unique solution.
For a unique solution, we need $\operatorname{rank}(A) = \operatorname{rank}([A|b]) = n$, where $n=3$ is the number of variables.
This requires that all three rows of the coefficient matrix $A$ (left of the bar) must be non-zero after reduction.
Therefore, the entry $(\lambda-6)$ in the third row must be non-zero.
So, $\lambda - 6 \ne 0 \implies \lambda \ne 6$.
If $\lambda \ne 6$, then $\operatorname{rank}(A) = 3$ and $\operatorname{rank}([A|b]) = 3$, irrespective of the value of $\mu$. Thus, the system is consistent and has a unique solution.

Step 4: Conclude the conditions.
The system has a unique solution if $\lambda \ne 6$ and $\mu$ can be any real number ($\mu \in \mathbb{R}$).

Answer: The system has a unique solution if $\lambda \ne 6$, $\mu \in \mathbb{R}$.

:::question type="MCQ" question="For the system $x+2y+z=3$, $2x+5y+2z=7$, $3x+my+3z=10$. If the system has infinitely many solutions, what is the value of $m$?" options=["7","8","9","10"] answer="7" hint="For infinitely many solutions, $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$. Reduce the augmented matrix to REF and find the condition on $m$ that leads to a row of zeros in both $A$ and $[A|b]$." solution="Step 1: Form the augmented matrix.
>

Step 2: Reduce the augmented matrix to Row Echelon Form.
$R_2 - 2R_1 \to R_2$
$R_3 - 3R_1 \to R_3$
>

$R_3 - (m-6)R_2 \to R_3$
>

>
>

Step 3: Apply conditions for infinitely many solutions.
For a system to have infinitely many solutions, it must be consistent, and $\operatorname{rank}(A) < n$.
Here $n=3$. From the REF, $\operatorname{rank}(A)$ will be 2 if the third row of $A$ is zero. This happens when $0=0$ in the third row.
For consistency, the entire third row of the augmented matrix must be zero. This means $7-m=0$.
Thus, $m=7$.
If $m=7$, then $\operatorname{rank}(A) = 2$ and $\operatorname{rank}([A|b]) = 2$. Since $n=3$, $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$, leading to infinitely many solutions.

Answer: $\boxed{7}$"
:::

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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Consider the system: $x+y+z=1$, $2x+3y+z=2$, $3x+4y+2z=k$. For what value of $k$ does the system have infinitely many solutions?" options=["1","2","3","4"] answer="3" hint="Reduce the augmented matrix to REF. For infinitely many solutions, $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$." solution="Step 1: Form the augmented matrix.
>

Step 2: Reduce to REF.
$R_2 - 2R_1 \to R_2$
$R_3 - 3R_1 \to R_3$
>

$R_3 - R_2 \to R_3$
>

Step 3: Analyze for infinitely many solutions.
For infinitely many solutions, the system must be consistent, and $\operatorname{rank}(A) < n$.
Here, $n=3$. From the REF, $\operatorname{rank}(A) = 2$ (since the third column of $A$ is zero in the last row).
For consistency, we need $\operatorname{rank}([A|b]) = \operatorname{rank}(A) = 2$.
This requires the last entry in the augmented matrix's third row to be zero, i.e., $k-3=0$.
Thus, $k=3$.

Answer: $\boxed{3}$"
:::

:::question type="NAT" question="For the homogeneous system $x+y+z=0$, $2x+2y+2z=0$, $3x+3y+3z=0$, find the number of linearly independent solutions." answer="2" hint="The number of linearly independent solutions for a homogeneous system is $n - \operatorname{rank}(A)$, where $n$ is the number of variables." solution="Step 1: Form the coefficient matrix $A$.
>

Step 2: Reduce $A$ to REF.
$R_2 - 2R_1 \to R_2$
$R_3 - 3R_1 \to R_3$
>

Step 3: Determine $\operatorname{rank}(A)$.
The number of non-zero rows in REF is 1. So, $\operatorname{rank}(A) = 1$.

Step 4: Calculate the number of linearly independent solutions.
The number of variables is $n=3$.
The number of linearly independent solutions (dimension of the null space) is $n - \operatorname{rank}(A) = 3 - 1 = 2$.

Answer: $\boxed{2}$"
:::

:::question type="MSQ" question="Which of the following statements are true regarding the system $Ax=b$ if $A$ is an $m \times n$ matrix?" options=["If $\operatorname{rank}(A) < \operatorname{rank}([A|b])$, the system has no solution.","If $\operatorname{rank}(A) = n$, the system always has a unique solution.","If $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$, the system has infinitely many solutions.","If $m < n$, the system must have infinitely many solutions."] answer="If $\operatorname{rank}(A) < \operatorname{rank}([A|b])$, the system has no solution.,If $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$, the system has infinitely many solutions." hint="Carefully apply the Rouché-Capelli Theorem and the conditions for unique/infinite solutions. Also consider cases where $m < n$ for consistency." solution="Let's analyze each option:
* 'If $\operatorname{rank}(A) < \operatorname{rank}([A|b])$, the system has no solution.' This is the direct statement of inconsistency from the Rouché-Capelli Theorem. This statement is TRUE.
* 'If $\operatorname{rank}(A) = n$, the system always has a unique solution.' This statement is only true if the system is consistent. If $\operatorname{rank}(A) = n$ but $\operatorname{rank}([A|b]) \ne n$, the system is inconsistent (no solution). For example, $x=1$, $x=2$. Here $\operatorname{rank}(A)=1$, $n=1$, but $\operatorname{rank}([A|b])=2$. So this statement is FALSE. A more precise statement would be: 'If $\operatorname{rank}(A) = \operatorname{rank}([A|b]) = n$, the system has a unique solution.'
* 'If $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$, the system has infinitely many solutions.' This is the correct condition for a consistent system to have infinitely many solutions, as there are $n - \operatorname{rank}(A)$ free variables. This statement is TRUE.
'If $m < n$, the system must have infinitely many solutions.' If the number of equations ($m$) is less than the number of variables ($n$), the system might have infinitely many solutions (if consistent) OR no solution (if inconsistent). It does not must* have infinitely many solutions. For example, $x+y+z=1$ (1 equation, 3 variables). This has infinite solutions. But consider $x+y+z=1$, $x+y+z=2$. Here $m=2, n=3$, but it's inconsistent (no solution). So this statement is FALSE.

Therefore, the correct statements are: 'If $\operatorname{rank}(A) < \operatorname{rank}([A|b])$, the system has no solution.' and 'If $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$, the system has infinitely many solutions.'"
:::

:::question type="MCQ" question="Given the system: $x+2y=c_1$, $3x+6y=c_2$. For what conditions on $c_1$ and $c_2$ will the system have infinitely many solutions?" options=["$c_2 = 3c_1$","$\operatorname{rank}(A) = 1$","Both $c_1, c_2$ are non-zero","The system is always inconsistent"] answer="$c_2 = 3c_1$" hint="Reduce the augmented matrix to REF. For infinitely many solutions, the last row must be all zeros." solution="Step 1: Form the augmented matrix.
>

Step 2: Reduce to REF.
$R_2 - 3R_1 \to R_2$
>

Step 3: Analyze for infinitely many solutions.
For infinitely many solutions, we need $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$.
Here, $n=2$. From the REF, $\operatorname{rank}(A) = 1$.
For $\operatorname{rank}([A|b])$ to be 1, the last entry in the augmented matrix must be zero.
So, $c_2 - 3c_1 = 0 \implies c_2 = 3c_1$.
If $c_2 = 3c_1$, then $\operatorname{rank}(A) = 1$ and $\operatorname{rank}([A|b]) = 1$. Since $1 < n=2$, the system has infinitely many solutions.

Answer: $\boxed{c_2 = 3c_1}$"
:::

:::question type="NAT" question="If the system $x+y=1$, $2x+ay=b$ has no solution, and $a, b$ are integers, find the smallest positive integer value of $a+b$." answer="3" hint="For no solution, $\operatorname{rank}(A) < \operatorname{rank}([A|b])$. Reduce the augmented matrix to REF and find the conditions on $a$ and $b$." solution="Step 1: Form the augmented matrix.
>

Step 2: Reduce to REF.
$R_2 - 2R_1 \to R_2$
>

Step 3: Analyze for no solution.
For no solution, we need $\operatorname{rank}(A) < \operatorname{rank}([A|b])$.
This occurs when the coefficient part of the last row is zero, but the constant part is non-zero.
So, $a-2=0 \implies a=2$.
And $b-2 \ne 0 \implies b \ne 2$.

Step 4: Find the smallest positive integer value of $a+b$.
We have $a=2$.
We need $b \ne 2$. Since $b$ is an integer, the smallest positive integer value for $b$ (that is not 2) is $b=1$.
Then $a+b = 2+1 = 3$.

Answer: $\boxed{3}$"
:::

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Summary

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What's Next?

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Part 3: Rank and Inverse of a Matrix

This section establishes the fundamental concepts of matrix rank and invertibility, which are crucial for understanding linear transformations, solving systems of linear equations, and analyzing the properties of vector spaces. We will explore methods for their computation and examine their essential properties relevant to competitive examinations.

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Core Concepts

1. Rank of a Matrix

We define the rank of a matrix $A$, denoted $\operatorname{rank}(A)$, as the maximum number of linearly independent column vectors in $A$, which is equivalent to the maximum number of linearly independent row vectors in $A$. Alternatively, it is the order of the largest non-singular square submatrix of $A$. A practical method for determining rank involves reducing the matrix to its row echelon form.

The rank of a matrix is the number of non-zero rows in its row echelon form.

Quick Example: Determine the rank of $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}$.

Step 1: Apply row operations to achieve row echelon form.
>

Step 2: Perform $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$.
>

Step 3: Count the number of non-zero rows.
The matrix in row echelon form has one non-zero row.

Answer: $\operatorname{rank}(A) = 1$.

:::question type="MCQ" question="What is the rank of the matrix $M = \begin{bmatrix} 1 & 2 & 0 & -1 \\ 2 & 6 & -3 & -3 \\ 3 & 10 & -6 & -5 \end{bmatrix}$?" options=["Rank ($M$) = 4","Rank ($M$) = 3","Rank ($M$) = 2","Rank ($M$) = 1"] answer="Rank ($M$) = 2" hint="Reduce the matrix to row echelon form using elementary row operations and count the number of non-zero rows." solution="Step 1: Start with the given matrix.
>

Step 2: Perform $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$.
>

Step 3: Perform $R_3 \to R_3 - 2R_2$.
>

Step 4: Count the number of non-zero rows.
The row echelon form has two non-zero rows.
Therefore, $\operatorname{rank}(M) = 2$."
:::

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2. Properties of Rank

The rank of a matrix possesses several important properties that are useful in various applications, particularly concerning matrix products and systems of linear equations.

Quick Example: If $A$ is a $5 \times 7$ matrix and $\operatorname{nullity}(A) = 3$, find $\operatorname{rank}(A)$.

Step 1: Apply the Rank-Nullity Theorem.
>

Step 2: Substitute the given values.
Here, $n=7$ (number of columns) and $\operatorname{nullity}(A) = 3$.
>

Step 3: Solve for $\operatorname{rank}(A)$.
>

Answer: $\operatorname{rank}(A) = 4$.

:::question type="MSQ" question="Let $A$ be an $m \times n$ matrix and $B$ be an $n \times p$ matrix. Which of the following statements about rank are always true?" options=["$\operatorname{rank}(A) \leq m$","$\operatorname{rank}(AB) \leq \operatorname{rank}(A)$","$\operatorname{rank}(A) + \operatorname{nullity}(A) = m$","$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$"] answer="$\operatorname{rank}(A) \leq m$,$\operatorname{rank}(AB) \leq \operatorname{rank}(A)$,$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$" hint="Recall the fundamental properties of matrix rank and the Rank-Nullity Theorem. Pay attention to the dimensions." solution="Let $A$ be an $m \times n$ matrix and $B$ be an $n \times p$ matrix.

* Option 1: $\operatorname{rank}(A) \leq m$
The rank of a matrix is at most its number of rows (or columns). Since $A$ has $m$ rows, $\operatorname{rank}(A) \leq m$ is always true.

* Option 2: $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$
The rank of a product of matrices is less than or equal to the rank of either factor. Specifically, the column space of $AB$ is a subspace of the column space of $A$. Thus, $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$ is always true.

* Option 3: $\operatorname{rank}(A) + \operatorname{nullity}(A) = m$
The Rank-Nullity Theorem states that $\operatorname{rank}(A) + \operatorname{nullity}(A) = n$, where $n$ is the number of columns of $A$. This statement claims it equals $m$ (number of rows), which is generally false unless $m=n$.

* Option 4: $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$
Similarly, the row space of $AB$ is a subspace of the row space of $B$. Thus, $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ is always true.

Therefore, the correct options are $\operatorname{rank}(A) \leq m$, $\operatorname{rank}(AB) \leq \operatorname{rank}(A)$, and $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$."
:::

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3. Inverse of a Matrix

An inverse matrix is a fundamental concept for solving linear systems and understanding matrix transformations. Not all matrices have an inverse.

Quick Example: Verify if $B = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$ is the inverse of $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$.

Step 1: Calculate the product $AB$.
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Step 2: Calculate the product $BA$.
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Step 3: Compare products to the identity matrix.
Since $AB = BA = I$, $B$ is indeed the inverse of $A$.

Answer: Yes, $B$ is the inverse of $A$.

:::question type="MCQ" question="Let $P$ be the inverse of a matrix $Q$. If $I$ denotes the identity matrix, which one of the following options is correct?" options=["$PQ = I$ but $QP \neq I$","$QP = I$ but $PQ \neq I$","$PQ = I$ and $QP = I$","$PQ - QP = I$"] answer="$PQ = I$ and $QP = I$" hint="Recall the definition of a matrix inverse. The definition requires commutativity with respect to multiplication by the inverse to produce the identity matrix." solution="By the definition of a matrix inverse, if $P$ is the inverse of $Q$, then their product must yield the identity matrix, regardless of the order of multiplication. That is, $PQ = I$ and $QP = I$. The inverse is unique and satisfies this property for both pre-multiplication and post-multiplication.
Therefore, the correct option is '$PQ = I$ and $QP = I$'."
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4. Methods to Find Inverse

We can compute the inverse of a square matrix using two primary methods: the adjoint method and the Gauss-Jordan elimination method.

4.1. Adjoint Method

For a square matrix $A$, its inverse $A^{-1}$ can be found using the formula:

Quick Example: Find the inverse of $A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$ using the adjoint method.

Step 1: Calculate the determinant of $A$.
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Step 2: Find the cofactor matrix $C$ of $A$.
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> $C_{11} = 2$
> $C_{12} = -3$
> $C_{21} = -1$
> $C_{22} = 2$
>

Step 3: Find the adjoint of $A$, which is $C^T$.
>

Step 4: Compute $A^{-1}$.
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Answer: $A^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}$.

4.2. Gauss-Jordan Elimination Method

This method involves augmenting the matrix $A$ with an identity matrix $I$ of the same order, forming $[A|I]$, and then performing elementary row operations to transform $A$ into $I$. The matrix that results on the right side will be $A^{-1}$.

Quick Example: Find the inverse of $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ using Gauss-Jordan elimination.

Step 1: Form the augmented matrix $[A|I]$.
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Step 2: Perform $R_2 \to R_2 - 3R_1$.
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Step 3: Perform $R_2 \to -\frac{1}{2}R_2$.
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Step 4: Perform $R_1 \to R_1 - 2R_2$.
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>

Answer: $A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$.

:::question type="NAT" question="Find the element in the first row and second column of the inverse of $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}$." answer="-0.5" hint="Use Gauss-Jordan elimination to find the inverse. The element $A^{-1}_{12}$ is required." solution="Step 1: Form the augmented matrix $[A|I]$.
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Step 2: Perform $R_2 \to R_2 - 2R_1$.
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Step 3: Perform $R_3 \to R_3 - R_2$.
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Step 4: Perform $R_3 \to \frac{1}{3}R_3$.
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Step 5: Perform $R_1 \to R_1 - R_3$ and $R_2 \to R_2 + 2R_3$.
>

>
The inverse matrix is $A^{-1} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} \\ -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \end{bmatrix}$.

The element in the first row and second column of $A^{-1}$ is $\frac{1}{3}$. This is approximately $0.333$.
Let's recheck with the adjoint method for verification:
$\det(A) = 1(1-0) - 0(2-0) + 1(2-0) = 1+2=3$.
Cofactor matrix:
$C_{11} = (1)(1) - (0)(1) = 1$
$C_{12} = -((2)(1) - (0)(0)) = -2$
$C_{13} = (2)(1) - (1)(0) = 2$
$C_{21} = -((0)(1) - (1)(1)) = 1$
$C_{22} = (1)(1) - (0)(1) = 1$
$C_{23} = -((1)(1) - (0)(0)) = -1$
$C_{31} = ((0)(0) - (1)(1)) = -1$
$C_{32} = -((1)(0) - (1)(2)) = 2$
$C_{33} = ((1)(1) - (0)(2)) = 1$
Cofactor matrix $C = \begin{bmatrix} 1 & -2 & 2 \\ 1 & 1 & -1 \\ -1 & 2 & 1 \end{bmatrix}$.
$\operatorname{adj}(A) = C^T = \begin{bmatrix} 1 & 1 & -1 \\ -2 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix}$.
$A^{-1} = \frac{1}{3} \begin{bmatrix} 1 & 1 & -1 \\ -2 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} \\ -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \end{bmatrix}$.

The element in the first row and second column is $1/3$."
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5. Properties of Inverse

The inverse operation has several algebraic properties that are useful for manipulating matrix expressions.

Quick Example: Given $A^{-1} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$, find $(AB)^{-1}$.

Step 1: Apply the property $(AB)^{-1} = B^{-1}A^{-1}$.
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Step 2: Substitute the given inverse matrices and perform multiplication.
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Answer: $(AB)^{-1} = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}$.

:::question type="MCQ" question="Let $A$ and $B$ be invertible matrices of order $n$. Which of the following statements is INCORRECT?" options=["$(A^T)^{-1} = (A^{-1})^T$","$((A^{-1})^T)^{-1} = A^T$","$(A+B)^{-1} = A^{-1} + B^{-1}$","$(5A)^{-1} = \frac{1}{5}A^{-1}$"] answer="$(A+B)^{-1} = A^{-1} + B^{-1}$" hint="Carefully review the properties of matrix inverse. Matrix addition inverse does not distribute." solution="Let us analyze each statement:

* Option 1: $(A^T)^{-1} = (A^{-1})^T$
This is a standard property of matrix inverses and transposes. The inverse of the transpose is the transpose of the inverse. This statement is correct.

* Option 2: $((A^{-1})^T)^{-1} = A^T$
Using the property from Option 1, $(A^{-1})^T = (A^T)^{-1}$. So, $((A^{-1})^T)^{-1} = ((A^T)^{-1})^{-1}$. Since $(X^{-1})^{-1} = X$, we have $((A^T)^{-1})^{-1} = A^T$. This statement is correct.

* Option 3: $(A+B)^{-1} = A^{-1} + B^{-1}$
This statement is generally false. There is no distributive property for matrix inversion over matrix addition. For example, consider $A=I$ and $B=I$. Then $(A+B)^{-1} = (2I)^{-1} = \frac{1}{2}I$. However, $A^{-1}+B^{-1} = I^{-1}+I^{-1} = I+I = 2I$. Since $\frac{1}{2}I \neq 2I$, this statement is incorrect.

* Option 4: $(5A)^{-1} = \frac{1}{5}A^{-1}$
This is a standard property of scalar multiplication with matrix inverse. If $k$ is a non-zero scalar, $(kA)^{-1} = \frac{1}{k}A^{-1}$. This statement is correct.

Therefore, the incorrect statement is $(A+B)^{-1} = A^{-1} + B^{-1}$."
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Advanced Applications

Rank and inverse concepts are foundational for understanding the solvability and nature of solutions to systems of linear equations.

Advanced Example: Consider the system of linear equations $Ax=b$. If $A$ is an $n \times n$ matrix, discuss the solution based on its rank.

Step 1: If $\operatorname{rank}(A) = n$.
This implies $A$ is invertible. The system $Ax=b$ has a unique solution given by $x = A^{-1}b$.

Step 2: If $\operatorname{rank}(A) < n$.
This implies $A$ is singular (not invertible). In this case, we must consider the augmented matrix $[A|b]$ and its rank.
If $\operatorname{rank}(A) \neq \operatorname{rank}([A|b])$, the system is inconsistent and has no solution.
If $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$, the system is consistent and has infinitely many solutions. The number of free variables will be $n - \operatorname{rank}(A)$.

:::question type="MCQ" question="Let $A$ be a $3 \times 3$ matrix such that $\det(A) = 0$. Consider the system of linear equations $Ax = b$. Which of the following statements is necessarily true?" options=["The system $Ax=b$ has a unique solution." ,"The system $Ax=b$ has no solution." ,"The system $Ax=b$ has infinitely many solutions." ,"The system $Ax=b$ has either no solution or infinitely many solutions."] answer="The system $Ax=b$ has either no solution or infinitely many solutions." hint="If $\det(A)=0$, the matrix $A$ is singular, meaning $\operatorname{rank}(A) < 3$. Recall the conditions for the existence and uniqueness of solutions to linear systems based on rank." solution="Given that $A$ is a $3 \times 3$ matrix and $\det(A) = 0$.
This implies that $A$ is a singular matrix, and its rank is less than 3 ($\operatorname{rank}(A) < 3$).

For a system of linear equations $Ax=b$:
* If $\operatorname{rank}(A) = \operatorname{rank}([A|b]) = n$ (where $n$ is the number of variables, here $n=3$), then the system has a unique solution.
* If $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < n$, then the system has infinitely many solutions.
* If $\operatorname{rank}(A) \neq \operatorname{rank}([A|b])$, then the system has no solution.

Since $\operatorname{rank}(A) < 3$, the first case (unique solution) is impossible.
Therefore, the system $Ax=b$ must either have no solution (if $\operatorname{rank}(A) \neq \operatorname{rank}([A|b])$) or infinitely many solutions (if $\operatorname{rank}(A) = \operatorname{rank}([A|b]) < 3$).
We cannot definitively say it has no solution or infinitely many solutions without knowing $\operatorname{rank}([A|b])$. However, we can definitively say it will be one of these two possibilities.

Therefore, the statement 'The system $Ax=b$ has either no solution or infinitely many solutions' is necessarily true."
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Problem-Solving Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="If $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{bmatrix}$, what is $\operatorname{rank}(A)$?" options=["0","1","2","3"] answer="2" hint="The matrix is already in row echelon form. Count the number of non-zero rows." solution="The given matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{bmatrix}$ is in row echelon form.
The number of non-zero rows is 2 (the first and second rows).
Therefore, $\operatorname{rank}(A) = 2$."
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:::question type="NAT" question="Let $A = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$. If $A^{-1} = \begin{bmatrix} x & y \\ z & w \end{bmatrix}$, find the value of $x+y+z+w$." answer="11" hint="First, find the inverse of $A$. Then sum its elements." solution="Step 1: Calculate the determinant of $A$.
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Step 2: Find the inverse of $A$ using the $2 \times 2$ inverse formula.
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Step 3: Sum the elements of $A^{-1}$.
>

"
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:::question type="MCQ" question="If $A$ is a $4 \times 5$ matrix and $\operatorname{nullity}(A) = 2$, what is $\operatorname{rank}(A)$?" options=["2","3","4","5"] answer="3" hint="Apply the Rank-Nullity Theorem: $\operatorname{rank}(A) + \operatorname{nullity}(A) = n$, where $n$ is the number of columns." solution="Given $A$ is a $4 \times 5$ matrix, so the number of columns $n=5$.
Given $\operatorname{nullity}(A) = 2$.
By the Rank-Nullity Theorem:
>

>
>
The rank of $A$ is 3."
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:::question type="MSQ" question="Let $A$ and $B$ be $n \times n$ matrices. Which of the following conditions imply that $A$ is invertible?" options=["$\det(A) \neq 0$","$\operatorname{rank}(A) = n$","$Ax = 0$ has only the trivial solution $x=0$","$A^T$ is invertible"] answer="$\det(A) \neq 0$,$\operatorname{rank}(A) = n$,$Ax = 0$ has only the trivial solution $x=0$,$A^T$ is invertible" hint="Recall the equivalent conditions for matrix invertibility." solution="A square matrix $A$ of order $n$ is invertible if and only if any of the following equivalent conditions hold:

* $\det(A) \neq 0$: This is the fundamental definition of a non-singular matrix, which is equivalent to invertibility. Correct.
* $\operatorname{rank}(A) = n$: A matrix is invertible if and only if its rank is equal to its order (full rank). Correct.
* $Ax = 0$ has only the trivial solution $x=0$: This means the null space of $A$ contains only the zero vector, implying $\operatorname{nullity}(A)=0$. By the Rank-Nullity Theorem ($\operatorname{rank}(A) + \operatorname{nullity}(A) = n$), this means $\operatorname{rank}(A)=n$, which implies invertibility. Correct.
* $A^T$ is invertible: If $A^T$ is invertible, then $\det(A^T) \neq 0$. Since $\det(A^T) = \det(A)$, this implies $\det(A) \neq 0$, meaning $A$ is also invertible. Correct.

All the given options are correct conditions for $A$ to be invertible."
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:::question type="NAT" question="Given $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix}$. Find the determinant of $(AB)^{-1}$." answer="0.25" hint="Use the properties of determinants: $\det(AB) = \det(A)\det(B)$ and $\det(A^{-1}) = 1/\det(A)$." solution="Step 1: Calculate $\det(A)$.
>

Step 2: Calculate $\det(B)$.
>

Step 3: Calculate $\det(AB)$.
Using the property $\det(AB) = \det(A)\det(B)$:
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Step 4: Calculate $\det((AB)^{-1})$.
Using the property $\det(X^{-1}) = 1/\det(X)$:
>

The value is $1/4 = 0.25$."
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Summary

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What's Next?

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="For what value of

is the matrix singular?" options=["0","1","2","5"] answer="0" hint="A matrix is singular if its determinant is zero." solution="The determinant of an upper triangular matrix is the product of its diagonal elements.

For to be singular, .
"
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:::question type="NAT" question="Consider the system of linear equations:

Find the value of for which the system has infinitely many solutions." answer="3" hint="For infinitely many solutions, the rank of the coefficient matrix must be equal to the rank of the augmented matrix, and both must be less than the number of variables. This implies a row of zeros in the row-reduced echelon form for both matrices." solution="The augmented matrix is:

Applying row operations:





For infinitely many solutions, the last row of the row-reduced echelon form must be entirely zeros.
Thus,
And
The question asks for , which is 3."
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:::question type="MCQ" question="What is the rank of the matrix

?" options=["1","2","3","0"] answer="1" hint="Observe the linear dependence between the rows or columns. The rank is the number of linearly independent rows (or columns)." solution="Notice that and . All rows are scalar multiples of the first row. Since the first row is non-zero, there is only one linearly independent row. Therefore, the rank of the matrix is 1."
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:::question type="NAT" question="If

is a matrix such that , what is the value of ?" answer="6.75" hint="Recall the determinant properties: for an matrix, and ." solution="Given that is a matrix and .
We need to find .
Using the property , where is the order of the matrix (here, ):


Now, using the property :


"
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What's Next?