Matrices Short Notes for GATE DA

GATE DA Matrices: 8 chapters, 13 previous year questions (65% of Linear Algebra), 685 practice questions and one solved question from each chapter.

About Matrices Short Notes

Quick revision sheets for Matrices in GATE DA. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 8 chapters fast before the exam.

Matrices Weightage in GATE DA

Matrices accounts for 13 of 20 Linear Algebra previous year questions in our bank (65%), about 4.3 per paper across 3 papers.

Matrices Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Chapter 1 — Matrices		0	0%	0
Matrix Operations, Determinants and Gaussian Elimination	Gaussian Elimination and Matrix Operation Complexity, Determinants of Matrix Expressions	2	15%	104
Chapter 2 — Matrices		0	0%	0
Rank, Invertibility and Linear Systems	Rank, Nullity and Matrix Polynomial Equations, Invertibility and Eigenvalues of Rank-One Updates	3	23%	158
Chapter 3 — Matrices		0	0%	0
Eigenvalues, Eigenvectors and Matrix Powers	Rotation Matrices, Matrix Powers and Trace-Eigenvalue Relations, Characteristic Polynomial and Nature of Eigenvalues	3	23%	154
Orthogonal, Projection and Special Matrices	Projection Matrices and Quadratic Forms, Orthogonal and Involutory Matrices	3	23%	157
Singular Values and Gram Matrices	Gram Matrices and Positive Definiteness, Singular Values and Spectral Properties of Special Matrices	2	15%	112

One Solved Question from Each Matrices Chapter

Question 1 · Matrix Operations, Determinants and Gaussian Elimination MCQ

For an $n \times n$ upper triangular system of linear equations, the forward elimination phase requires no multiplications, and the back substitution phase requires a specific number of multiplications. If $n = 14$ , what is the exact total number of multiplications required to completely solve the system?

A.
91
B.
910
C.
1001
D.
105

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct formula application question that tests your understanding of how matrix structure changes the cost of Gaussian elimination. An upper triangular matrix already has zeros below the diagonal, so the forward elimination phase requires no work.

Step 1: Identify the cost of forward elimination for an upper triangular matrix. Since there are no entries below the diagonal to eliminate, the number of multiplications is 0.

Step 2: Identify the cost of back substitution for an $n \times n$ system. The exact formula for the number of multiplications is $\frac{n ( n - 1 )}{2}$ .

Step 3: Substitute $n = 14$ into the back substitution formula.

Total multiplications = $\frac{14 \times 13}{2} = \frac{182}{2} = 91$ .

Answer: 91

Question 2 · Rank, Invertibility and Linear Systems MCQ

Let $A$ be a $4 \times 4$ real matrix satisfying $A^{2} = 2 A$ with $Tr (A) = 2$ . Let $B = A - I$ . Which of the following correctly orders $Nullity (B)$ , $Rank (A)$ , and $Nullity (A)$ ?

A.
$e x t N u l l i t y (B) < e x t R ank (A) < e x t N u l l i t y (A)$
B.
$e x t R ank (A) < e x t N u l l i t y (B) < e x t N u l l i t y (A)$
C.
$e x t N u l l i t y (A) < e x t R ank (A) < e x t N u l l i t y (B)$
D.
$e x t R ank (A) < e x t N u l l i t y (B) = e x t N u l l i t y (A)$

Correct Answer:

Step-by-Step Solution

Key idea: This is a matrix polynomial eigenvalue extraction question. The equation $A^{2} = 2 A$ constrains the eigenvalues, and the trace determines their multiplicities. The shift $B = A - I$ then transforms these eigenvalues, changing the null space structure.

Step 1: Extract eigenvalue constraints from $A^{2} = 2 A$ .

Rewrite as $A^{2} - 2 A = 0$ , so $A (A - 2 I) = 0$ . The minimal polynomial divides $x (x - 2)$ , which has distinct roots $0$ and $2$ . Therefore, the eigenvalues of $A$ are a subset of ${0, 2}$ .

Step 2: Use the trace to determine multiplicities.

$Tr (A) = 2$ means the sum of eigenvalues is $2$ . Since eigenvalues are $0$ or $2$ , and the matrix is $4 \times 4$ , exactly one eigenvalue is $2$ and three are $0$ .

Step 3: Compute $Rank (A)$ and $Nullity (A)$ .

$Rank (A) =$ number of non-zero eigenvalues $= 1$ .

$Nullity (A) =$ number of zero eigenvalues $= 3$ .

Step 4: Compute eigenvalues of $B = A - I$ .

If $λ$ is an eigenvalue of $A$ , then $λ - 1$ is an eigenvalue of $B$ .

Eigenvalues of $B$ : $2 - 1 = 1$ (one time), $0 - 1 = - 1$ (three times).

All eigenvalues of $B$ are non-zero, so $Rank (B) = 4$ and $Nullity (B) = 0$ .

Step 5: Order the values.

$Nullity (B) = 0$ , $Rank (A) = 1$ , $Nullity (A) = 3$ .

Therefore, $Nullity (B) < Rank (A) < Nullity (A)$ .

The trap is overcounting $Nullity (B)$ by assuming the shift $B = A - I$ preserves the null space structure of $A$ . A student might incorrectly conclude $Nullity (B) = Nullity (A) = 3$ , leading to option D.

Answer: A

Question 3 · Eigenvalues, Eigenvectors and Matrix Powers MCQ

Let $A$ be a $2 \times 2$ real matrix satisfying $A^{T} A = I$ and $det (A) = 1$ . Which one of the following is IMPOSSIBLE?

A.
$A^{3} = I$ and $A \neq = I$
B.
$Tr (A) = 3$ and $A^{12} = I$
C.
$A^{6} = I$ and $Tr (A) = - 1$
D.
$A^{2} = - I$ and $Tr (A) = 1$

Correct Answer:

Step-by-Step Solution

Key idea: This is a boundary-case elimination question. Since $A^{T} A = I$ and $det (A) = 1$ , we know $A = R (θ)$ for some $θ$ . Each option imposes two conditions on $θ$ ; we check whether any $θ$ can satisfy both simultaneously.

Why this method applies: The question asks which combination is impossible. The fastest route is to translate each condition into a constraint on $θ$ and check for consistency.

Step-by-step:

Option A: $A^{3} = I$ means $R (3 θ) = I$ , so $3 θ = 2 k π$ , giving $θ = 2 k π /3$ . For $A \neq = I$ , we need $θ \neq = 0$ , so $θ = 2 π /3$ or $4 π /3$ . Both are valid. Possible. ✓

Option B: $Tr (A) = 3$ means $2 cos θ = 3$ , so $cos θ = 3 /2$ , giving $θ = π /6$ or $11 π /6$ . Check $A^{12} = I$ : $R (12 \cdot π /6) = R (2 π) = I$ . ✓ Possible. ✓

Option C: $A^{6} = I$ means $6 θ = 2 k π$ , so $θ = k π /3$ . We need $Tr (A) = 2 cos (k π /3) = - 1$ , i.e., $cos (k π /3) = - 1/2$ . For $k = 2$ : $cos (2 π /3) = - 1/2$ . ✓ Possible. ✓

Option D: $A^{2} = - I$ means $R (2 θ) = R (π)$ , so $2 θ = π + 2 k π$ , giving $θ = π /2 + k π$ . Then $Tr (A) = 2 cos (π /2 + k π)$ . For any integer $k$ : $cos (π /2 + k π) = 0$ . So $Tr (A) = 0$ always. It can never equal $1$ . Impossible. ✗

Answer: D

Trap walk (Option C): A student checks $A^{6} = I$ and lists $θ \in {0, π /3, 2 π /3, π, 4 π /3, 5 π /3}$ . They then compute $Tr (A) = 2 cos θ$ for each and might overcount the constraints, thinking "I need to check all 6 values and none might give $- 1$ ." But $θ = 2 π /3$ gives $2 cos (2 π /3) = 2 (- 1/2) = - 1$ . The student who overcounts might skip this value or miscalculate, incorrectly marking C as impossible.

Verification for D: If $A^{2} = - I$ , eigenvalues of $A$ satisfy $λ^{2} = - 1$ , so $λ = \pm i$ . The trace is $i + (- i) = 0$ (conjugate pair for real matrix). $Tr (A) = 1$ would require eigenvalues summing to 1, which contradicts $λ \in {i, - i}$ . Confirmed impossible.

Question 4 · Orthogonal, Projection and Special Matrices NAT

Let $A$ and $B$ be $3 \times 3$ real symmetric matrices. The eigenvalues of $A$ are $1, 2, 3$ and the eigenvalues of $B$ are $- 1, 0, 1$ . What is the minimum possible value of the trace of $(A + B)^{2}$ ?

Correct Answer:

12.00

Step-by-Step Solution

Key idea: The trace of a matrix square is the sum of the squares of its eigenvalues, which can be expanded using the trace properties.

Step 1: Expand the expression: $tr ((A + B)^{2}) = tr (A^{2} + A B + B A + B^{2})$ .

Step 2: Use the linearity and cyclic property of trace: $tr (A B) = tr (B A)$ . Thus, $tr ((A + B)^{2}) = tr (A^{2}) + tr (B^{2}) + 2 tr (A B)$ .

Step 3: Calculate $tr (A^{2})$ and $tr (B^{2})$ using their eigenvalues:

$tr (A^{2}) = 1^{2} + 2^{2} + 3^{2} = 14$ .

$tr (B^{2}) = (- 1)^{2} + 0^{2} + 1^{2} = 2$ .

Step 4: The expression becomes $14 + 2 + 2 tr (A B) = 16 + 2 tr (A B)$ .

Step 5: To minimize this, we must minimize $tr (A B)$ . By von Neumann's trace inequality, the minimum of $tr (A B)$ is achieved by pairing the eigenvalues of $A$ and $B$ in opposite sorted orders.

Step 6: Pair largest with smallest: $1 (1) + 2 (0) + 3 (- 1) = 1 + 0 - 3 = - 2$ .

Step 7: The minimum trace is $16 + 2 (- 2) = 12$ .

Answer: 12.00

Question 5 · Singular Values and Gram Matrices MCQ

Let $v_{1}, v_{2}, v_{3}$ be vectors in $R^{3}$ such that $v_{1}$ and $v_{2}$ are linearly independent, and $v_{3} = 2 v_{1} - v_{2}$ . Let $G$ be the $3 \times 3$ Gram matrix defined by $G_{ij} = v_{i}^{T} v_{j}$ .

Consider the following statements:

P: $det (G) > 0$

Q: $rank (G) = 2$

R: All eigenvalues of $G$ are non-negative

S: There exists a non-zero vector $c \in R^{3}$ such that $c^{T} G c = 0$

Which one of these statements is IMPOSSIBLE?

A.
P
B.
Q
C.
R
D.
S

Correct Answer:

Step-by-Step Solution

Key idea: This is a linear dependence question disguised in Gram matrix language. The relation $v_{3} = 2 v_{1} - v_{2}$ immediately tells us the vectors are dependent, which forces $det (G) = 0$ .

Exam route:

Translate the condition: $v_{3} = 2 v_{1} - v_{2}$ means $2 v_{1} - v_{2} - v_{3} = 0$ , so the vectors are linearly dependent.
For dependent vectors, the Gram matrix is singular: $det (G) = 0$ . Statement P claims $det (G) > 0$ , which is impossible.
Verify other statements are possible:

Q: Since $v_{1}, v_{2}$ are independent and $v_{3}$ depends on them, $rank (G) = 2$ . Possible.
R: All Gram matrices are positive semi-definite, so eigenvalues $\geq 0$ . Possible.
S: Since $G$ is singular, there exists non-zero $c$ with $G c = 0$ , so $c^{T} G c = 0$ . Possible.

Learning route:

The Gram matrix $G = X^{T} X$ where $X$ has the vectors as columns. Linear dependence means $X c = 0$ for some non-zero $c$ . Then $G c = X^{T} X c = X^{T} 0 = 0$ , so $G$ is singular with $det (G) = 0$ .

The rank of $G$ equals the rank of $X$ , which is the dimension of the span of the vectors. Here, $span (v_{1}, v_{2}, v_{3}) = span (v_{1}, v_{2})$ has dimension 2.

Trap analysis: Students often misread " $v_{3} = 2 v_{1} - v_{2}$ " as " $v_{3}$ is independent" or focus on " $v_{1}, v_{2}$ are independent" and forget about $v_{3}$ . The key is to translate the equation into the dependence relation first.

Verification: If $v_{1} = [1, 0, 0]^{T}$ , $v_{2} = [0, 1, 0]^{T}$ , then $v_{3} = [2, - 1, 0]^{T}$ . The matrix $X = 100010 2 - 1 0$ has rank 2, so $G = X^{T} X$ has rank 2 and $det (G) = 0$ .

Answer: Option A (P is impossible)

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Matrices Short Notes for GATE DA

About Matrices Short Notes

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Matrices Chapter Matrix

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One Solved Question from Each Matrices Chapter

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