GATE CSE Linear Algebra – Engineering Mathematics

GATE CSE Linear Algebra – Engineering Mathematics | Set 1

Questions 1-5 of 100

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Question 1

MCQ · Single Correct 1 mark

Let $A =$$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B =$$\begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}$. What is the determinant of matrix $C = A \cdot B$?

A 4 B -8 C -4 D 8

Question 2

MCQ · Single Correct 1 mark

Consider the following system of linear equations:

x+y+$z = 6$
2x+2y+2z = 12
x-y+$z = 2$

A The system has infinitely many solutions. B The system has no solution. C The system has a unique solution. D The system has exactly two solutions.

📘 Detailed Solution

Explanation:
The second equation is exactly twice the first equation, so it is not independent.
Subtracting the third equation from the first:
$2y=4\implies y=2$
Substituting into the first equation:
$x+z=4$
Therefore, infinitely many solutions exist because $x$ and $z$ can take infinitely many values satisfying $x + z = 4$.
$\boxed{\text{Infinitely\ many\ solutions}}$

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Further explanation:
The first and third equations are independent. Their subtraction gives 2y = 4, so y = 2; then x + z = 4. Taking z = t gives the complete family (x,y,z) = (4−t, 2, t), where t is any real number. This explicitly shows infinitely many solutions.

Key Points to Remember:
• For Ax=b, consistency requires rank(A)=rank([A|b]).
• A consistent system has a unique solution when the common rank equals the number of variables.
• If the common rank is smaller than the number of variables, free variables produce infinitely many solutions.

Question 3

MCQ · Single Correct 1 mark

If A is a 3x3 matrix with $\det(A) = 4, what$ is $\det(adj(adj(A)))$?

A 16 B 1024 C 256 D 64

📘 Solution & Detailed Explanation

🔢 Step-by-Step Solution

Step-by-Step Derivation

Recall the Key Properties:
• For any square matrix $A$ of order $n \times n$:
$\det (\text{adj}(A))=[\det (A)]^{n-1}$
• Applying this property a second time for the double adjoint gives:
$\det (\text{adj}(\text{adj}(A)))=[\det (A)]^{(n-1)^{2}}$

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Substitute the Given Values:
• Order of the matrix (\(n\)) = \(3\)
• Determinant of \(A\) (\(\det(A)\)) = \(4\)
\(\det (\text{adj}(\text{adj}(A)))=4^{(3-1)^{2}}\)
\(\det (\text{adj}(\text{adj}(A)))=4^{(2)^{2}}\)
\(\det (\text{adj}(\text{adj}(A)))=4^{4}\)

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Final Calculation:
\(4^{4}=4\times 4\times 4\times 4=256\)

Further explanation:
For a 3 x 3 matrix, $\det(adj(M)) = [\det(M)]^2$. First set M=A: $\det(adj(A)) = 4^2 = 16$. Apply the same identity again to M=adj(A): $\det(adj(adj(A))) = 16^2 = 256$. Equivalently, the exponent is (3-1)^2 = 4.

Key Points to Remember:
• Use determinant identities carefully: $\det(AB)=\det(A)\det(B), \det(A^{-1})=1/\det(A)$, and $\det(cA)=c^n \det(A)$ for an n x n matrix.
• A square matrix is invertible exactly when its determinant is nonzero.
• For triangular matrices, the determinant is the product of the diagonal entries.

Question 4

MCQ · Single Correct 1 mark

Let $A$ be a $3 \times 3$ real matrix with eigenvalues $1$, $-1$, and $3$. What is the trace of the matrix $A^2 + 2I$, where $I$ is the $3 \times 3$ identity matrix?

A 17 B 20 C 22 D 27

📘 Detailed Solution

Correct! If $\lambda $ is an eigenvalue of $A$, then $\lambda^2 + 2$ is an eigenvalue of $A^2 + 2I$.

For eigenvalues $1, -1, 3$, the eigenvalues of $A^2 + 2I$ are $1^2+2 = 3$, $(-1)^2+2 = 3$, and $3^2+2 = 11$. The trace is the sum of these eigenvalues: $3 + 3 + 11 = 17$.

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Further explanation:
The eigenvalue mapping rule follows directly from Av=λv: (A^2+2I)v = (λ^2+2)v. Thus the transformed eigenvalues are 3, 3, and 11. The trace is their sum, 17.

Key Points to Remember:
• The trace is the sum of eigenvalues and the determinant is their product, counting algebraic multiplicity.
• For a polynomial p, eigenvalues of p(A) are p($\lambda$), where $\lambda ranges over$ the eigenvalues of A.
• The characteristic polynomial encodes the eigenvalues, trace, and determinant.

Question 5

MSQ · Multiple Correct 1 mark

Let A be an n x n real matrix. Which of the following statements are ALWAYS TRUE? (Select all that apply)

A If A is symmetric, all its eigenvalues are real. B If A is skew-symmetric, its non-zero eigenvalues are purely imaginary. C If A is orthogonal, the absolute value of all its eigenvalues is 1. D If $\det(A) = 0$, then 0 is an eigenvalue of A.

📘 Solution & Detailed Explanation

🔢 Step-by-Step Solution

All four options are fundamental theorems of linear algebra:

Real symmetric matrices have purely real eigenvalues.

Real skew-symmetric matrices have zero or purely imaginary eigenvalues.

Orthogonal matrices preserve length, so |$\lambda| = 1$.

$\det(A) = product of eigenvalues$, so $\det(A) = 0 implies$ at least one eigenvalue is 0.

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Further explanation:
Option-by-option: (1) is the spectral theorem for real symmetric matrices; (2) follows because a real skew-symmetric matrix has eigenvalues that are zero or purely imaginary; (3) follows from preservation of vector length by an orthogonal matrix; (4) follows because det(A)=0 makes A singular, so 0 is an eigenvalue.

Key Points to Remember:
• Identify the definition or theorem that directly applies to the given data.
• Keep dimensions and matrix order consistent throughout the calculation.
• Check the result against the relevant algebraic property before selecting the answer.