GATE 2024 EE Q.11: solution and answer

Official answer: CSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Rule: a square matrix has an inverse if and only if $\det A\ne 0$. Equivalently, its rows must be linearly independent, so a row that is a multiple of another row makes the matrix singular.

2. (A): Row 3 is $[0.524]=0.5[148]=0.5\times \text{Row 1}$. So $\det =0$ and the matrix is singular. (Check by expanding along Row 1: $1(4\cdot 4-2\cdot 2)-4(0\cdot 4-2\cdot 0.5)+8(0\cdot 2-4\cdot 0.5)=12+4-16=0$.)

3. (B): Row 2 is $[246]=2\times \text{Row 1}$. So $\det =0$ and the matrix is singular.

4. (D): Row 3 is $[31224]=3\times \text{Row 1}$. So $\det =0$ and the matrix is singular.

5. (C): No row is a multiple of another. That alone does not prove the matrix is invertible, so compute the determinant by cofactor expansion along Row 1: $$\det =1(4\cdot 4-2\cdot 2)-4(0\cdot 4-2\cdot 1)+8(0\cdot 2-4\cdot 1)=12+8-32=-12$$

6. $\det =-12\ne 0$, so matrix (C) is non-singular and has an inverse. The answer is (C).