Q62 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
If $A=\left[\begin{array}{ccc}4 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3\end{array}\right]$ then $A^{-1}$ exists if
- A$\lambda \neq 2$
- B$\lambda=0$
- C$\lambda=2$
- Correct answerD$\lambda \neq-2$
Solution
A matrix $A$ is invertible if and only if its determinant is non-zero, i.e., $|A| \neq 0$.
The determinant of matrix $A = \begin{bmatrix} 4 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix}$ is calculated by expanding along the first column:
$|A| = 4(2 \times 3 - 5 \times 1) - 0(\lambda \times 3 - (-3) \times 1) + 1(\lambda \times 5 - (-3) \times 2)$
$|A| = 4(6 - 5) - 0 + 1(5\lambda + 6)$
$|A| = 4(1) + 5\lambda + 6 = 5\lambda + 10$
For $A^{-1}$ to exist, $|A| \neq 0$.
$5\lambda + 10 \neq 0$
$5\lambda \neq -10$
$\lambda \neq -2$
Answer: $\lambda \neq-2$
The determinant of matrix $A = \begin{bmatrix} 4 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{bmatrix}$ is calculated by expanding along the first column:
$|A| = 4(2 \times 3 - 5 \times 1) - 0(\lambda \times 3 - (-3) \times 1) + 1(\lambda \times 5 - (-3) \times 2)$
$|A| = 4(6 - 5) - 0 + 1(5\lambda + 6)$
$|A| = 4(1) + 5\lambda + 6 = 5\lambda + 10$
For $A^{-1}$ to exist, $|A| \neq 0$.
$5\lambda + 10 \neq 0$
$5\lambda \neq -10$
$\lambda \neq -2$
Answer: $\lambda \neq-2$
Hence, the correct answer is option: 'D'.