Q93 / 180 · COMEDK 2023 (Evening Shift) · Mathematics
If $\left[\begin{array}{ccc}2+x & 3 & 4 \\ 1 & -1 & 2 \\ x & 1 & -5\end{array}\right]$ is a singular matrix, then $x$ is
- Correct answerA$-\dfrac{25}{13}$
- B$\dfrac{5}{13}$
- C$\dfrac{25}{13}$
- D$\dfrac{13}{25}$
Solution
A matrix is singular if its determinant is equal to zero. Let $A = \begin{bmatrix} 2+x & 3 & 4 \\ 1 & -1 & 2 \\ x & 1 & -5 \end{bmatrix}$.
The determinant $|A|$ is given by:
$|A| = (2+x) \begin{vmatrix} -1 & 2 \\ 1 & -5 \end{vmatrix} - 3 \begin{vmatrix} 1 & 2 \\ x & -5 \end{vmatrix} + 4 \begin{vmatrix} 1 & -1 \\ x & 1 \end{vmatrix} = 0$
Calculating the $2 \times 2$ determinants:
$|A| = (2+x)(5 - 2) - 3(-5 - 2x) + 4(1 - (-x)) = 0$
$|A| = (2+x)(3) - 3(-5 - 2x) + 4(1 + x) = 0$
Expanding the terms:
$6 + 3x + 15 + 6x + 4 + 4x = 0$
Combining like terms:
$13x + 25 = 0$
$13x = -25$
$x = -\dfrac{25}{13}$
Answer: $-\dfrac{25}{13}$
The determinant $|A|$ is given by:
$|A| = (2+x) \begin{vmatrix} -1 & 2 \\ 1 & -5 \end{vmatrix} - 3 \begin{vmatrix} 1 & 2 \\ x & -5 \end{vmatrix} + 4 \begin{vmatrix} 1 & -1 \\ x & 1 \end{vmatrix} = 0$
Calculating the $2 \times 2$ determinants:
$|A| = (2+x)(5 - 2) - 3(-5 - 2x) + 4(1 - (-x)) = 0$
$|A| = (2+x)(3) - 3(-5 - 2x) + 4(1 + x) = 0$
Expanding the terms:
$6 + 3x + 15 + 6x + 4 + 4x = 0$
Combining like terms:
$13x + 25 = 0$
$13x = -25$
$x = -\dfrac{25}{13}$
Answer: $-\dfrac{25}{13}$
Hence, the correct answer is option: 'A'.