Q60 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
If $A = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}$ and $A^2 + xA + yI_2 = O_2$, where $I_2$ and $O_2$ are the identity matrix and null matrix of order 2 respectively, then:
- A$x = 4,\ y = 4$
- Correct answerB$x = -4,\ y = 4$
- C$x = -4,\ y = -2$
- D$x = 4,\ y = -4$
Solution
$A^2 = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 4 & -4 \\ 0 & 4 \end{bmatrix}$
Substituting $A^2$, $A$, and $I_2$ into the given equation $A^2 + xA + yI_2 = O_2$:
$\begin{bmatrix} 4 & -4 \\ 0 & 4 \end{bmatrix} + x \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
$\begin{bmatrix} 4 + 2x + y & -4 - x \\ 0 & 4 + 2x + y \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Equating the corresponding elements to zero:
$-4 - x = 0 \Rightarrow x = -4$
$4 + 2x + y = 0$
Substituting $x = -4$ into the second equation:
$4 + 2(-4) + y = 0 \Rightarrow y = 4$
Answer: $x = -4,\ y = 4$
Substituting $A^2$, $A$, and $I_2$ into the given equation $A^2 + xA + yI_2 = O_2$:
$\begin{bmatrix} 4 & -4 \\ 0 & 4 \end{bmatrix} + x \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix} + y \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
$\begin{bmatrix} 4 + 2x + y & -4 - x \\ 0 & 4 + 2x + y \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Equating the corresponding elements to zero:
$-4 - x = 0 \Rightarrow x = -4$
$4 + 2x + y = 0$
Substituting $x = -4$ into the second equation:
$4 + 2(-4) + y = 0 \Rightarrow y = 4$
Answer: $x = -4,\ y = 4$
Hence, the correct answer is option: 'B'.