Q64 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics

Let $A = \begin{bmatrix} -3 & 2 \\ 1 & 4 \end{bmatrix}$ and if $A^2 - 2A + I = \begin{bmatrix} 18 & p \\ q & 11 \end{bmatrix}$, then $\ldots$

  • Correct answer
    A
    $p = -2,\ q = -1$
  • B
    $p = 2,\ q = 1$
  • C
    $p = -1,\ q = -2$
  • D
    $p = 1,\ q = 2$

Solution

Given $A = \begin{bmatrix} -3 & 2 \\ 1 & 4 \end{bmatrix}$

First, we calculate $A^2$:
$A^2 = \begin{bmatrix} -3 & 2 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 9 + 2 & -6 + 8 \\ -3 + 4 & 2 + 16 \end{bmatrix} = \begin{bmatrix} 11 & 2 \\ 1 & 18 \end{bmatrix}$

Next, we calculate $2A$:
$2A = 2 \begin{bmatrix} -3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -6 & 4 \\ 2 & 8 \end{bmatrix}$

Now, substituting these into $A^2 - 2A + I$:
$A^2 - 2A + I = \begin{bmatrix} 11 & 2 \\ 1 & 18 \end{bmatrix} - \begin{bmatrix} -6 & 4 \\ 2 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 11 - (-6) + 1 & 2 - 4 + 0 \\ 1 - 2 + 0 & 18 - 8 + 1 \end{bmatrix}$

$= \begin{bmatrix} 18 & -2 \\ -1 & 11 \end{bmatrix}$

We are given that $A^2 - 2A + I = \begin{bmatrix} 18 & p \\ q & 11 \end{bmatrix}$.

Comparing the corresponding elements of the matrices, we get:
$p = -2$
$q = -1$

Answer: $p = -2,\ q = -1$

Hence, the correct answer is option: 'A'.

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