Q64 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics

Let $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & -13 \\ 5 & -10 \end{bmatrix}$ be two matrices. If the variables $x$ and $y$ satisfy the matrix equation $((A^{-1})^2 + B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, then the ordered pair $(x, y) = $

  • A
    $(3, 5)$
  • B
    $(10, 7)$
  • C
    $(4, 6)$
  • Correct answer
    D
    $(5, 3)$

Solution

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

Determinant of $A$ is $|A| = (3)(-1) - (-4)(1) = 1$

The inverse of $A$ is $A^{-1} = \dfrac{1}{|A|} \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix}$

Now, $(A^{-1})^2 = \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 1 - 4 & -4 + 12 \\ 1 - 3 & -4 + 9 \end{bmatrix} = \begin{bmatrix} -3 & 8 \\ -2 & 5 \end{bmatrix}$

Adding matrix $B$:

$(A^{-1})^2 + B = \begin{bmatrix} -3 & 8 \\ -2 & 5 \end{bmatrix} + \begin{bmatrix} 6 & -13 \\ 5 & -10 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ 3 & -5 \end{bmatrix}$

The given matrix equation is $((A^{-1})^2 + B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

$\begin{bmatrix} 3 & -5 \\ 3 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

This gives the equation $3x - 5y = 0 \Rightarrow 3x = 5y$

Checking the given options, the ordered pair $(5, 3)$ satisfies this equation since $3(5) - 5(3) = 0$.

Answer: $(5, 3)$

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

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