If A = bmatrix 3 & 1 \\ -1 & 2 bmatrix and I = bmatrix 1 & 0 \\ 0 & 1 bmatrix, then the value of A^2 - 5A + 6I is Option 1: bmatrix -1 & 0 \\ 0 & -1 bmatrix Option 2: bmatrix 0 & 0 \\ 0 & 0 bmatrix Option 3: bmatrix 1 & 1 \\ 1 & 1 bmatrix Option 4: bmatrix 0 & -1 \\ -1 & 0 bmatrix

Answer: bmatrix -1 & 0 \\ 0 & -1 bmatrix. Solution: A = bmatrix 3 & 1 \\ -1 & 2 bmatrix I = bmatrix 1 & 0 \\ 0 & 1 bmatrix A^2 = A x A = bmatrix 3 & 1 \\ -1 & 2 bmatrix x bmatrix 3 & 1 \\ -1 & 2 bmatrix Position (1,1): (3)(3) + (1)(-1) = 9 - 1 = 8 Position (1,2): (3)(1) + (1)(2) = 3 + 2 = 5 Position (2,1): (-1)(3) + (2)(-1) = -3 - 2 = -5 Position (2,2): (-1)(1) + (2)(2) = -1 + 4 = 3 A^2 = bmatrix 8 & 5 \\ -5 & 3 bmatrix 5A = 5 x bmatrix 3 & 1 \\ -1 & 2 bmatrix 5A = bmatrix 15 & 5 \\ -5 & 10 bmatrix 6I = 6 x bmatrix 1 & 0 \\ 0 & 1 bmatrix 6I = bmatrix 6 & 0 \\ 0 & 6 bmatrix A^2 - 5A + 6I = bmatrix 8 & 5 \\ -5 & 3 bmatrix - bmatrix 15 & 5 \\ -5 & 10 bmatrix + bmatrix 6 & 0 \\ 0 & 6 bmatrix Position (1,1): 8 - 15 + 6 = -1 Position (1,2): 5 - 5 + 0 = 0 Position (2,1): -5 - (-5) + 0 = 0 Position (2,2): 3 - 10 + 6 = -1 A^2 - 5A + 6I = bmatrix -1 & 0 \\ 0 & -1 bmatrix

Algebra

Matrices & Determinants

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If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ , then the value of $A^2 - 5A + 6I$ is

$\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$

$\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

$\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$

✅ Correct Option: 1

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

$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 = A \times A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

Position (1,1): $(3)(3) + (1)(-1) = 9 - 1 = 8$

Position (1,2): $(3)(1) + (1)(2) = 3 + 2 = 5$

Position (2,1): $(-1)(3) + (2)(-1) = -3 - 2 = -5$

Position (2,2): $(-1)(1) + (2)(2) = -1 + 4 = 3$

$A^2 = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

$5A = 5 \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$

$6I = 6 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$6I = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

$A^2 - 5A + 6I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

Position (1,1): $8 - 15 + 6 = -1$

Position (1,2): $5 - 5 + 0 = 0$

Position (2,1): $-5 - (-5) + 0 = 0$

Position (2,2): $3 - 10 + 6 = -1$

$A^2 - 5A + 6I = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$

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CUET Mathematics Past Year Questions (Topic-Wise)

If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ , then the value of $A^2 - 5A + 6I$ is