If $A = \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix}$, then the matrix $A^2 - 6A + 14$ I is (where I is an identity matrix of order 2)

Given $A = \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix}$

Calculate $A^2$:

$A^2 = \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix}$

Position (1,1): $(5)(5) + (3)(2) = 25 + 6 = 31$

Position (1,2): $(5)(3) + (3)(4) = 15 + 12 = 27$

Position (2,1): $(2)(5) + (4)(2) = 10 + 8 = 18$

Position (2,2): $(2)(3) + (4)(4) = 6 + 16 = 22$

$A^2 = \begin{bmatrix} 31 & 27 \\ 18 & 22 \end{bmatrix}$

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Calculate $6A$:

$6A = 6 \times \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix}$

$6A = \begin{bmatrix} 30 & 18 \\ 12 & 24 \end{bmatrix}$

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Calculate $14I$ where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$:

$14I = 14 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$14I = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$

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Calculate $A^2 - 6A + 14I$:

$A^2 - 6A + 14I = \begin{bmatrix} 31 & 27 \\ 18 & 22 \end{bmatrix} - \begin{bmatrix} 30 & 18 \\ 12 & 24 \end{bmatrix} + \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$

Position (1,1): $31 - 30 + 14 = 15$

Position (1,2): $27 - 18 + 0 = 9$

Position (2,1): $18 - 12 + 0 = 6$

Position (2,2): $22 - 24 + 14 = 12$

$A^2 - 6A + 14I = \begin{bmatrix} 15 & 9 \\ 6 & 12 \end{bmatrix}$