If $A = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$ and $I$ is an identity matrix of order 2, then $A - 3I$ equals

Given $A = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$ and $I$ is an identity matrix of order 2.

The identity matrix for a 2×2 matrix is $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

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To find $3I$:

$3I = 3 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$3I = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

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To find $A - 3I$:

$A - 3I = \begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

$A - 3I = \begin{bmatrix} 4-3 & 5-0 \\ 2-0 & 1-3 \end{bmatrix}$

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

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To find the inverse of $A$, calculate the determinant:

$\det(A) = (4)(1) - (5)(2)$

$\det(A) = 4 - 10$

$\det(A) = -6$

Using the inverse formula for a 2×2 matrix:

$A^{-1} = \frac{1}{-6}\begin{bmatrix} 1 & -5 \\ -2 & 4 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -\frac{1}{6} & \frac{5}{6} \\ \frac{1}{3} & -\frac{2}{3} \end{bmatrix}$

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To find $3A^{-1}$:

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

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To find $I + 3A^{-1}$:

$I + 3A^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -\frac{1}{2} & \frac{5}{2} \\ 1 & -2 \end{bmatrix}$

$I + 3A^{-1} = \begin{bmatrix} \frac{1}{2} & \frac{5}{2} \\ 1 & -1 \end{bmatrix}$

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To find $2(I + 3A^{-1})$:

$2(I + 3A^{-1}) = 2 \times \begin{bmatrix} \frac{1}{2} & \frac{5}{2} \\ 1 & -1 \end{bmatrix}$

$2(I + 3A^{-1}) = \begin{bmatrix} 1 & 5 \\ 2 & -2 \end{bmatrix}$

Therefore, $A - 3I = 2(I + 3A^{-1}) = \begin{bmatrix} 1 & 5 \\ 2 & -2 \end{bmatrix}$