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)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\begin{bmatrix}15&9\\6&12\end{bmatrix}$
Given: $A = \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix},\quad I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Compute $A^2$:
$A^2 = A \cdot A = \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} (5)(5)+(3)(2) & (5)(3)+(3)(4) \\ (2)(5)+(4)(2) & (2)(3)+(4)(4) \end{bmatrix} = \begin{bmatrix} 25 + 6 & 15 + 12 \\ 10 + 8 & 6 + 16 \end{bmatrix} = \begin{bmatrix} 31 & 27 \\ 18 & 22 \end{bmatrix} $
Compute $6A$:
$6A = 6 \cdot \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 30 & 18 \\ 12 & 24 \end{bmatrix}$
Compute $14I$:
$14I = 14 \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$
Now compute $A^2 - 6A + 14I$:
$\begin{bmatrix} 31 & 27 \\ 18 & 22 \end{bmatrix} - \begin{bmatrix} 30 & 18 \\ 12 & 24 \end{bmatrix} + \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix} = \begin{bmatrix} 31 - 30 + 14 & 27 - 18 + 0 \\ 18 - 12 + 0 & 22 - 24 + 14 \end{bmatrix} = \begin{bmatrix} 15 & 9 \\ 6 & 12 \end{bmatrix} $