Question

If $A = \begin{bmatrix} 2 & -3\\ -4 & 1 \end{bmatrix}$, then $(A^T)^2 + (12A)^T = $

• A. $\begin{bmatrix} 5 & 8 \\ -9 & 5 \end{bmatrix}$
• B. $\begin{bmatrix} 5 & 8 \\ -12 & 5 \end{bmatrix}$
• C. $\begin{bmatrix} 40 & -45 \\ 60 & 25 \end{bmatrix}$
• D. $\begin{bmatrix} 40 & -60 \\ -45 & 25 \end{bmatrix}$

Hint:

Transpose distributes over matrix operations but not over products; evaluate each step carefully.

Answer 1

The Correct Option is D

Compute $A^T = \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix}$
Then $(A^T)^2 = A^T \cdot A^T = \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix}$
$= \begin{bmatrix} 4 + 12 & -8 - 4 \\ -6 - 3 & 12 + 1 \end{bmatrix} = \begin{bmatrix} 16 & -12 \\ -9 & 13 \end{bmatrix}$
Now compute $(12A)^T = 12A^T = 12 \cdot \begin{bmatrix} 2 & -4 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} 24 & -48 \\ -36 & 12 \end{bmatrix}$
Add both: $\begin{bmatrix} 16 & -12 \\ -9 & 13 \end{bmatrix} + \begin{bmatrix} 24 & -48 \\ -36 & 12 \end{bmatrix} = \begin{bmatrix} 40 & -60 \\ -45 & 25 \end{bmatrix}$