Question

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

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

Hint:

Remember that: \[ (kA)^T=kA^T \] and matrix multiplication must always follow row-column multiplication rules.

Answer 1

The Correct Option is D

Step 1: Find the transpose of matrix \(A\).
Given, \[ A= \begin{bmatrix} 2 & -3\\ -4 & 1 \end{bmatrix} \] Therefore, \[ A^T= \begin{bmatrix} 2 & -4\\ -3 & 1 \end{bmatrix} \]

Step 2: Calculate \((A^T)^2\).
We compute \[ (A^T)^2=A^T \cdot A^T \] So, \[ (A^T)^2= \begin{bmatrix} 2 & -4\\ -3 & 1 \end{bmatrix} \begin{bmatrix} 2 & -4\\ -3 & 1 \end{bmatrix} \] Multiplying the matrices,
First row first column: \[ (2)(2)+(-4)(-3)=4+12=16 \] First row second column: \[ (2)(-4)+(-4)(1)=-8-4=-12 \] Second row first column: \[ (-3)(2)+(1)(-3)=-6-3=-9 \] Second row second column: \[ (-3)(-4)+(1)(1)=12+1=13 \] Hence, \[ (A^T)^2= \begin{bmatrix} 16 & -12\\ -9 & 13 \end{bmatrix} \]

Step 3: Calculate \((12A)^T\).
First, \[ 12A= \begin{bmatrix} 24 & -36\\ -48 & 12 \end{bmatrix} \] Taking transpose, \[ (12A)^T= \begin{bmatrix} 24 & -48\\ -36 & 12 \end{bmatrix} \]

Step 4: Add the two matrices.
Now, \[ (A^T)^2+(12A)^T = \begin{bmatrix} 16 & -12\\ -9 & 13 \end{bmatrix} + \begin{bmatrix} 24 & -48\\ -36 & 12 \end{bmatrix} \] Adding corresponding entries, \[ = \begin{bmatrix} 16+24 & -12-48\\ -9-36 & 13+12 \end{bmatrix} \] \[ = \begin{bmatrix} 40 & -60\\ -45 & 25 \end{bmatrix} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{ \begin{bmatrix} 40 & -60\\ -45 & 25 \end{bmatrix} } \] which corresponds to option (4).