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Mathematics Question on Matrices

If A=[32 −11]andB=[−10 25 34],A = \begin{bmatrix} 3 & 2 \\\ -1 & 1 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} -1 & 0 \\\ 2 & 5 \\\ 3 & 4 \end{bmatrix},then (BA)T(BA)^T is equal to:

A

[−315 −2910]\begin{bmatrix} -3 & 1 & 5 \\\ -2 & 9 & 10 \end{bmatrix}

B

[315 2910]\begin{bmatrix} 3 & 1 & 5 \\\ 2 & 9 & 10 \end{bmatrix}

C

[−3−2 19 510]\begin{bmatrix} -3 & -2 \\\ 1 & 9 \\\ 5 & 10 \end{bmatrix}

D

[32 19 510]\begin{bmatrix} 3 & 2 \\\ 1 & 9 \\\ 5 & 10 \end{bmatrix}

Answer

[−315 −2910]\begin{bmatrix} -3 & 1 & 5 \\\ -2 & 9 & 10 \end{bmatrix}

Explanation

Solution

To find (BA)T(BA)^T, we compute BABA first, then take its transpose.
Multiplying BB and AA The matrices BB and AA are:
B=[−10 25 34],A=[32 −11]B = \begin{bmatrix} -1 & 0 \\\ 2 & 5 \\\ 3 & 4 \end{bmatrix}, \quad A = \begin{bmatrix} 3 & 2 \\\ -1 & 1 \end{bmatrix}
The product BABA is calculated as:
BA=[−10 25 34][32 −11]BA = \begin{bmatrix} -1 & 0 \\\ 2 & 5 \\\ 3 & 4 \end{bmatrix} \begin{bmatrix} 3 & 2 \\\ -1 & 1 \end{bmatrix}
Perform the multiplication row by row:
1.First row of BB with both columns of AA:
[−1⋅3+0⋅(−1), −1⋅2+0⋅1]=[−3,−2][-1 \cdot 3 + 0 \cdot (-1), \ -1 \cdot 2 + 0 \cdot 1] = [-3, -2]
2.Second row of BB with both columns of AA:
[2⋅3+5⋅(−1), 2⋅2+5⋅1]=[6−5,4+5]=[1,9][2 \cdot 3 + 5 \cdot (-1), \ 2 \cdot 2 + 5 \cdot 1] = [6 - 5, 4 + 5] = [1, 9]
3.Third row of BB with both columns of AA:
[3⋅3+4⋅(−1), 3⋅2+4⋅1]=[9−4,6+4]=[5,10][3 \cdot 3 + 4 \cdot (-1), \ 3 \cdot 2 + 4 \cdot 1] = [9 - 4, 6 + 4] = [5, 10]
Thus,
BA=[−3−2 19 510]BA = \begin{bmatrix} -3 & -2 \\\ 1 & 9 \\\ 5 & 10 \end{bmatrix}
Transposing BABA The transpose of BABA is obtained by interchanging rows and columns:
(BA)T=[−315 −2910](BA)^T = \begin{bmatrix} -3 & 1 & 5 \\\ -2 & 9 & 10 \end{bmatrix}
Final Answer: The matrix (BA)T(BA)^T is:
[−315 −2910]\begin{bmatrix} -3 & 1 & 5 \\\ -2 & 9 & 10 \end{bmatrix}