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Mathematics Question on Matrices

If A=[51 −20]A = \begin{bmatrix} 5 & 1 \\\ -2 & 0 \end{bmatrix} and BT=[110 −2−1]B^T = \begin{bmatrix} 1 & 10 \\\ -2 & -1 \end{bmatrix}, then the matrix ABAB is:

A

[110 −10]\begin{bmatrix} 1 & 10 \\\ -1 & 0 \end{bmatrix}

B

[15−11 −24]\begin{bmatrix} 15 & -11 \\\ -2 & 4 \end{bmatrix}

C

[349 −2−20]\begin{bmatrix} 3 & 49 \\\ -2 & -20 \end{bmatrix}

D

[19 −2−20]\begin{bmatrix} 1 & 9 \\\ -2 & -20 \end{bmatrix}

Answer

[15−11 −24]\begin{bmatrix} 15 & -11 \\\ -2 & 4 \end{bmatrix}

Explanation

Solution

First, compute BB by transposing BTB^T:

B=[1−2 10−1].B = \begin{bmatrix} 1 & -2 \\\ 10 & -1 \end{bmatrix}.

Now compute ABAB using matrix multiplication:

AB=[51 −20]⋅[1−2 10−1].AB = \begin{bmatrix} 5 & 1 \\\ -2 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & -2 \\\ 10 & -1 \end{bmatrix}.

Perform the row-column multiplication:

First row:

[51]⋅[1 10]=(5)(1)+(1)(10)=5+10=15,\begin{bmatrix} 5 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 \\\ 10 \end{bmatrix} = (5)(1) + (1)(10) = 5 + 10 = 15,

First row:

[51]⋅[−2 −1]=(5)(−2)+(1)(−1)=−10−1=−11.\begin{bmatrix} 5 & 1 \end{bmatrix} \cdot \begin{bmatrix} -2 \\\ -1 \end{bmatrix} = (5)(-2) + (1)(-1) = -10 - 1 = -11.

Second row:

[−20]⋅[1 10]=(−2)(1)+(0)(10)=−2,\begin{bmatrix} -2 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 \\\ 10 \end{bmatrix} = (-2)(1) + (0)(10) = -2,

Second row:

[−20]⋅[−2 −1]=(−2)(−2)+(0)(−1)=4.\begin{bmatrix} -2 & 0 \end{bmatrix} \cdot \begin{bmatrix} -2 \\\ -1 \end{bmatrix} = (-2)(-2) + (0)(-1) = 4.

Thus, the resulting matrix is:

AB=[15−11 −24].AB = \begin{bmatrix} 15 & -11 \\\ -2 & 4 \end{bmatrix}.

Hence, the correct answer is:

(2)[15−11 −24].\text{(2)} \begin{bmatrix} 15 & -11 \\\ -2 & 4 \end{bmatrix}.