Question

question 16 points 2

for the matrices \\(a = \

$$\begin{bmatrix} -9 & 1 \\\\ 2 & -2 \\\\ 3 & 0 \\end{bmatrix}$$

\\) and \\(b = \

$$\begin{bmatrix} -1 & 8 \\\\ 2 & 5 \\end{bmatrix}$$

\\), find \\(a \times b\\).

\\(\circ \

$$\begin{bmatrix} 11 & -67 \\\\ -6 & 6 \\\\ -3 & 24 \\end{bmatrix}$$

\\)

\\(\circ \

$$\begin{bmatrix} 9 & 8 \\\\ 4 & -10 \\\\ 3 & 0 \\end{bmatrix}$$

\\)

Explanation:

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<reasoning>

Determine matrix dimensions

Using the Matrix Dimensions knowledge point, we first identify the sizes of the given matrices to see if multiplication is defined.
Matrix \(A\) has 3 rows and 2 columns, so its dimension is \(3 \times 2\).
Matrix \(B\) has 2 rows and 2 columns, so its dimension is \(2 \times 2\).
Since the number of columns in \(A\) (which is 2) equals the number of rows in \(B\) (which is 2), the product \(A \times B\) is defined and will result in a \(3 \times 2\) matrix.

Set up the product matrix elements

Let the product matrix be \(C = A \times B\), where \(C\) is a \(3 \times 2\) matrix:
\[
C =

$$\begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \\ c_{31} & c_{32} \end{bmatrix}$$

\]
We calculate each element \(c_{ij}\) by taking the dot product of row \(i\) of matrix \(A\) and column \(j\) of matrix \(B\).
Given:
\[
A =

$$\begin{bmatrix} -9 & 1 \\ 2 & -2 \\ 3 & 0 \end{bmatrix}$$

, \quad B =

$$\begin{bmatrix} -1 & 8 \\ 2 & 5 \end{bmatrix}$$

\]

Calculate the first row of the product

For row 1 of \(C\):

• \(c_{11} = (-9)(-1) + (1)(2) = 9 + 2 = 11\)
• \(c_{12} = (-9)(8) + (1)(5) = -72 + 5 = -67\)

Calculate the second row of the product

For row 2 of \(C\):

• \(c_{21} = (2)(-1) + (-2)(2) = -2 - 4 = -6\)
• \(c_{22} = (2)(8) + (-2)(5) = 16 - 10 = 6\)

Calculate the third row of the product

For row 3 of \(C\):

• \(c_{31} = (3)(-1) + (0)(2) = -3 + 0 = -3\)
• \(c_{32} = (3)(8) + (0)(5) = 24 + 0 = 24\)

Combining these elements gives:
\[
A \times B =

$$\begin{bmatrix} 11 & -67 \\ -6 = & 6 \\ -3 & 24 \end{bmatrix}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) \(

$$\begin{bmatrix} 11 & -67 \\ -6 & 6 \\ -3 & 24 \end{bmatrix}$$

\)</mcq-correct>
<mcq-option>(B) \(

$$\begin{bmatrix} 9 & 8 \\ 4 & -10 \\ 3 & 0 \end{bmatrix}$$

\)</mcq-option>
</answer>

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Answer:

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</pre_analysis>

<reasoning>

Determine matrix dimensions

Using the Matrix Dimensions knowledge point, we first identify the sizes of the given matrices to see if multiplication is defined.
Matrix \(A\) has 3 rows and 2 columns, so its dimension is \(3 \times 2\).
Matrix \(B\) has 2 rows and 2 columns, so its dimension is \(2 \times 2\).
Since the number of columns in \(A\) (which is 2) equals the number of rows in \(B\) (which is 2), the product \(A \times B\) is defined and will result in a \(3 \times 2\) matrix.

Set up the product matrix elements

Let the product matrix be \(C = A \times B\), where \(C\) is a \(3 \times 2\) matrix:
\[
C =

$$\begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \\ c_{31} & c_{32} \end{bmatrix}$$

\]
We calculate each element \(c_{ij}\) by taking the dot product of row \(i\) of matrix \(A\) and column \(j\) of matrix \(B\).
Given:
\[
A =

$$\begin{bmatrix} -9 & 1 \\ 2 & -2 \\ 3 & 0 \end{bmatrix}$$

, \quad B =

$$\begin{bmatrix} -1 & 8 \\ 2 & 5 \end{bmatrix}$$

\]

Calculate the first row of the product

For row 1 of \(C\):

• \(c_{11} = (-9)(-1) + (1)(2) = 9 + 2 = 11\)
• \(c_{12} = (-9)(8) + (1)(5) = -72 + 5 = -67\)

Calculate the second row of the product

For row 2 of \(C\):

• \(c_{21} = (2)(-1) + (-2)(2) = -2 - 4 = -6\)
• \(c_{22} = (2)(8) + (-2)(5) = 16 - 10 = 6\)

Calculate the third row of the product

For row 3 of \(C\):

• \(c_{31} = (3)(-1) + (0)(2) = -3 + 0 = -3\)
• \(c_{32} = (3)(8) + (0)(5) = 24 + 0 = 24\)

Combining these elements gives:
\[
A \times B =

$$\begin{bmatrix} 11 & -67 \\ -6 = & 6 \\ -3 & 24 \end{bmatrix}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) \(

$$\begin{bmatrix} 11 & -67 \\ -6 & 6 \\ -3 & 24 \end{bmatrix}$$

\)</mcq-correct>
<mcq-option>(B) \(

$$\begin{bmatrix} 9 & 8 \\ 4 & -10 \\ 3 & 0 \end{bmatrix}$$

\)</mcq-option>
</answer>

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