Let

$$A = \left [\begin{matrix}-1 & 1 & 3\-2 & 5 & -4\-3 & -2 & 3\end{matrix}\right ], \text { and } B = \left [\begin{matrix}0 & -5 & -1\4 & 1 & 0\1 & 4 & 2\end{matrix}\right ].$$

Which of the following options is $- 2 A B^{T}$?

Question

Let

$$A = \left [\begin{matrix}-1 & 1 & 3\-2 & 5 & -4\-3 & -2 & 3\end{matrix}\right ], \text { and } B = \left [\begin{matrix}0 & -5 & -1\4 & 1 & 0\1 & 4 & 2\end{matrix}\right ].$$

Which of the following options is $- 2 A B^{T}$?

BlackTom AI · Accepted Answer

Matrix product AB^T is found by multiplying A by the transpose of B. A key fact: (AB^T)_{ij} = row_i(A) · row_j(B). After computing AB^T, multiply the result by -2.

1) Relevant concepts:
- Matrix transpose: B^T has rows that are the columns of B.
- Matrix multiplication: (AB^T)_{ij} = sum_k A_{ik} (B^T)_{kj} = sum_k A_{ik} B_{jk}, i.e., dot product of row i of A with row j of B.
- Scalar multiplication: multiply each entry by -2.

2) Step-by-step computation (showing all intermediate steps):

Rows of A:
row1(A) = [-1, 1, 3]
row2(A) = [-2, 5, -4]
row3(A) = [-3, -2, 3]

Rows of B:
row1(B) = [0, -5, -1]
row2(B) = [4, 1, 0]
row3(B) = [1, 4, 2]

Compute AB^T entries as dot products:

First row of AB^T:
c11 = (-1)*0 + 1*(-5) + 3*(-1) = 0 -5 -3 = -8
c12 = (-1)*4 + 1*1 + 3*0 = -4 +1 +0 = -3
c13 = (-1)*1 + 1*4 + 3*2 = -1 +4 +6 = 9
=> row1(AB^T) = [-8, -3, 9]

Second row:
c21 = (-2)*0 + 5*(-5) + (-4)*(-1) = 0 -25 +4 = -21
c22 = (-2)*4 + 5*1 + (-4)*0 = -8 +5 +0 = -3
c23 = (-2)*1 + 5*4 + (-4)*2 = -2 +20 -8 = 10
=> row2(AB^T) = [-21, -3, 10]

Third row:
c31 = (-3)*0 + (-2)*(-5) + 3*(-1) = 0 +10 -3 = 7
c32 = (-3)*4 + (-2)*1 + 3*0 = -12 -2 +0 = -14
c33 = (-3)*1 + (-2)*4 + 3*2 = -3 -8 +6 = -5
=> row3(AB^T) = [7, -14, -5]

So AB^T = [[-8, -3, 9], [-21, -3, 10], [7, -14, -5]].

Multiply each entry by -2:
-2 AB^T = [[16, 6, -18], [42, 6, -20], [-14, 28, 10]].

3) Why this is correct:
Each entry follows directly from the dot-product formula; signs and coefficients were handled stepwise. The final matrix matches the computed -2 times AB^T.

4) Why other options would be incorrect:
Any alternative differing in sign or magnitude results from arithmetic errors (e.g., forgetting transpose rules, swapping rows/columns, or sign mistakes in dot products). Those lead to at least one entry not matching the correct computations above.

Conclusion: The correct matrix is [[16, 6, -18], [42, 6, -20], [-14, 28, 10]], which matches the provided answer.

Let \[A = \left [\begin{matrix}-1 & 1 & 3\\-2 & 5 & -4\\-3 & -2 & 3\end{matrix}\right ], \text { and } B = \left [\begin{matrix}0 & -5 & -1\\4 & 1 & 0\\1 & 4 & 2\end{matrix}\right ].\] Which of the following options is \(- 2 A B^{T}\)?单项选择题

类似问题

设 \[A = \左 [\begin{matrix}-2 & -2\\-3 & -2\\4 & 4\end{matrix}\right ]， \text { 和} B = \左 [\begin{matrix}-5 & 2 & 1\\2 & -3 & 0\\2 & 4 & 4\4\end{matrix}\right ].\] 以下哪种选项是\（- 3 A B^{T}\）？

If 𝐴 = [ 5 − 5 5 1 − 2 − 1 1 − 4 − 4 ] , then which of the following is 𝐴 2 ?

Let 𝐴 = [ 3 5 3 1 − 4 0 − 2 3 1 ] , and 𝐵 = [ − 3 4 2 0 1 1 0 − 1 3 ] . Which of the following options is − 3 𝐴 𝐵 𝑇 ?

If 𝐴 = [ − 4 − 3 − 2 5 2 − 5 − 1 − 2 − 2 ] , then which of the following is 𝐴 2 ?

设 A = \左 [\begin{matrix}-2 & -2\\-3 & -2\\4 & 4\end{matrix}\right ]， \text { 和} B = \左 [\begin{matrix}-5 & 2 & 1\\2 & -3 & 0\\2 & 4 & 4\4\end{matrix}\right ]。以下哪种选项是\（- 3 A B^{T}\）？

设 \[A = \左 [\begin{matrix}-3 & 0 & 2\\-4 & 1 & -4\\1 & 2 & -4\end{matrix}\right ]， \text { 和 } B = \left [\begin{matrix}-5 & 1 & -4\\4 & -4\\2 & 2 & 2\end{matrix}\right ].\] 以下哪种选项是\（- 3 A B^{T}\）？

Let 𝐴 = [ − 3 0 2 − 4 1 − 4 1 2 − 4 ] , and 𝐵 = [ − 5 1 − 4 4 4 − 4 2 2 2 ] . Which of the following options is − 3 𝐴 𝐵 𝑇 ?

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