Norris Cross
01/04/2024 · Middle School

Find the Following matrices where \( A=\left[\begin{array}{cc}6 & -7 \\ -4 & 6 \\ -1 & 4\end{array}\right] \) and \( B=\left[\begin{array}{cc}-9 & 8 \\ 3 & 0 \\ -3 & 8\end{array}\right] \) Select the correct chace below O. A \( -3 A-9 B=\square \) simplify your answer 0 B. This matrix operation is not passible. O.

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To find the matrix \( -3A - 9B \), we need to perform scalar multiplication on matrices \( A \) and \( B \) and then subtract the results. Given: \( A=\left[\begin{array}{cc}6 & -7 \\ -4 & 6 \\ -1 & 4\end{array}\right] \) \( B=\left[\begin{array}{cc}-9 & 8 \\ 3 & 0 \\ -3 & 8\end{array}\right] \) First, we perform scalar multiplication on matrix \( A \) by multiplying each element by \( -3 \): \( -3A = -3 \left[\begin{array}{cc}6 & -7 \\ -4 & 6 \\ -1 & 4\end{array}\right] = \left[\begin{array}{cc}-18 & 21 \\ 12 & -18 \\ 3 & -12\end{array}\right] \) Next, we perform scalar multiplication on matrix \( B \) by multiplying each element by \( -9 \): \( -9B = -9 \left[\begin{array}{cc}-9 & 8 \\ 3 & 0 \\ -3 & 8\end{array}\right] = \left[\begin{array}{cc}81 & -72 \\ -27 & 0 \\ 27 & -72\end{array}\right] \) Now, we subtract the two resulting matrices: \( -3A - 9B = \left[\begin{array}{cc}-18 & 21 \\ 12 & -18 \\ 3 & -12\end{array}\right] - \left[\begin{array}{cc}81 & -72 \\ -27 & 0 \\ 27 & -72\end{array}\right] \) To subtract the matrices, we subtract the corresponding elements: \( -3A - 9B = \left[\begin{array}{cc}-18 - 81 & 21 - (-72) \\ 12 - (-27) & -18 - 0 \\ 3 - 27 & -12 - (-72)\end{array}\right] \) \( -3A - 9B = \left[\begin{array}{cc}-99 & 93 \\ 39 & -18 \\ -24 & 60\end{array}\right] \) So, the simplified answer is: \( -3A - 9B = \left[\begin{array}{cc}-99 & 93 \\ 39 & -18 \\ -24 & 60\end{array}\right] \) This is the correct answer, and it corresponds to option O. A \( -3A - 9B = \left[\begin{array}{cc}-99 & 93 \\ 39 & -18 \\ -24 & 60\end{array}\right] \).

Quick Answer

\( -3A - 9B = \left[\begin{array}{cc}-99 & 93 \\ 39 & -18 \\ -24 & 60\end{array}\right] \)
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