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Ch. 6 - Matrices and Determinants
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 6 - Matrices and DeterminantsProblema 12d
Capitolo 7, Problema 12d

Find the following matrices: - 3A + 2B
A=[311−125],B=[2−36−31−4]A = \(\begin{bmatrix}\) 3 & 1 & 1 \\ -1 & 2 & 5 \(\end{bmatrix}\) , B = \(\begin{bmatrix}\) 2 & -3 & 6 \\ -3 & 1 & -4 \(\end{bmatrix}\)

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First, write down the matrices A and B clearly: \(A = \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 1 & 2 \\ -3 & 6 \end{bmatrix}\).
Next, multiply matrix A by the scalar -3. This means multiplying each element of A by -3: \(-3A = -3 \times \begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} -3 \times 3 & -3 \times 1 \\ -3 \times 1 & -3 \times 2 \end{bmatrix}\).
Then, multiply matrix B by the scalar 2. Multiply each element of B by 2: \(2B = 2 \times \begin{bmatrix} 1 & 2 \\ -3 & 6 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times (-3) & 2 \times 6 \end{bmatrix}\).
After finding \(-3A\) and \$2B$, add the two resulting matrices element-wise: \(-3A + 2B = \begin{bmatrix} (-3 \times 3) + (2 \times 1) & (-3 \times 1) + (2 \times 2) \\ (-3 \times 1) + (2 \times (-3)) & (-3 \times 2) + (2 \times 6) \end{bmatrix}\).
Finally, simplify each element in the resulting matrix to get the final matrix for \(-3A + 2B\).

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