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If A=[(1,2,3),(-1,0,2),(1,-8,-1)],B=[(4,...

If `A=[(1,2,3),(-1,0,2),(1,-8,-1)],B=[(4,5,6),(-1,0,1),(2,1,2)],C=[(-1,-2,1),(-1,2,3),(-1,-2,2)]`, find (i) `2B-3C` (ii) `A-2B+3C`.

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To solve the given problem, we will follow these steps: ### Given Matrices: - \( A = \begin{pmatrix} 1 & 2 & 3 \\ -1 & 0 & 2 \\ 1 & -8 & -1 \end{pmatrix} \) - \( B = \begin{pmatrix} 4 & 5 & 6 \\ -1 & 0 & 1 \\ 2 & 1 & 2 \end{pmatrix} \) - \( C = \begin{pmatrix} -1 & -2 & 1 \\ -1 & 2 & 3 \\ -1 & -2 & 2 \end{pmatrix} \) ### (i) Finding \( 2B - 3C \) 1. **Calculate \( 2B \)**: \[ 2B = 2 \times \begin{pmatrix} 4 & 5 & 6 \\ -1 & 0 & 1 \\ 2 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 8 & 10 & 12 \\ -2 & 0 & 2 \\ 4 & 2 & 4 \end{pmatrix} \] 2. **Calculate \( 3C \)**: \[ 3C = 3 \times \begin{pmatrix} -1 & -2 & 1 \\ -1 & 2 & 3 \\ -1 & -2 & 2 \end{pmatrix} = \begin{pmatrix} -3 & -6 & 3 \\ -3 & 6 & 9 \\ -3 & -6 & 6 \end{pmatrix} \] 3. **Now, calculate \( 2B - 3C \)**: \[ 2B - 3C = \begin{pmatrix} 8 & 10 & 12 \\ -2 & 0 & 2 \\ 4 & 2 & 4 \end{pmatrix} - \begin{pmatrix} -3 & -6 & 3 \\ -3 & 6 & 9 \\ -3 & -6 & 6 \end{pmatrix} \] \[ = \begin{pmatrix} 8 - (-3) & 10 - (-6) & 12 - 3 \\ -2 - (-3) & 0 - 6 & 2 - 9 \\ 4 - (-3) & 2 - (-6) & 4 - 6 \end{pmatrix} \] \[ = \begin{pmatrix} 8 + 3 & 10 + 6 & 12 - 3 \\ -2 + 3 & 0 - 6 & 2 - 9 \\ 4 + 3 & 2 + 6 & 4 - 6 \end{pmatrix} \] \[ = \begin{pmatrix} 11 & 16 & 9 \\ 1 & -6 & -7 \\ 7 & 8 & -2 \end{pmatrix} \] ### (ii) Finding \( A - 2B + 3C \) 1. **Use the previously calculated \( 2B - 3C \)**: \[ A - 2B + 3C = A + (2B - 3C) \] \[ = A + \begin{pmatrix} 11 & 16 & 9 \\ 1 & -6 & -7 \\ 7 & 8 & -2 \end{pmatrix} \] 2. **Calculate \( A - 2B + 3C \)**: \[ A = \begin{pmatrix} 1 & 2 & 3 \\ -1 & 0 & 2 \\ 1 & -8 & -1 \end{pmatrix} \] \[ = \begin{pmatrix} 1 + 11 & 2 + 16 & 3 + 9 \\ -1 + 1 & 0 - 6 & 2 - 7 \\ 1 + 7 & -8 + 8 & -1 - 2 \end{pmatrix} \] \[ = \begin{pmatrix} 12 & 18 & 12 \\ 0 & -6 & -5 \\ 8 & 0 & -3 \end{pmatrix} \] ### Final Results: - \( 2B - 3C = \begin{pmatrix} 11 & 16 & 9 \\ 1 & -6 & -7 \\ 7 & 8 & -2 \end{pmatrix} \) - \( A - 2B + 3C = \begin{pmatrix} 12 & 18 & 12 \\ 0 & -6 & -5 \\ 8 & 0 & -3 \end{pmatrix} \)
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