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Given matrices A, B, and C, calculate C = A2 − 6B.

Mathematics

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  1. Calculate A2A^2.

    A2=[11−1121−112][11−1121−112]A^2 = \begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 1 \\ -1 & 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 1 \\ -1 & 1 & 2 \end{bmatrix} A2=[(1)(1)+(1)(1)+(−1)(−1)(1)(1)+(1)(2)+(−1)(1)(1)(−1)+(1)(1)+(−1)(2)(1)(1)+(2)(1)+(1)(−1)(1)(1)+(2)(2)+(1)(1)(1)(−1)+(2)(1)+(1)(2)(−1)(1)+(1)(1)+(2)(−1)(−1)(1)+(1)(2)+(2)(1)(−1)(−1)+(1)(1)+(2)(2)]A^2 = \begin{bmatrix} (1)(1)+(1)(1)+(-1)(-1) & (1)(1)+(1)(2)+(-1)(1) & (1)(-1)+(1)(1)+(-1)(2) \\ (1)(1)+(2)(1)+(1)(-1) & (1)(1)+(2)(2)+(1)(1) & (1)(-1)+(2)(1)+(1)(2) \\ (-1)(1)+(1)(1)+(2)(-1) & (-1)(1)+(1)(2)+(2)(1) & (-1)(-1)+(1)(1)+(2)(2) \end{bmatrix} A2=[1+1+11+2−1−1+1−21+2−11+4+1−1+2+2−1+1−2−1+2+21+1+4]A^2 = \begin{bmatrix} 1+1+1 & 1+2-1 & -1+1-2 \\ 1+2-1 & 1+4+1 & -1+2+2 \\ -1+1-2 & -1+2+2 & 1+1+4 \end{bmatrix} A2=[32−2263−236]A^2 = \begin{bmatrix} 3 & 2 & -2 \\ 2 & 6 & 3 \\ -2 & 3 & 6 \end{bmatrix}

  2. Calculate $6B$.

    6B=6[1−12−1110−13]=[6×16×(−1)6×26×(−1)6×16×16×06×(−1)6×3]6B = 6 \begin{bmatrix} 1 & -1 & 2 \\ -1 & 1 & 1 \\ 0 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 6 \times 1 & 6 \times (-1) & 6 \times 2 \\ 6 \times (-1) & 6 \times 1 & 6 \times 1 \\ 6 \times 0 & 6 \times (-1) & 6 \times 3 \end{bmatrix} 6B=[6−612−6660−618]6B = \begin{bmatrix} 6 & -6 & 12 \\ -6 & 6 & 6 \\ 0 & -6 & 18 \end{bmatrix}

  3. Calculate C=A2−6BC = A^2 - 6B.

    C=[32−2263−236]−[6−612−6660−618]C = \begin{bmatrix} 3 & 2 & -2 \\ 2 & 6 & 3 \\ -2 & 3 & 6 \end{bmatrix} - \begin{bmatrix} 6 & -6 & 12 \\ -6 & 6 & 6 \\ 0 & -6 & 18 \end{bmatrix} C=[3−62−(−6)−2−122−(−6)6−63−6−2−03−(−6)6−18]C = \begin{bmatrix} 3-6 & 2-(-6) & -2-12 \\ 2-(-6) & 6-6 & 3-6 \\ -2-0 & 3-(-6) & 6-18 \end{bmatrix} C=[−32+6−142+60−3−23+6−12]C = \begin{bmatrix} -3 & 2+6 & -14 \\ 2+6 & 0 & -3 \\ -2 & 3+6 & -12 \end{bmatrix} C=[−38−1480−3−29−12]C = \boxed{\begin{bmatrix} -3 & 8 & -14 \\ 8 & 0 & -3 \\ -2 & 9 & -12 \end{bmatrix}}
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