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Given matrices A and B, find their product AB.

Mathematics

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Given the matrices:

A=[134211134]A = \begin{bmatrix} 1 & 3 & 4 \\ 2 & 1 & 1 \\ 1 & 3 & 4 \end{bmatrix} B=[−3−211312−13]B = \begin{bmatrix} -3 & -2 & 1 \\ 1 & 3 & 1 \\ 2 & -1 & 3 \end{bmatrix}

We need to find the product ABAB. Both A and B are 3×33 \times 3 matrices, so the resulting matrix ABAB will also be a 3×33 \times 3 matrix.

  1. Calculate the elements for the first row of ABAB.

    Multiply the first row of A by each column of B.

    c11=(1×−3)+(3×1)+(4×2)=−3+3+8=8c_{11} = (1 \times -3) + (3 \times 1) + (4 \times 2) = -3 + 3 + 8 = 8 c12=(1×−2)+(3×3)+(4×−1)=−2+9−4=3c_{12} = (1 \times -2) + (3 \times 3) + (4 \times -1) = -2 + 9 - 4 = 3 c13=(1×1)+(3×1)+(4×3)=1+3+12=16c_{13} = (1 \times 1) + (3 \times 1) + (4 \times 3) = 1 + 3 + 12 = 16

  2. Calculate the elements for the second row of ABAB.

    Multiply the second row of A by each column of B.

    c21=(2×−3)+(1×1)+(1×2)=−6+1+2=−3c_{21} = (2 \times -3) + (1 \times 1) + (1 \times 2) = -6 + 1 + 2 = -3 c22=(2×−2)+(1×3)+(1×−1)=−4+3−1=−2c_{22} = (2 \times -2) + (1 \times 3) + (1 \times -1) = -4 + 3 - 1 = -2 c23=(2×1)+(1×1)+(1×3)=2+1+3=6c_{23} = (2 \times 1) + (1 \times 1) + (1 \times 3) = 2 + 1 + 3 = 6

  3. Calculate the elements for the third row of ABAB.

    Multiply the third row of A by each column of B.

    c31=(1×−3)+(3×1)+(4×2)=−3+3+8=8c_{31} = (1 \times -3) + (3 \times 1) + (4 \times 2) = -3 + 3 + 8 = 8 c32=(1×−2)+(3×3)+(4×−1)=−2+9−4=3c_{32} = (1 \times -2) + (3 \times 3) + (4 \times -1) = -2 + 9 - 4 = 3 c33=(1×1)+(3×1)+(4×3)=1+3+12=16c_{33} = (1 \times 1) + (3 \times 1) + (4 \times 3) = 1 + 3 + 12 = 16

  4. Combine the calculated elements to form the product matrix ABAB.

    AB=[8316−3−268316]AB = \begin{bmatrix} 8 & 3 & 16 \\ -3 & -2 & 6 \\ 8 & 3 & 16 \end{bmatrix}

    The product of the matrices is:

    [8316−3−268316]\boxed{\begin{bmatrix} 8 & 3 & 16 \\ -3 & -2 & 6 \\ 8 & 3 & 16 \end{bmatrix}}
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