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Multiply matrices A and B, where A = [[2, 1, -1], [0, 3, 2]] and B = [[1, 0], [3, -1], [-1, 2]].

Mathematics

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Answer

[6−371]\begin{bmatrix} 6 & -3 \\ 7 & 1 \end{bmatrix}

Step-by-step solution

5 steps

Alright, Hurry, let's multiply these matrices.

We are given two matrices:

A=[21−1032]A = \begin{bmatrix} 2 & 1 & -1 \\ 0 & 3 & 2 \end{bmatrix} B=[103−1−12]B = \begin{bmatrix} 1 & 0 \\ 3 & -1 \\ -1 & 2 \end{bmatrix}

To find the product ABAB, we multiply the rows of matrix AA by the columns of matrix BB. The resulting matrix will be a 2×22 \times 2 matrix.

  1. Calculate the element in the first row, first column (c11c_{11}).

    Multiply the first row of AA by the first column of BB:

    c11=(2)(1)+(1)(3)+(−1)(−1)c_{11} = (2)(1) + (1)(3) + (-1)(-1) c11=2+3+1=6c_{11} = 2 + 3 + 1 = 6
  2. Calculate the element in the first row, second column (c12c_{12}).

    Multiply the first row of AA by the second column of BB:

    c12=(2)(0)+(1)(−1)+(−1)(2)c_{12} = (2)(0) + (1)(-1) + (-1)(2) c12=0−1−2=−3c_{12} = 0 - 1 - 2 = -3
  3. Calculate the element in the second row, first column (c21c_{21}).

    Multiply the second row of AA by the first column of BB:

    c21=(0)(1)+(3)(3)+(2)(−1)c_{21} = (0)(1) + (3)(3) + (2)(-1) c21=0+9−2=7c_{21} = 0 + 9 - 2 = 7
  4. Calculate the element in the second row, second column (c22c_{22}).

    Multiply the second row of AA by the second column of BB:

    c22=(0)(0)+(3)(−1)+(2)(2)c_{22} = (0)(0) + (3)(-1) + (2)(2) c22=0−3+4=1c_{22} = 0 - 3 + 4 = 1
  5. Form the product matrix ABAB.

    AB=[c11c12c21c22]=[6−371]AB = \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix} = \begin{bmatrix} 6 & -3 \\ 7 & 1 \end{bmatrix}

    The product ABAB is [6−371]\boxed{\begin{bmatrix} 6 & -3 \\ 7 & 1 \end{bmatrix}}.

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