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Given matrices A and B, calculate the matrix product BA.

Mathematics

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Given the matrices:

B=[123456]B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} A=[123456]A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}

We need to find the product BABA.
Matrix B is a 3×23 \times 2 matrix and Matrix A is a 2×32 \times 3 matrix. The resulting matrix BABA will be a 3×33 \times 3 matrix.

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

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

    (1×1)+(2×4)=1+8=9(element c11)(1 \times 1) + (2 \times 4) = 1 + 8 = 9 \quad (\text{element } c_{11}) (1×2)+(2×5)=2+10=12(element c12)(1 \times 2) + (2 \times 5) = 2 + 10 = 12 \quad (\text{element } c_{12}) (1×3)+(2×6)=3+12=15(element c13)(1 \times 3) + (2 \times 6) = 3 + 12 = 15 \quad (\text{element } c_{13})

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

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

    (3×1)+(4×4)=3+16=19(element c21)(3 \times 1) + (4 \times 4) = 3 + 16 = 19 \quad (\text{element } c_{21}) (3×2)+(4×5)=6+20=26(element c22)(3 \times 2) + (4 \times 5) = 6 + 20 = 26 \quad (\text{element } c_{22}) (3×3)+(4×6)=9+24=33(element c23)(3 \times 3) + (4 \times 6) = 9 + 24 = 33 \quad (\text{element } c_{23})

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

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

    (5×1)+(6×4)=5+24=29(element c31)(5 \times 1) + (6 \times 4) = 5 + 24 = 29 \quad (\text{element } c_{31}) (5×2)+(6×5)=10+30=40(element c32)(5 \times 2) + (6 \times 5) = 10 + 30 = 40 \quad (\text{element } c_{32}) (5×3)+(6×6)=15+36=51(element c33)(5 \times 3) + (6 \times 6) = 15 + 36 = 51 \quad (\text{element } c_{33})

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

    BA=[91215192633294051]BA = \begin{bmatrix} 9 & 12 & 15 \\ 19 & 26 & 33 \\ 29 & 40 & 51 \end{bmatrix}

    The product of the matrices is:

    [91215192633294051]\boxed{\begin{bmatrix} 9 & 12 & 15 \\ 19 & 26 & 33 \\ 29 & 40 & 51 \end{bmatrix}}
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