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Given that AB = [[-2, 3, -2], [-4, -1, 2], [0, 1, -4]], find the matrix B, where A = [[3, 2, -1], [-4, 8, 4], [-1, 2, 3]]

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let's find matrix B.

We are given matrix AA and the product ABAB. To find matrix BB, we need to calculate B=A−1(AB)B = A^{-1}(AB).

Given matrices:

A=(32−1−484−123)A = \begin{pmatrix} 3 & 2 & -1 \\ -4 & 8 & 4 \\ -1 & 2 & 3 \end{pmatrix} AB=(−23−2−4−1201−4)AB = \begin{pmatrix} -2 & 3 & -2 \\ -4 & -1 & 2 \\ 0 & 1 & -4 \end{pmatrix}

  1. Calculate the determinant of matrix AA.

    det⁡(A)=3∣8423∣−2∣−44−13∣+(−1)∣−48−12∣\det(A) = 3 \begin{vmatrix} 8 & 4 \\ 2 & 3 \end{vmatrix} - 2 \begin{vmatrix} -4 & 4 \\ -1 & 3 \end{vmatrix} + (-1) \begin{vmatrix} -4 & 8 \\ -1 & 2 \end{vmatrix} det⁡(A)=3(8×3−4×2)−2(−4×3−4×(−1))−1(−4×2−8×(−1))\det(A) = 3(8 \times 3 - 4 \times 2) - 2(-4 \times 3 - 4 \times (-1)) - 1(-4 \times 2 - 8 \times (-1)) det⁡(A)=3(24−8)−2(−12+4)−1(−8+8)\det(A) = 3(24 - 8) - 2(-12 + 4) - 1(-8 + 8) det⁡(A)=3(16)−2(−8)−1(0)\det(A) = 3(16) - 2(-8) - 1(0) det⁡(A)=48+16−0=64\det(A) = 48 + 16 - 0 = 64

  2. Calculate the cofactor matrix of AA.

    C11=∣8423∣=16C12=−∣−44−13∣=8C13=∣−48−12∣=0C_{11} = \begin{vmatrix} 8 & 4 \\ 2 & 3 \end{vmatrix} = 16 \quad C_{12} = -\begin{vmatrix} -4 & 4 \\ -1 & 3 \end{vmatrix} = 8 \quad C_{13} = \begin{vmatrix} -4 & 8 \\ -1 & 2 \end{vmatrix} = 0 C21=−∣2−123∣=−8C22=∣3−1−13∣=8C23=−∣32−12∣=−8C_{21} = -\begin{vmatrix} 2 & -1 \\ 2 & 3 \end{vmatrix} = -8 \quad C_{22} = \begin{vmatrix} 3 & -1 \\ -1 & 3 \end{vmatrix} = 8 \quad C_{23} = -\begin{vmatrix} 3 & 2 \\ -1 & 2 \end{vmatrix} = -8 C31=∣2−184∣=16C32=−∣3−1−44∣=−8C33=∣32−48∣=32C_{31} = \begin{vmatrix} 2 & -1 \\ 8 & 4 \end{vmatrix} = 16 \quad C_{32} = -\begin{vmatrix} 3 & -1 \\ -4 & 4 \end{vmatrix} = -8 \quad C_{33} = \begin{vmatrix} 3 & 2 \\ -4 & 8 \end{vmatrix} = 32

    The cofactor matrix CC is:

    C=(1680−88−816−832)C = \begin{pmatrix} 16 & 8 & 0 \\ -8 & 8 & -8 \\ 16 & -8 & 32 \end{pmatrix}

  3. Calculate the adjugate matrix of AA, which is the transpose of the cofactor matrix.

    adj(A)=CT=(16−81688−80−832)\text{adj}(A) = C^T = \begin{pmatrix} 16 & -8 & 16 \\ 8 & 8 & -8 \\ 0 & -8 & 32 \end{pmatrix}

  4. Calculate the inverse of matrix AA.

    A−1=1det⁡(A)adj(A)=164(16−81688−80−832)A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{64} \begin{pmatrix} 16 & -8 & 16 \\ 8 & 8 & -8 \\ 0 & -8 & 32 \end{pmatrix} A−1=(1664−8641664864864−864064−8643264)=(14−18141818−180−1812)A^{-1} = \begin{pmatrix} \frac{16}{64} & \frac{-8}{64} & \frac{16}{64} \\ \frac{8}{64} & \frac{8}{64} & \frac{-8}{64} \\ \frac{0}{64} & \frac{-8}{64} & \frac{32}{64} \end{pmatrix} = \begin{pmatrix} \frac{1}{4} & -\frac{1}{8} & \frac{1}{4} \\ \frac{1}{8} & \frac{1}{8} & -\frac{1}{8} \\ 0 & -\frac{1}{8} & \frac{1}{2} \end{pmatrix}

  5. Calculate matrix BB by multiplying A−1A^{-1} by ABAB.

    B=A−1(AB)=(14−18141818−180−1812)(−23−2−4−1201−4)B = A^{-1}(AB) = \begin{pmatrix} \frac{1}{4} & -\frac{1}{8} & \frac{1}{4} \\ \frac{1}{8} & \frac{1}{8} & -\frac{1}{8} \\ 0 & -\frac{1}{8} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} -2 & 3 & -2 \\ -4 & -1 & 2 \\ 0 & 1 & -4 \end{pmatrix}

    Performing the matrix multiplication:

    B11=14(−2)−18(−4)+14(0)=−12+12+0=0B_{11} = \frac{1}{4}(-2) - \frac{1}{8}(-4) + \frac{1}{4}(0) = -\frac{1}{2} + \frac{1}{2} + 0 = 0 B12=14(3)−18(−1)+14(1)=34+18+14=6+1+28=98B_{12} = \frac{1}{4}(3) - \frac{1}{8}(-1) + \frac{1}{4}(1) = \frac{3}{4} + \frac{1}{8} + \frac{1}{4} = \frac{6+1+2}{8} = \frac{9}{8} B13=14(−2)−18(2)+14(−4)=−12−14−1=−2+1+44=−74B_{13} = \frac{1}{4}(-2) - \frac{1}{8}(2) + \frac{1}{4}(-4) = -\frac{1}{2} - \frac{1}{4} - 1 = -\frac{2+1+4}{4} = -\frac{7}{4} B21=18(−2)+18(−4)−18(0)=−28−48+0=−68=−34B_{21} = \frac{1}{8}(-2) + \frac{1}{8}(-4) - \frac{1}{8}(0) = -\frac{2}{8} - \frac{4}{8} + 0 = -\frac{6}{8} = -\frac{3}{4} B22=18(3)+18(−1)−18(1)=3−1−18=18B_{22} = \frac{1}{8}(3) + \frac{1}{8}(-1) - \frac{1}{8}(1) = \frac{3-1-1}{8} = \frac{1}{8} B23=18(−2)+18(2)−18(−4)=−28+28+48=48=12B_{23} = \frac{1}{8}(-2) + \frac{1}{8}(2) - \frac{1}{8}(-4) = -\frac{2}{8} + \frac{2}{8} + \frac{4}{8} = \frac{4}{8} = \frac{1}{2} B31=0(−2)−18(−4)+12(0)=0+48+0=12B_{31} = 0(-2) - \frac{1}{8}(-4) + \frac{1}{2}(0) = 0 + \frac{4}{8} + 0 = \frac{1}{2} B32=0(3)−18(−1)+12(1)=0+18+12=1+48=58B_{32} = 0(3) - \frac{1}{8}(-1) + \frac{1}{2}(1) = 0 + \frac{1}{8} + \frac{1}{2} = \frac{1+4}{8} = \frac{5}{8} B33=0(−2)−18(2)+12(−4)=0−28−2=−14−2=−1+84=−94B_{33} = 0(-2) - \frac{1}{8}(2) + \frac{1}{2}(-4) = 0 - \frac{2}{8} - 2 = -\frac{1}{4} - 2 = -\frac{1+8}{4} = -\frac{9}{4}

    Thus, matrix BB is:

    B=(098−74−3418121258−94)\boxed{B = \begin{pmatrix} 0 & \frac{9}{8} & -\frac{7}{4} \\ -\frac{3}{4} & \frac{1}{8} & \frac{1}{2} \\ \frac{1}{2} & \frac{5}{8} & -\frac{9}{4} \end{pmatrix}}
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