Question

Let \[ A = \begin{pmatrix} 1 & 4 \\ -2 & 1 \end{pmatrix} \quad \text{and} \quad C = \begin{pmatrix} 3 & 4 & 2 \\ 12 & 16 & 8 \\ -6 & -8 & -4 \end{pmatrix}. \] Then, find the matrix $B$ if $AB = C$.

Hint:

To find the matrix $B$ in an equation like $AB = C$, calculate the inverse of $A$ and multiply it by $C$.

Answer 1

We are given the equation $AB = C$. To find matrix $B$, we need to multiply matrix $A^{-1}$ (the inverse of matrix $A$) with matrix $C$. Hence, we first need to calculate $A^{-1}$. The determinant of $A$ is: \[ \text{det}(A) = (1)(1) - (-2)(4) = 1 + 8 = 9. \] Since the determinant is non-zero, $A$ is invertible. The inverse of $A$ is given by: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} 1 & -4 \\ 2 & 1 \end{pmatrix} = \frac{1}{9} \begin{pmatrix} 1 & -4 \\ 2 & 1 \end{pmatrix}. \] Now, to find matrix $B$, we multiply both sides of $AB = C$ by $A^{-1}$: \[ A^{-1} AB = A^{-1} C \quad \Rightarrow \quad B = A^{-1} C. \] Multiplying $A^{-1}$ with $C$: \[ B = \frac{1}{9} \begin{pmatrix} 1 & -4 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 3 & 4 & 2 \\ 12 & 16 & 8 \\ -6 & -8 & -4 \end{pmatrix}. \] Performing the matrix multiplication: \[ B = \frac{1}{9} \begin{pmatrix} (1)(3) + (-4)(12) & (1)(4) + (-4)(16) & (1)(2) + (-4)(8) \\ (2)(3) + (1)(12) & (2)(4) + (1)(16) & (2)(2) + (1)(8) \end{pmatrix}. \] \[ B = \frac{1}{9} \begin{pmatrix} 3 - 48 & 4 - 64 & 2 - 32 \\ 6 + 12 & 8 + 16 & 4 + 8 \end{pmatrix} = \frac{1}{9} \begin{pmatrix} -45 & -60 & -30 \\ 18 & 24 & 12 \end{pmatrix}. \] Thus, the matrix $B$ is: \[ B = \begin{pmatrix} -5 & -\frac{20}{3} & -\frac{10}{3} \\ 2 & \frac{8}{3} & \frac{4}{3} \end{pmatrix}. \]