Question

Find \( (AB)^{-1} \), if

\[ A = \begin{bmatrix} 3 & 4 \\ 1 & 1 \end{bmatrix}, \quad B^{-1} = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} \]

Hint:

To find the inverse of a matrix product \( AB \), use the property \( (AB)^{-1} = B^{-1} A^{-1} \). Make sure to compute the inverse of each matrix separately before multiplying.

Answer 1

Step 1: Recall the property of inverses.

We know that the inverse of the product of two matrices \(A\) and \(B\) is: \[ (AB)^{-1} = B^{-1} A^{-1} \] This property holds for invertible matrices.

Step 2: Finding \(A^{-1}\).

The inverse of a \(2 \times 2\) matrix \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] For \[ A = \begin{bmatrix} 3 & 4 \\ 1 & 1 \end{bmatrix} \] the determinant is: \[ \det(A) = (3)(1) - (4)(1) = -1 \] Hence, \[ A^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & -4 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ 1 & -3 \end{bmatrix} \]

Step 3: Compute \( (AB)^{-1} \).

Using \[ (AB)^{-1} = B^{-1} A^{-1} \] Given: \[ B^{-1} = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix}, \quad A^{-1} = \begin{bmatrix} -1 & 4 \\ 1 & -3 \end{bmatrix} \] Now multiply: \[ (AB)^{-1} = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} -1 & 4 \\ 1 & -3 \end{bmatrix} \] \[ (AB)^{-1} = \begin{bmatrix} 4(-1)+3(1) & 4(4)+3(-3) \\ 2(-1)+1(1) & 2(4)+1(-3) \end{bmatrix} \] \[ (AB)^{-1} = \begin{bmatrix} -1 & 7 \\ -1 & 5 \end{bmatrix} \]

Step 4: Conclusion.

\[ (AB)^{-1} = \begin{bmatrix} -1 & 7 \\ -1 & 5 \end{bmatrix} \]