Question

Matrices \(A\) and \(B\) are given. Compute \((A B)^{-1}\) and \(B^{-1} A^{-1}\). $$ A=\left[\begin{array}{ll} 2 & 5 \\ 3 & 8 \end{array}\right], \quad B=\left[\begin{array}{cc} 1 & -1 \\ 1 & 4 \end{array}\right] $$

Short answer

\((AB)^{-1} = B^{-1}A^{-1}\).

Step-by-step solution

Compute the Determinant of Matrix A

First, let's find the determinant of matrix \( A \). The determinant for a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( ad - bc \). For matrix \( A \), \( a = 2 \), \( b = 5 \), \( c = 3 \), and \( d = 8 \). So, \( \text{det}(A) = 2 \cdot 8 - 5 \cdot 3 = 16 - 15 = 1 \).

Find the Inverse of Matrix A

The inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \). Since \( \text{det}(A) = 1 \), the inverse of \( A \) is \( A^{-1} = \begin{bmatrix} 8 & -5 \ -3 & 2 \end{bmatrix} \).

Compute the Determinant of Matrix B

Now, let's calculate the determinant of matrix \( B \). For matrix \( B \), \( a = 1 \), \( b = -1 \), \( c = 1 \), and \( d = 4 \). The determinant \( \text{det}(B) = 1 \cdot 4 - (-1) \cdot 1 = 4 + 1 = 5 \).

Find the Inverse of Matrix B

Using the determinant \( \text{det}(B) = 5 \), the inverse of \( B \) is \( B^{-1} = \frac{1}{5} \begin{bmatrix} 4 & 1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{5} & \frac{1}{5} \ -\frac{1}{5} & \frac{1}{5} \end{bmatrix} \).

Verify \((AB)^{-1} = B^{-1}A^{-1}\)

The property of matrix multiplication states \((AB)^{-1} = B^{-1}A^{-1}\). We previously found \( A^{-1} = \begin{bmatrix} 8 & -5 \ -3 & 2 \end{bmatrix} \) and \( B^{-1} = \begin{bmatrix} \frac{4}{5} & \frac{1}{5} \ -\frac{1}{5} & \frac{1}{5} \end{bmatrix} \). Compute \( B^{-1}A^{-1} \) to confirm the equivalence.

Key concepts

Determinant Calculation

Understanding determinants is crucial when working with matrix inverses. The determinant of a 2x2 matrix is a scalar value that provides important information about the matrix, including whether it is invertible. To compute the determinant of a matrix \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\], use the formula:

• Determinant = \( ad - bc \)

This simple equation involves multiplying the top-left and bottom-right elements, then subtracting the product of the top-right and bottom-left elements.
For example, the determinant of matrix \( A \) given as \[\begin{bmatrix} 2 & 5 \ 3 & 8 \end{bmatrix}\]is calculated as follows:

• \( \text{det}(A) = 2 \times 8 - 5 \times 3 \)
• \( \text{det}(A) = 16 - 15 = 1 \)

This non-zero determinant indicates that matrix \( A \) is invertible.
Likewise, for matrix \( B \):

• Given \( B = \begin{bmatrix} 1 & -1 \ 1 & 4 \end{bmatrix} \)
• \( \text{det}(B) = 1 \times 4 - (-1) \times 1 \)
• \( \text{det}(B) = 4 + 1 = 5 \)

Again, since the determinant is nonzero, matrix \( B \) is invertible.

Matrix Multiplication

Matrix multiplication is a fundamental concept for understanding how matrices interact. It allows us to combine different matrices to form new systems of equations. When you multiply matrices, the number of columns in the first matrix must match the number of rows in the second matrix.
For instance, multiplying a 2x2 matrix by another 2x2 matrix results in another 2x2 matrix.Each element in the resulting matrix is obtained by taking the dot product of rows from the first matrix and columns from the second matrix. More concretely, for matrices \[A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix}\] and \[B = \begin{bmatrix} b_{11} & b_{12} \ b_{21} & b_{22} \end{bmatrix}\], the product \( AB \) is calculated as:

• Element (1, 1): \( a_{11}b_{11} + a_{12}b_{21} \)
• Element (1, 2): \( a_{11}b_{12} + a_{12}b_{22} \)
• Element (2, 1): \( a_{21}b_{11} + a_{22}b_{21} \)
• Element (2, 2): \( a_{21}b_{12} + a_{22}b_{22} \)

This calculation gives you each entry in the new matrix by fetching information from both matrices being multiplied.

Inverse of a 2x2 Matrix

Finding the inverse of a matrix is akin to finding a matrix which can undo the transformation represented by the original matrix. For a 2x2 matrix \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\], it exists only if the determinant \( ad-bc \) is non-zero. The inverse is calculated using the formula:

• Inverse = \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \)

Let’s compute the inverse of matrix \( A \) and matrix \( B \):
For matrix \( A = \begin{bmatrix} 2 & 5 \ 3 & 8 \end{bmatrix} \)

• \( \text{det}(A) = 1 \)
• Inverse of \( A \) = \( \begin{bmatrix} 8 & -5 \ -3 & 2 \end{bmatrix} \)

And for matrix \( B = \begin{bmatrix} 1 & -1 \ 1 & 4 \end{bmatrix} \)

• \( \text{det}(B) = 5 \)
• Inverse of \( B \) = \( \frac{1}{5} \begin{bmatrix} 4 & 1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{4}{5} & \frac{1}{5} \ -\frac{1}{5} & \frac{1}{5} \end{bmatrix} \)

Thus, each element of the inverse is carefully derived through this pattern, ensuring you have a matrix that can reverse the transformation of the original.