Problem 27 - Advanced Engineering Mathematics

Question

Use the given matrices to find \((\mathbf{A B})^{-1}\). $$ \mathbf{A}^{-1}=\left(\begin{array}{rr} 1 & -\frac{5}{2} \\ -\frac{1}{2} & \frac{3}{2} \end{array}\right), \quad \mathbf{B}^{-1}=\left(\begin{array}{rr} 2 & \frac{4}{3} \\ -\frac{1}{3} & \frac{5}{2} \end{array}\right) $$

Step-by-Step Solution

Answer

\((\mathbf{AB})^{-1} = \begin{pmatrix} \frac{4}{3} & -3 \\ -\frac{17}{12} & \frac{65}{12} \end{pmatrix}\)."

Step 1: Understanding Matrix Inverses and Multiplication

The problem requires us to find \((\mathbf{AB})^{-1}\) given \(\mathbf{A}^{-1}\) and \(\mathbf{B}^{-1}\). Recall that for two invertible matrices \(\mathbf{A}\) and \(\mathbf{B}\), the inverse of their product \((\mathbf{AB})^{-1}\) is given by \((\mathbf{B}^{-1}\mathbf{A}^{-1})\). This means we need to multiply \(\mathbf{B}^{-1}\) by \(\mathbf{A}^{-1}\).

Step 2: Multiply B^{-1} and A^{-1}

Compute the product \(\mathbf{B}^{-1}\mathbf{A}^{-1}\). Using the given matrices: \[ \mathbf{B}^{-1} = \begin{pmatrix} 2 & \frac{4}{3} \ -\frac{1}{3} & \frac{5}{2} \end{pmatrix}, \quad \mathbf{A}^{-1} = \begin{pmatrix} 1 & -\frac{5}{2} \ -\frac{1}{2} & \frac{3}{2} \end{pmatrix} \]Calculate the product: \[ \mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{pmatrix} 2 & \frac{4}{3} \ -\frac{1}{3} & \frac{5}{2} \end{pmatrix} \begin{pmatrix} 1 & -\frac{5}{2} \ -\frac{1}{2} & \frac{3}{2} \end{pmatrix}\]

Step 3: Calculate First Row of Result

Calculate the first row of the resulting matrix by summing the products of corresponding elements from the first row of \(\mathbf{B}^{-1}\) and the columns of \(\mathbf{A}^{-1}\): First element: \(2\times 1 + \frac{4}{3} \times -\frac{1}{2} = 2 - \frac{2}{3} = \frac{4}{3}\) Second element: \(2 \times -\frac{5}{2} + \frac{4}{3} \times \frac{3}{2} = -5 + 2 = -3\) So the first row is \(\left( \frac{4}{3}, -3 \right)\).

Step 4: Calculate Second Row of Result

Calculate the second row of the resulting matrix by summing the products of corresponding elements from the second row of \(\mathbf{B}^{-1}\) and the columns of \(\mathbf{A}^{-1}\): First element: \(-\frac{1}{3} \times 1 + \frac{5}{2} \times -\frac{1}{2} = -\frac{1}{3} - \frac{5}{4} = -\frac{17}{12}\) Second element: \(-\frac{1}{3} \times -\frac{5}{2} + \frac{5}{2} \times \frac{3}{2} = \frac{5}{6} + \frac{15}{4} = \frac{65}{12}\) So the second row is \(\left(-\frac{17}{12}, \frac{65}{12}\right)\).

Step 5: Combine the Calculations

Combine the first and second row results to form the inverse matrix: \[\mathbf{B}^{-1}\mathbf{A}^{-1} = \begin{pmatrix} \frac{4}{3} & -3 \ -\frac{17}{12} & \frac{65}{12} \end{pmatrix}\] Hence, \((\mathbf{AB})^{-1} = \begin{pmatrix} \frac{4}{3} & -3 \ -\frac{17}{12} & \frac{65}{12} \end{pmatrix}\).

Key Concepts

Matrix Multiplication

When matrices are multiplied, each element of the resulting matrix is a sum of products of elements from the rows and columns of the matrices being multiplied. Consider two matrices, \( \mathbf{X} \) and \( \mathbf{Y} \), such that \( \mathbf{X} \) is of size \( m \times n \) and \( \mathbf{Y} \) is of size \( n \times p \). The resulting matrix, after multiplication, will be of size \( m \times p \). Each element in the resulting matrix is calculated as follows:

• Select a row from the first matrix and a column from the second matrix.
• Multiply corresponding elements from the row and column and sum these products.
• The sum is the value for that element in the result matrix.

Matrix multiplication is not commutative, meaning \( \mathbf{XY} eq \mathbf{YX} \) generally. This rule is essential to ensure calculations for specific tasks, such as finding the inverse of a product of matrices, are accurate.

Matrix Algebra

Matrix algebra involves various operations including addition, subtraction, scalar multiplication, and inverses. In matrix algebra:

• Matrices must be of the same dimensions for addition or subtraction.
• Scalar multiplication involves multiplying each element by the scalar value.
• An inverse of a matrix \( \mathbf{M} \) is a matrix \( \mathbf{M}^{-1} \) such that \( \mathbf{M} \mathbf{M}^{-1} = \mathbf{I} \), where \( \mathbf{I} \) is the identity matrix.

Understanding the interaction between these operations allows for solving complex linear equations. Only square matrices have inverses, and not all square matrices are invertible. An invertible matrix is also known as a non-singular or full-rank matrix.

Inverse of Product of Matrices

To find the inverse of a product of two matrices produced by matrix multiplication, you use the rule for inverses of product matrices. If \( \mathbf{A} \) and \( \mathbf{B} \) are two invertible matrices, then the inverse of their product \( \mathbf{AB} \) is given by \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\). Notice that both the order of multiplication and the order of the matrices in the product are reversed.
This principle arises from associating the inverse operation with division in traditional algebra:

• First, invert the multiplicands \( \mathbf{A} \) and \( \mathbf{B} \).
• Then, multiply them in reverse order, \( \mathbf{B}^{-1} \) followed by \( \mathbf{A}^{-1} \).

Hence, for any matrices \( \mathbf{A} \) and \( \mathbf{B} \) with known inverses, you can find the inverse of their product by performing these operations.