Problem 21 - College Algebra with Corequisite Support

Question

For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$ \left[\begin{array}{ccc} 0 & 1 & -3 \\ 4 & 1 & 0 \\ 1 & 0 & 5 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse is \(\begin{bmatrix} -\frac{5}{17} & \frac{3}{17} & \frac{1}{17} \\ \frac{20}{17} & -\frac{5}{17} & \frac{12}{17} \\ -\frac{4}{17} & -\frac{1}{17} & \frac{4}{17} \end{bmatrix}\).

Step 1: Calculate the Determinant

To find the inverse of a matrix, we must first check if the determinant is non-zero. Compute the determinant of the matrix: \[\text{det} = \begin{vmatrix} 0 & 1 & -3 \ 4 & 1 & 0 \ 1 & 0 & 5 \end{vmatrix} = 0(1 \times 5 - 0 \times 0) - 1(4 \times 5 - 0 \times 1) - 3(4 \times 0 - 1 \times 1).\] Simplify the expression: \[= 0 - 1(20) - 3(-1) = -20 + 3 = -17.\] Since the determinant is \(-17\), which is non-zero, an inverse exists.

Step 2: Calculate the Adjugate Matrix

The adjugate is the transpose of the cofactor matrix. First calculate the cofactor matrix. Cofactors are calculated by eliminating the corresponding row and column, then taking the determinant of the resulting 2x2 matrix and multiplying by \((-1)^{i+j}\). For example, calculate the cofactor for the first element:\[C_{11} = (-1)^{1+1}\begin{vmatrix} 1 & 0 \ 0 & 5 \end{vmatrix} = 5.\] Calculate all necessary cofactors and form the matrix of cofactors:\[\begin{bmatrix} 5 & -20 & 4 \ -3 & 5 & 1 \ -1 & -12 & -4 \end{bmatrix}.\]Now take the transpose of this cofactor matrix to get the adjugate:\[\text{adjugate} = \begin{bmatrix} 5 & -3 & -1 \ -20 & 5 & -12 \ 4 & 1 & -4 \end{bmatrix}.\]

Step 3: Calculate the Inverse Matrix

The inverse of a matrix is given by the formula:\[A^{-1} = \frac{1}{\text{det}} \cdot \text{adjugate}.\]Thus, the inverse is calculated as:\[A^{-1} = \frac{1}{-17} \begin{bmatrix} 5 & -3 & -1 \ -20 & 5 & -12 \ 4 & 1 & -4 \end{bmatrix}.\] This simplifies to:\[A^{-1} = \begin{bmatrix} -\frac{5}{17} & \frac{3}{17} & \frac{1}{17} \ \frac{20}{17} & -\frac{5}{17} & \frac{12}{17} \ -\frac{4}{17} & -\frac{1}{17} & \frac{4}{17} \end{bmatrix}.\]

Key Concepts

Determinant Calculation

The calculation of the determinant is the first step to finding out if a matrix has an inverse. The determinant is a special scalar value that helps us understand certain properties of a matrix, such as invertibility. To find the determinant of a 3x3 matrix, you can use the rule of Sarrus or the cofactor expansion. Any method will show that in a matrix\[\begin{pmatrix}a & b & c\d & e & f\g & h & i\end{pmatrix}\]the determinant is calculated by the formula:

• \(\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\)

For our example matrix, the given determinant turns out to be \(-17\). Since it's not zero, you can proceed to find an inverse.

Cofactor Matrix

The cofactor matrix is a step towards finding the adjugate matrix, which is key in calculating the inverse of a matrix. Each element in the cofactor matrix is a minor of the corresponding element multiplied by \((-1)^{i+j}\) to account for the sign based on its position. For a matrix \(A_{ij}\), find the cofactor \(C_{ij}\) by temporarily removing the \(i\)th row and \(j\)th column and then compute the determinant of the remaining 2x2 matrix. If you place these cofactors correctly, you obtain a full cofactor matrix:

• For instance, \(C_{11}\) of our matrix equals \((-1)^(1+1)\begin{vmatrix}1 & 0\ 0 & 5\end{vmatrix}=5\).
• The complete cofactor matrix looks like this:\[\begin{bmatrix}5 & -20 & 4 \ -3 & 5 & 1 \ -1 & -12 & -4\end{bmatrix}\]

Adjugate Matrix

After obtaining the cofactor matrix, the adjugate is formed by taking the transpose of this matrix. Transposing involves swapping rows for columns,

• For instance, if we had a cofactor matrix\[\begin{bmatrix}a & b & c\ d & e & f\ g & h & i\end{bmatrix}\],
• the transpose (or adjugate) will be \[\begin{bmatrix}a & d & g\ b & e & h \ c & f & i\end{bmatrix}\].

This method yields an organized rearrangement of elements. In this exercise, you use the transposed cofactor matrix to further calculations. The resulting adjugate matrix \(\ ext{adj}(A)\) helps in the subsequent calculation of the inverse matrix.

Inverse Matrix Formula

The inverse of a matrix \(A\) is symbolized as \(A^{-1}\) and given by the formula:

• \(A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adjugate}(A)\)

This formula requires both a non-zero determinant and the adjugate matrix. Applying these to our exercise:

• Since the determinant is \(-17\), while the adjugate is \[\begin{bmatrix}5 & -3 & -1\ -20 & 5 & -12\ 4 & 1 & -4\end{bmatrix}\], substitute these values into the formula to find \(A^{-1}\)
• This results in \[A^{-1}=\begin{bmatrix}-\frac{5}{17} & \frac{3}{17} & \frac{1}{17}\ \frac{20}{17} & -\frac{5}{17} & \frac{12}{17}\ -\frac{4}{17} & -\frac{1}{17} & \frac{4}{17}\end{bmatrix}\].

The inverse provides a matrix, which when multiplied by the original, yields the identity matrix, affirming its correctness.