Problem 23 - A Graphical Approach to College Algebra

Question

For each matrix, find \(A^{-1}\) if it exists. $$A=\left[\begin{array}{rrr}1 & 0 & 0 \\\0 & -1 & 0 \\\1 & 0 & 1\end{array}\right]$$

Step-by-Step Solution

Answer

The inverse is \(A^{-1} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\).

Step 1: Determine if the Matrix is Invertible

To find the inverse of a matrix, the matrix must be square and have a non-zero determinant. We need to calculate the determinant of matrix \(A\).

Step 2: Calculate the Determinant of A

The formula for the determinant \(|A|\) of a 3x3 matrix \( \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \) is \( a(ei - fh) - b(di - fg) + c(dh - eg) \). For matrix \(A\), substitute values: \[|A| = 1((-1)(1) - (0)(0)) - 0(0 - 0) + 0(0 - (-1)) = -1\]. Thus, the determinant is \(-1\).

Step 3: Adjugate and Transpose the Matrix

Since the determinant is non-zero, matrix \(A\) is invertible. The inverse of a 3x3 matrix \(A\) is given by \(A^{-1} = \frac{1}{|A|} \, \text{adj}(A)^T\). First, find the cofactors to form the adjugate matrix and then transpose it.

Step 4: Finding the Cofactor Matrix

Calculate the cofactor for each element of the matrix. Since \(A\) is mostly zeros except for three ones and a minus one, many cofactors will be zero. Calculate the remaining relevant cofactors: - Co-factor for \(1\): \(C_{11} = (-1)^{1+1} \left((-1)(1) - (0)(0)\right) = 1\)- Co-factor for \(-1\): \(C_{22} = (-1)^{2+2} \left(1(1) - 1(0)\right) = 1\) - Co-factor for \(1\) in (3, 3) position: \(C_{33} = (-1)^{3+3} \left(1(-1) - 0(0)\right) = -1 \)

Step 5: Assemble the Adjugate Matrix

The adjugate matrix is the transpose of the cofactor matrix. It will look like: \[\text{adj}(A) = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix}\]

Step 6: Find the Inverse Matrix

Now, using the inverse matrix formula \(A^{-1} = \frac{1}{|A|} \, \text{adj}(A)\), we find: \[A^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix}\] Thus, \(A^{-1} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix}\).

Key Concepts

Determinant Calculation

Calculating the determinant is one of the first steps in determining if a matrix is invertible. A matrix is invertible if it is square (same number of rows and columns) and its determinant is not zero. For a 3x3 matrix, the determinant can be calculated using the formula:

• For matrix \(A\) where \(A = \left[ \begin{array}{ccc} a & b & c \ d & e & f \ g & h & i \end{array} \right]\):\[|A| = a(ei - fh) - b(di - fg) + c(dh - eg)\]

In the exercise, applying this formula to matrix \(A = \left[ \begin{array}{ccc} 1 & 0 & 0 \ 0 & -1 & 0 \ 1 & 0 & 1 \end{array} \right]\), the calculated determinant is \(-1\). This non-zero determinant confirms that the matrix is invertible.

Cofactor Expansion

Cofactor expansion is a method of calculating determinants or finding the adjugate of a matrix. It uses specific minors of the matrix. A minor is the determinant of the matrix that remains after removing a particular row and column.

• Each cofactor \(C_{ij}\) is given by \((-1)^{i+j}\) times the determinant of the minor obtained by deleting the \(i\)-th row and \(j\)-th column.

In this exercise, the only non-zero cofactors are:

• For the element \(1\) at position (1,1): \[C_{11} = (-1)^{2}( -1 ) = 1 \]
• For the element \(-1\) at position (2,2): \[C_{22} = (-1)^{4}( 1 ) = 1 \]
• For the element \(1\) at position (3,3): \[C_{33} = (-1)^{6}( -1 ) = -1 \]

Using these cofactors, we build the cofactor matrix necessary for determining the adjugate matrix.

Adjugate Matrix

The adjugate matrix, sometimes called the adjoint, is a matrix formed by taking the transposition of the cofactor matrix. It is crucial for finding the inverse of a matrix.

• To construct it, calculate the cofactor for each position in the original matrix.
• Then, reorganize these cofactors by transposing the matrix of cofactors, which swaps the rows and columns.

For the given exercise, after computing the necessary cofactors:

• The cofactor matrix is: \[\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix} \]
• Thus, the adjugate matrix (\(\text{adj}(A)\)) is the same as the cofactor matrix because it is already in its transposed form.

Inverse of a Matrix

The inverse of a matrix is a matrix that, when multiplied by the original matrix, yields the identity matrix. Not all matrices have an inverse, but when they do, the inverse is vital for solving systems of linear equations and other applications.The formula to find the inverse of a matrix \(A\) is:

• \[A^{-1} = \frac{1}{|A|} \times \text{adj}(A)\]

Where \(|A|\) is the determinant of \(A\), and \(\text{adj}(A)\) is the adjugate of \(A\).

• In this problem, the determinant \(|A|\) was \(-1\), and the adjugate matrix was \[\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix}\].
• So, the inverse is calculated as:\[A^{-1} = \frac{1}{-1} \times \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]

This inverse checks out because if you multiply it by the original matrix \(A\), you would get the identity matrix back.