Problem 28 - College Algebra with Modeling and Visualization

Question

( Refer to Examples 3-5.) LetA be the given matrix. Find \(A^{-1}\) without a calculator. $$ \left[\begin{array}{rrr} 1 & -1 & 1 \\ -1 & 2 & 1 \\ 0 & 2 & 1 \end{array}\right] $$

Step-by-Step Solution

Answer

The inverse of the matrix is \( \begin{bmatrix} 0 & -3 & 3 \\ -1 & -1 & 2 \\ 2 & 2 & -1 \end{bmatrix} \).

Step 1: Validate Matrix Dimensions

A matrix must be square to have an inverse. Our matrix \( A \) is a 3x3 matrix, which means it's square and, potentially invertible.

Step 2: Compute the Determinant

The determinant of a matrix must be non-zero for the inverse to exist. For the 3x3 matrix \( A \), the determinant is calculated as follows:\[\text{det}(A) = \begin{vmatrix}1 & -1 & 1 \-1 & 2 & 1 \0 & 2 & 1 \end{vmatrix}= 1 \times (2 \times 1 - 1 \times 2) + 1 \times (-1 \times 1 - 1 \times 0) +1 \times (-1 \times 2 - 2 \times 0) \]Simplify each term:\[\text{det}(A) = 1(2 - 2) + (-1)(-1) + (1)(-2)\]\[\text{det}(A) = 0 + 1 - 2 = -1\]Since the determinant is -1, which is non-zero, the matrix is invertible.

Step 3: Find the Adjugate Matrix

To find \( A^{-1} \), we use the formula \( A^{-1} = \frac{1}{\text{det}(A)}\text{Adj}(A) \). The adjugate matrix, \( \text{Adj}(A) \), is the transpose of the cofactor matrix of \( A \).First, we find the cofactors:- \( \text{Cofactor}(1,1) = \begin{vmatrix} 2 & 1 \ 2 & 1 \end{vmatrix} = (2 \times 1 - 1 \times 2) = 0 \)- \( \text{Cofactor}(1,2) = -\begin{vmatrix} -1 & 1 \ 0 & 1 \end{vmatrix} = -(-1 \times 1 - 1 \times 0) = 1 \)- \( \text{Cofactor}(1,3) = \begin{vmatrix} -1 & 2 \ 0 & 2 \end{vmatrix} = (-1 \times 2 - 2 \times 0) = -2 \)- \( \text{Cofactor}(2,1) = -\begin{vmatrix} -1 & 1 \ 2 & 1 \end{vmatrix} = -(-1 \times 1 - 1 \times 2) = 3 \)- \( \text{Cofactor}(2,2) = \begin{vmatrix} 1 & 1 \ 0 & 1 \end{vmatrix} = (1 \times 1 - 1 \times 0) = 1 \)- \( \text{Cofactor}(2,3) = -\begin{vmatrix} 1 & -1 \ 0 & 2 \end{vmatrix} = -(1 \times 2 - (-1) \times 0) = -2 \)- \( \text{Cofactor}(3,1) = \begin{vmatrix} -1 & 1 \ 2 & 1 \end{vmatrix} = (-1 \times 1 - 1 \times 2) = -3 \)- \( \text{Cofactor}(3,2) = -\begin{vmatrix} 1 & 1 \ -1 & 1 \end{vmatrix} = -(1 \times 1 - 1 \times (-1)) = -2 \)- \( \text{Cofactor}(3,3) = \begin{vmatrix} 1 & -1 \ -1 & 2 \end{vmatrix} = (1 \times 2 - (-1) \times (-1)) = 1 \)Putting this into a matrix gives the cofactor matrix:\[\begin{bmatrix}0 & 1 & -2 \3 & 1 & -2 \-3 & -2 & 1\end{bmatrix}\]The adjugate is the transpose of this matrix:\[\begin{bmatrix}0 & 3 & -3 \1 & 1 & -2 \-2 & -2 & 1\end{bmatrix}\]

Step 4: Compute the Inverse

The inverse of \( A \) is given by the formula:\[A^{-1} = \frac{1}{\text{det}(A)} \times \text{Adj}(A)\]Since \( \text{det}(A) = -1 \), we have:\[A^{-1} = -1 \cdot \begin{bmatrix}0 & 3 & -3 \1 & 1 & -2 \-2 & -2 & 1\end{bmatrix} = \begin{bmatrix}0 & -3 & 3 \-1 & -1 & 2 \2 & 2 & -1\end{bmatrix}\]

Step 5: Verify Solution

Verify the inverse by checking that \( A \times A^{-1} = I \):\[\begin{bmatrix} 1 & -1 & 1 \ -1 & 2 & 1 \ 0 & 2 & 1 \end{bmatrix} \times \begin{bmatrix} 0 & -3 & 3 \ -1 & -1 & 2 \ 2 & 2 & -1 \end{bmatrix} = \begin{bmatrix} (1\times0 + (-1)\times(-1) + 1\times2) & (1\times(-3) + (-1)\times(-1) + 1\times2) & (1\times3 + (-1)\times2 + 1\times(-1)) \ (-1\times0 + 2\times(-1) + 1\times2) & ((-1)\times(-3) + 2\times(-1) + 1\times2) & ((-1)\times3 + 2\times2 + 1\times(-1)) \ (0\times0 + 2\times(-1) + 1\times2) & (0\times(-3) + 2\times(-1) + 1\times2) & (0\times3 + 2\times2 + 1\times(-1)) \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]Since this results in the identity matrix, our solution is verified.

Key Concepts

Determinant Calculation

Determining whether a matrix has an inverse starts with calculating its determinant. For a square matrix, the determinant is a special number that can help assess whether the matrix is invertible. The presence of non-zero determinant signals the matrix can indeed have an inverse.
To calculate the determinant of a 3x3 matrix, such as: \\[\begin{bmatrix}1 & -1 & 1 \ -1 & 2 & 1 \ 0 & 2 & 1 \end{bmatrix}\]follow these steps:

• Choose a row or column to expand. It's often easiest to pick the row or column with the most zeros.
• Compute the determinant by using minor matrices: remove the row and column of each element selected, then calculate the determinant of the resulting 2x2 matrix.
• Add the computed values, being careful to alternate signs according to the cofactor expansion method.

In our example, after computation, the determinant was found to be -1. This non-zero result means the given matrix is invertible.

Cofactor Matrix

The next step in finding the matrix inverse involves constructing the cofactor matrix. Every element in a matrix has an associated minor, which is the determinant of a smaller matrix formed by deleting the current row and column. A cofactor adjusts this minor by a sign, following the rule \((-1)^{i+j}\) based on the element's position.
Cofactors are used to build the cofactor matrix. For instance, let's look at the cofactor of element in position (1,1):

• Remove the first row and column, form the 2x2 matrix and calculate its determinant.
• Apply the corresponding sign from \((-1)^{1+1}=1\), which means no change in sign for this element.

Once you apply this method to all elements of the matrix, you assemble the cofactor matrix. This matrix is needed to find the final inverse by assisting in forming the adjugate.

Adjugate Matrix

The adjugate, or adjoint, of a matrix is close to the final step in finding its inverse. The adjugate is formed by transposing the cofactor matrix. Transposing a matrix, simply swaps its rows and columns, turning the first row into the first column, the second row into the second column, and so on.

• First, construct the cofactor matrix as explained previously.
• Then transpose this matrix to get the adjugate.
• In the context of inverting a matrix, the adjugate provides a convenient way to ensure all needed positions are correctly inverted.

Remember, confidently forming the adjugate matrix is essential; it is multiplied by the reciprocal of the determinant to arrive at the final matrix inversion.

Identity Matrix

An identity matrix behaves much like the number 1 in regular multiplication for matrices. Multiplying any square matrix by an identity matrix of compatible size leaves it unchanged. It's symbolized as \(I\) for simplicity.
When finding inverses, achieving an identity matrix from a product, like \(A \times A^{-1}\), is the ultimate verification step. This means you've accurately found the inverse, as the product exactly results in:\[\begin{bmatrix}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]for a 3x3 identity matrix, aligning each identity element with its position. Performing this step confirms the accuracy of your inverse and can reassure you of your correctness in previous calculations. It is an essential part of validating inverses in the matrix inversion process, constructing a foundational concept in linear algebra.