Chapter 9: Problem 5
Use the Gauss-Jordan method to find \(\mathbf{A}^{-1}\), if it exists. Check your answers by using a graphing calculator to find \(\mathbf{A}^{-1} \mathbf{A}\) and \(\mathbf{A} \mathbf{A}^{-1}\). $$\mathbf{A}=\left[\begin{array}{ll}3 & 2 \\ 5 & 3\end{array}\right]$$
Short Answer
Expert verified
The inverse of \( \textbf{A} \) is \( \begin{pmatrix} 1 & -\frac{1}{2} \ -\frac{5}{4} & \frac{3}{4} \ \end{pmatrix} \). Verified using a graphing calculator.
Step by step solution
01
- Set Up the Augmented Matrix
Write down the augmented matrix combining matrix \(\textbf{A}\) and the identity matrix \( \textbf{I} \): \[ \begin{pmatrix} 3 & 2 & | & 1 & 0 \ 5 & 3 & | & 0 & 1 \ \end{pmatrix} \]
02
- Perform Row Operations to Get a Row-Equivalent Matrix
First, make the element in the first row and first column a 1 by dividing the first row by 3:\[ R1 \to \frac{1}{3}R1 \ \begin{pmatrix} 1 & \frac{2}{3} & | & \frac{1}{3} & 0 \ 5 & 3 & | & 0 & 1 \ \end{pmatrix} \]
03
- Eliminate the First Column in the Second Row
Use the first row to make the first column of the second row zero by replacing the second row with \( R2 - 5R1 \):\[ R2 \to R2 - 5R1 \ \begin{pmatrix} 1 & \frac{2}{3} & | & \frac{1}{3} & 0 \ 0 & \frac{4}{3} & | & -\frac{5}{3} & 1 \ \end{pmatrix} \]
04
- Make the Second Row Leading Coefficient a 1
Divide the second row by \( \frac{4}{3} \):\[ R2 \to \frac{3}{4}R2 \ \begin{pmatrix} 1 & \frac{2}{3} & | & \frac{1}{3} & 0 \ 0 & 1 & | & -\frac{5}{4} & \frac{3}{4} \ \end{pmatrix} \]
05
- Eliminate the Second Column in the First Row
Use the second row to make the second column of the first row zero by replacing the first row with \( R1 - \frac{2}{3}R2 \):\[ R1 \to R1 - \frac{2}{3}R2 \ \begin{pmatrix} 1 & 0 & | & 1 & -\frac{1}{2} \ 0 & 1 & | & -\frac{5}{4} & \frac{3}{4} \ \end{pmatrix} \]
06
- Check the Result
The augmented portion is now the inverse of matrix \( \textbf{A} \): \[ \textbf{A}^{-1} = \begin{pmatrix} 1 & -\frac{1}{2} \ -\frac{5}{4} & \frac{3}{4} \ \end{pmatrix} \]
07
- Verify Using a Graphing Calculator
Use a graphing calculator to check \( \textbf{A}^{-1} \textbf{A} \) and \( \textbf{A} \textbf{A}^{-1} \). Both should equal the identity matrix:\(\begin{pmatrix} 3 & 2 \ 5 & 3 \ \end{pmatrix} \begin{pmatrix} 1 & -\frac{1}{2} \ -\frac{5}{4} & \frac{3}{4} \ \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \ \end{pmatrix} \) and \(\begin{pmatrix} 1 & -\frac{1}{2} \ -\frac{5}{4} & \frac{3}{4} \ \end{pmatrix} \begin{pmatrix} 3 & 2 \ 5 & 3 \ \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \ \end{pmatrix} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Inverse
To understand the Gauss-Jordan method, we first need to know what a matrix inverse is. The inverse of a matrix \(\textbf{A}\) is another matrix, denoted as \(\textbf{A}^{-1}\), such that when you multiply them together, you get the identity matrix \(\textbf{I}\). The identity matrix is a square matrix with ones on the diagonal and zeros elsewhere. For a matrix to have an inverse, it must be square (same number of rows and columns) and its determinant must be non-zero. The equation looks like this: \[ \textbf{A} \textbf{A}^{-1} = \textbf{I} \] In this exercise, we use the Gauss-Jordan method to find \(\textbf{A}^{-1}\).
Row Operations
Row operations are transformations we perform on the rows of a matrix to simplify it. There are three main types:
- Swapping two rows
- Multiplying a row by a non-zero scalar
- Adding or subtracting the multiple of one row to/from another row
- Dividing the first row by 3
- Using the first row to eliminate the first column in the second row
- Dividing the second row by \(\frac{4}{3}\)
- Using the second row to eliminate the second column in the first row
Augmented Matrix
An augmented matrix is a combination of two matrices. In this case, we combine the original matrix \(\textbf{A}\) with the identity matrix \(\textbf{I}\). The augmented matrix looks like this:
\[ \begin{pmatrix} 3 & 2 & | & 1 & 0 \ 5 & 3 & | & 0 & 1 \ \end{pmatrix} \] This setup allows us to apply row operations simultaneously to both \(\textbf{A}\) and \(\textbf{I}\) in order to transform the identity matrix into the inverse \(\textbf{A}^{-1}\). As we perform row operations to get \(\textbf{A}\) into its reduced row echelon form, we also transform \(\textbf{I}\) into \(\textbf{A}^{-1}\). This method ensures that when \(\textbf{A}\) becomes \(\textbf{I}\), the augmented identity matrix becomes the inverse of \(\textbf{A}\) on the right side.
\[ \begin{pmatrix} 3 & 2 & | & 1 & 0 \ 5 & 3 & | & 0 & 1 \ \end{pmatrix} \] This setup allows us to apply row operations simultaneously to both \(\textbf{A}\) and \(\textbf{I}\) in order to transform the identity matrix into the inverse \(\textbf{A}^{-1}\). As we perform row operations to get \(\textbf{A}\) into its reduced row echelon form, we also transform \(\textbf{I}\) into \(\textbf{A}^{-1}\). This method ensures that when \(\textbf{A}\) becomes \(\textbf{I}\), the augmented identity matrix becomes the inverse of \(\textbf{A}\) on the right side.