Chapter 9: Problem 10
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 \(\mathbf{A} \mathbf{A}^{-1}\). $$\mathbf{A}=\left[\begin{array}{rrr}1 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & -1 & 1\end{array}\right]$$
Short Answer
Expert verified
\[\mathbf{A}^{-1} = \left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \-1 & \frac{3}{2} & 1 \ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{array}\right]\]
Step by step solution
01
- Augment the Matrix
Represent the matrix \(\mathbf{A}\) with the identity matrix \(\mathbf{I}\) to form an augmented matrix: \[\left[\mathbf{A} | \mathbf{I}\right] = \left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 2 & 1 & 0 & 0 & 1 & 0 \ 1 & -1 & 1 & 0 & 0 & 1\end{array}\right]\]
02
- Make the First Column's First Element 1
The first element is already 1, so no changes are needed. \[\left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 2 & 1 & 0 & 0 & 1 & 0 \ 1 & -1 & 1 & 0 & 0 & 1\end{array}\right]\]
03
- Eliminate Below Pivot in First Column
Subtract 2 times the first row from the second row, and subtract the first row from the third row: \[\begin{array}{l}\left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 0 & 1 & -2 & -2 & 1 & 0 \ 0 & -1 & 0 & -1 & 0 & 1\end{array}\right] \ \left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 0 & 1 & -2 & -2 & 1 & 0 \ 0 & -1 & 0 & -1 & 0 & 1\end{array}\right] \rightarrow \left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 0 & 1 & -2 & -2 & 1 & 0 \ 0 & 0 & 2 & 1 & 1 & 1\end{array}\right] \ \text{(add second row to third row)}\end{array}\]
04
- Normalize the Second Row
The pivot element of the second row (second column) is already 1. The matrix is: \[\left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 0 & 1 & -2 & -2 & 1 & 0 \ 0 & 0 & 2 & 1 & 1 & 1\end{array}\right]\]
05
- Eliminate Other Elements in the Second Column
Subtract -1 times the second row from the third row and add it to the first row: \[\begin{array}{l}\left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 0 & 1 & -2 & -2 & 1 & 0 \ 0 & 0 & 2 & 1 & 1 & 1\end{array}\right] \rightarrow \left[\begin{array}{rrr|rrr}1 & 2 & -1 & -3 & 2 & 0 \ 0 & 1 & -2 & -2 & 1 & 0 \ 0 & 0 & 2 & 1 & 1 & 1\end{array}\right] \rightarrow \left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 0 & 1 & -2 & -2 & 1 & 0 \ 0 & 0 & 2 & 1 & 1\end{array}\right] \text{(add second row to third row)}\end{array}\]
06
- Normalize the Third Row
Divide the third row by 2 so that the pivot element becomes 1: \[\left[\begin{array}{rrr|rrr}1 & 0 & 1 & 1 & 0 & 0 \ 0 & 1 & -2 & -2 & 1 & 0 \ 0 & 0 & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{array}\right]\]
07
- Eliminate Other Elements in the Third Column
Subtract the third row multiplied by the third element of the first and second rows to eliminate those elements: \[\begin{array}{l}\left[\begin{array}{rrr|rrr}1 & 0 & 0 & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \ 0 & 1 & 0 & -1 & \frac{3}{2} & 1 \ 0 & 0 & 1 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{array}\right]\end{array}\]
08
- Resulting Inverse Matrix
The matrix on the right-hand side of the augmented matrix is the inverse of the matrix \(\mathbf{A}\right)\): \[\begin{array}{ccc}\mathbf{A}^{-1} = \left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \ -1 & \frac{3}{2} & 1 \ \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{array}\right]\end{array}\]
09
- Verify Your Result
Use a graphing calculator to check \( \mathbf{A}^{-1} \mathbf{A} \) and \( \mathbf{A} \mathbf{A}^{-1} \) both result in the identity matrix: \[\begin{array}{l}\mathbf{A} \left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \ -1 & \frac{3}{2} & 1 \ \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{array}\right] = \mathbf{I} \ isheye studio\end{document}\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \ -1 & \frac{3}{2} & 1 \ \frac{1}{2} & \frac{1}{2} & \frac{1}{2}\end{array}\right] \mathbf{A} = \mathbf{I} \end{array}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Inversion
Matrix inversion is the process of finding the inverse of a given square matrix. The inverse of a matrix \( \mathbf{A} \ \) is denoted as \( \mathbf{A}^{-1} \ \) and is a matrix that, when multiplied with \( \mathbf{A} \ \), results in the identity matrix \( \mathbf{I} \ \). This is expressed as: \[ \mathbf{A} \mathbf{A}^{-1} = \mathbf{I} \] \[ \mathbf{A}^{-1} \mathbf{A} = \mathbf{I} \] The identity matrix, \( \mathbf{I} \ \), is a special matrix with ones on the diagonal and zeros elsewhere. Not all matrices have inverses. A matrix must be square (same number of rows and columns) and its determinant must not be zero. The determinant is a scalar value derived from the matrix elements that provides important properties about the matrix.
Inverse Matrix
Finding the inverse of a matrix manually is often done using methods like the Gauss-Jordan elimination method or by using the formula involving adjugate and determinant for 2x2 matrices. The Gauss-Jordan method systematically eliminates elements of a matrix to transform it into the identity matrix while performing the same row operations on an augmented identity matrix. For the matrix \( \mathbf{A} \ \) given in the exercise: \[ \mathbf{A} = \left[\begin{array}{rrr} 1 & 0 & 1 \ 2 & 1 & 0 \ 1 & -1 & 1 \end{array}\right] \] The Gauss-Jordan method involves creating an augmented matrix \( \[ \mathbf{A} \mathbf{|} \mathbf{I} \] \), and then performing a series of row operations to convert \( \mathbf{A} \mathbf{|} \mathbf{I} \) to \( \mathbf{I} \mathbf{|} \mathbf{A}^{-1} \ \). By following the steps outlined in the original solution, you can verify that \( \mathbf{A}^{-1} \ \) is: \[ \mathbf{A}^{-1} = \left[\begin{array}{ccc} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \ -1 & \frac{3}{2} & 1 \ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{array}\right] \] It’s a good practice to check your work using a graphing calculator by multiplying \( \mathbf{A} \mathbf{A}^{-1} \ \) and \( \mathbf{A}^{-1} \mathbf{A} \ \) to ensure they both yield the identity matrix.
Linear Algebra
Linear Algebra is a branch of mathematics that deals with vectors, vector spaces (or linear spaces), linear transformations, and systems of linear equations. Key concepts in linear algebra include:
- Matrices and Determinants
- Vector Spaces
- Linear Transformations
- Systems of Linear Equations