Problem 21
Question
In Exercises \(19-28,\) find \(A^{-1}\) by forming \([A | I]\) and then using row operations to obtain \([I | B],\) where \(A^{-1}=[B]\). Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $$A=\left[\begin{array}{rrr}1 & 2 & -1 \\\\-2 & 0 & 1 \\\1 & -1 & 0\end{array}\right]$$
Step-by-Step Solution
Verified Answer
The inverse matrix \( A^{-1} \) is:\[\left[\begin{array}{ccc}2 & 1 & -1 \1 & 1 & 1 \2 & 2 & 4\end{array}\right]\]This verifies as \( AA^{-1} \) and \( A^{-1}A \) both equal the identity matrix \( I \).
1Step 1: Forming the Augmented matrix
Combine the given matrix A with the identity matrix I to form the augmented matrix [A | I]:\[\left[\begin{array}{ccc|ccc}1 & 2 & -1 & 1 & 0 & 0 \-2 & 0 & 1 & 0 & 1 & 0 \1 & -1 & 0 & 0 & 0 & 1\end{array}\right]\]
2Step 2: Apply Row Operations
Apply suitable row operations to modify the left half of the augmented matrix into an identity matrix, rendering the right side to be [B]. Begin by adding two times the first row to the second row and subtracting the first row from the third row to get:\[\left[\begin{array}{ccc|ccc}1 & 2 & -1 & 1 & 0 & 0 \0 & 4 & -1 & 2 & 1 & 0 \0 & -3 & 1 & -1 & 0 & 1\end{array}\right]\]Proceed by adding three quarters of the second row to the third row and multiplying the second row by one quarter to yield the following matrix:\[\left[\begin{array}{ccc|ccc}1 & 2 & -1 & 1 & 0 & 0 \0 & 1 & -1/4 & 1/2 & 1/4 & 0 \0 & 0 & 1/4 & 1/2 & 1/4 & 1\end{array}\right]\]Lastly, add four times the third row to the second row, add the first row to the third row, and multiply the third row by four:\[\left[\begin{array}{ccc|ccc}1 & 0 & 0 & 2 & 1 & -1 \0 & 1 & 0 & 1 & 1 & 1 \0 & 0 & 1 & 2 & 2 & 4\end{array}\right]\]
3Step 3: Verify matrix A^-1
Matrix B in the right side of the augmented matrix is the inverse of A, denoted A^-1. It is:\[A^{-1}=\left[\begin{array}{ccc}2 & 1 & -1 \1 & 1 & 1 \2 & 2 & 4\end{array}\right]\]Now, verify A^-1 by checking that AA^-1 = I and A^-1A = I. Compute the matrix product and simplify:\[AA^{-1}=\left[\begin{array}{ccc}1 & 2 & -1 \-2 & 0 & 1 \1 & -1 & 0\end{array}\right]\left[\begin{array}{ccc}2 & 1 & -1 \1 & 1 & 1 \2 & 2 & 4\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{array}\right]=I\]Similarly: \[A^{-1}A=\left[\begin{array}{ccc}2 & 1 & -1 \1 & 1 & 1 \2 & 2 & 4\end{array}\right]\left[\begin{array}{ccc}1 & 2 & -1 \-2 & 0 & 1 \1 & -1 & 0\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{array}\right]=I\]So indeed, AA^-1 = I and A^-1A = I, so A^-1 is the correct inverse of A.
Key Concepts
Augmented MatrixRow OperationsIdentity Matrix
Augmented Matrix
An augmented matrix serves as a powerful tool for solving systems of equations or finding the inverse of a matrix. It essentially merges two matrices side by side. In the process to find the inverse of a matrix \(A\), the first step is to create an augmented matrix by placing the identity matrix \(I\) next to \(A\), forming \[A | I\]. This creates a new matrix where the left-hand side holds the original matrix \(A\), and the right-hand side holds the identity matrix.
The beauty of the augmented matrix lies in its arrangement, which prepares the ground for systematic row operations to transform \(A\) into \(I\), and consequently, the identity matrix transforms into the inverse of \(A\). Thus, the augmented matrix helps to clearly track the transformations required to achieve the inverse.
The beauty of the augmented matrix lies in its arrangement, which prepares the ground for systematic row operations to transform \(A\) into \(I\), and consequently, the identity matrix transforms into the inverse of \(A\). Thus, the augmented matrix helps to clearly track the transformations required to achieve the inverse.
Row Operations
Row operations, which include row switching, row multiplication, and row addition, are the manipulative steps applied to the rows of a matrix to achieve a desired form. In our case, they are used to turn the left side of the augmented matrix into the identity matrix. These operations are elemental because they maintain the equivalence of the matrix system, meaning they won't alter the solution or properties of the original matrix.
- Row Switching: This involves swapping the positions of two rows.
- Row Multiplication: This means multiplying all the elements of a row by a non-zero constant.
- Row Addition: This implies adding the elements of one row to another, often multiplied by a constant.
Identity Matrix
The identity matrix, denoted \(I\), is the linchpin in the quest to find a matrix inverse. It is a square matrix with ones on the diagonal and zeros everywhere else. In matrix operations, it plays a similar role to the number 1 in arithmetic, wherein multiplying any matrix by \(I\) leaves the original matrix unchanged.
Here lies its importance in finding the inverse: the goal of the row operations on the augmented matrix \[A | I\] is to transform \(A\) into the identity matrix, \(I\), because only then can we ensure that the corresponding transformation on the identity matrix side of the augmented matrix has resulted in \(A^{-1}\), the inverse of \(A\). Thus, the identity matrix is both the target in the transformation process and a critical component of verifying that \(A^{-1}A = I\) and \(AA^{-1} = I\), which confirm the correctness of the inverse obtained.
Here lies its importance in finding the inverse: the goal of the row operations on the augmented matrix \[A | I\] is to transform \(A\) into the identity matrix, \(I\), because only then can we ensure that the corresponding transformation on the identity matrix side of the augmented matrix has resulted in \(A^{-1}\), the inverse of \(A\). Thus, the identity matrix is both the target in the transformation process and a critical component of verifying that \(A^{-1}A = I\) and \(AA^{-1} = I\), which confirm the correctness of the inverse obtained.
Other exercises in this chapter
Problem 20
In Exercises \(19-28,\) find \(A^{-1}\) by forming \([A | I]\) and then using row operations to obtain \([I | B],\) where \(A^{-1}=[B]\). Check that \(A A^{-1}=
View solution Problem 20
Use Cramer's Rule to solve each system. $$\left\\{\begin{array}{l}3 x=7 y+1 \\\2 x=3 y-1\end{array}\right.$$
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Use Cramer's Rule to solve each system. $$\left\\{\begin{array}{l}2 x=3 y+2 \\\5 x=51-4 y\end{array}\right.$$
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Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{array}{rr} w+x-y+z= & -2 \\ 2 w-
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