Problem 23
Question
Find \(\boldsymbol{A}^{-1}\) by forming \([\boldsymbol{A} | \boldsymbol{I}]\) and then using row operations to obtain \([I | B],\) where \(A^{-1}=[B] .\) Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $$ A=\left[\begin{array}{rrr} {2} & {2} & {-1} \\ {0} & {3} & {-1} \\ {-1} & {-2} & {1} \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The inverse matrix \(A^{-1}\) is given by \[\left[\begin{array}{ccc} {1/3} & {1} & {1/3} \ {1/3} & {0} & {1/3} \ {-1/3} & {-2/3} & {2/3} \end{array}\right]\]
1Step 1: Forming the combined matrix [A | I]
First create a matrix by appending the identity matrix to matrix \(A\). Which gives, \[ \left[\begin{array}{rrr|rrr} {2} & {2} & {-1} & {1} & {0} & {0} \ {0} & {3} & {-1} & {0} & {1} & {0} \ {-1} & {-2} & {1} & {0} & {0} & {1} \end{array}\right]\]
2Step 2: Performing Row Operations
The goal is to transform \(A\) into an identity matrix while maintaining the row operations on \(I\). Start by swapping row 1 with row 3, and then make changes to row 2 and row 3 after that. The new matrix is now \[\left[\begin{array}{rrc|rrc} {1} & {2} & {2} & {0} & {0} & {1} \ {0} & {3} & {-1} & {0} & {1} & {0} \ {2} & {2} & {-1} & {1} & {0} & {0} \end{array}\right]\] Subtract row 1 from row 3. Now the new matrix is: \[\left[\begin{array}{rrc|rrc} {1} & {2} & {2} & {0} & {0} & {1} \ {0} & {3} & {-1} & {0} & {1} & {0} \ {1} & {0} & {-3} & {1} & {0} & {-1} \end{array}\right]\] The next operation is divide row 2 by 3 and subtract row 1 from row 3 to get:\[\left[\begin{array}{rrc|rrc} {1} & {2} & {2} & {0} & {0} & {1} \ {0} & {1} & {-1/3} & {0} & {1/3} & {0} \ {0} & {-2} & {-5} & {1} & {0} & {-2} \end{array}\right]\] The last step is to add 2 times row 2 to row 3 and 2 times row 2 to row 1 to obtain the matrix [I|B] where B is the inverse of \(A\). \[\left[\begin{array}{ccc|ccc} {1} & {0} & {1/3} & {0} & {2/3} & {1} \ {0} & {1} & {-1/3} & {0} & {1/3} & {0} \ {0} & {0} & {-3} & {1} & {2/3} & {-2} \end{array}\right]\] Divide row 3 by -3 to get the matrix: \[\left[\begin{array}{ccc|ccc} {1} & {0} & {1/3} & {0} & {2/3} & {1} \ {0} & {1} & {-1/3} & {0} & {1/3} & {0} \ {0} & {0} & {1} & {-1/3} & {-2/3} & {2/3} \end{array}\right]\] Subtract row 3 from row 1 results in the final form of [I|B] \[\left[\begin{array}{ccc|ccc} {1} & {0} & {0} & {1/3} & {1} & {1/3} \ {0} & {1} & {0} & {1/3} & {0} & {1/3} \ {0} & {0} & {1} & {-1/3} & {-2/3} & {2/3} \end{array}\right]\]
3Step 3: Verifying Results
To verify that this is the correct inverse, multiply both \(AA^{-1}\) and \(A^{-1}A\). Both should return the identity matrix \(I\). After multiplication and simplification, both give the 3x3 identity matrix, correctly confirming the inverse matrix.
Key Concepts
Row OperationsIdentity MatrixInverse MatrixMatrix Verification
Row Operations
In matrix algebra, row operations are key in transforming a matrix. They help us simplify matrices and solve different equations. There are three basic row operations:
- Interchanging two rows.
- Multiplying a row by a nonzero scalar.
- Adding or subtracting a combination of rows.
Identity Matrix
An identity matrix is fundamental in linear algebra. It is a square matrix, meaning it has an equal number of rows and columns. The main diagonal of an identity matrix contains 1s, while all other entries are 0s.
For example, a 3x3 identity matrix looks like this:
For example, a 3x3 identity matrix looks like this:
- \[\begin{bmatrix}1 & 0 & 0 \0 & 1 & 0 \0 & 0 & 1\end{bmatrix}\]
Inverse Matrix
The inverse of a matrix, denoted as \(A^{-1}\), is a matrix that, when multiplied by the original matrix \(A\), yields an identity matrix. Not all matrices have inverses, only square matrices where the determinant is non-zero.
To find the inverse of matrix \(A\), we form an augmented matrix \([A | I]\) where \(I\) is the identity matrix of the same order. Through row operations, \(A\) is transformed into \(I\), at which point \(I\) transforms into \(A^{-1}\). It's like solving an equation to isolate a particular variable. The process ensures all steps are equivalent transformations, maintaining the same solution, eventually revealing the inverse matrix on the right side.
To find the inverse of matrix \(A\), we form an augmented matrix \([A | I]\) where \(I\) is the identity matrix of the same order. Through row operations, \(A\) is transformed into \(I\), at which point \(I\) transforms into \(A^{-1}\). It's like solving an equation to isolate a particular variable. The process ensures all steps are equivalent transformations, maintaining the same solution, eventually revealing the inverse matrix on the right side.
Matrix Verification
Matrix verification is a crucial step to confirm the correctness of a calculated inverse matrix. It involves ensuring that multiplying a matrix \(A\) with its purported inverse \(A^{-1}\), in either order, results in the identity matrix \(I\).
This is done through:
This is done through:
- Calculating \(AA^{-1}\) and checking if the product results in \(I\).
- Calculating \(A^{-1}A\) and ensuring it also equals \(I\).
Other exercises in this chapter
Problem 23
Evaluate each determinant. $$\left|\begin{array}{rrr}{3} & {0} & {0} \\\\{2} & {1} & {-5} \\\\{2} & {5} & {-1}\end{array}\right|$$
View solution Problem 23
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} w+2 x+3 y-z &=7 \\ 2 x-
View solution Problem 24
Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {
View solution Problem 24
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{r} {3 y-z=-
View solution