Problem 20
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 \(A^{-1} A=I\) $$ A=\left[\begin{array}{lll} {3} & {0} & {0} \\ {0} & {6} & {0} \\ {0} & {0} & {9} \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The inverse (\(A^{-1}\) or B) of the matrix A is: \[ \left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \ 0 & \frac{1}{6} & 0 \ 0 & 0 & \frac{1}{9} \end{array}\right] \]
1Step 1: Concatenate A with the Identity matrix
Form a new matrix by combining the given matrix A and the Identity matrix of the same size. This formed matrix \([A|I]\) would be: \[ \left[\begin{array}{ccc|ccc} 3 & 0 & 0 & 1 & 0 & 0 \ 0 & 6 & 0 & 0 & 1 & 0 \ 0 & 0 & 9 & 0 & 0 & 1 \end{array}\right] \]
2Step 2: Applying Gauss-Jordan Elimination
Since A is a diagonal matrix, the primary goal is converting the diagonal entries into 1. This is done by dividing each row by the respective diagonal element. Here, the first row is divided by 3, the second by 6, and the third by 9.
3Step 3: Observe new matrix
After performing the row operations, the new matrix is \([I|B]\), where B is the right-side matrix. \[ \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \frac{1}{3} & 0 & 0 \ 0 & 1 & 0 & 0 & \frac{1}{6} & 0 \ 0 & 0 & 1 & 0 & 0 & \frac{1}{9} \end{array}\right] \] Accordingly, B can be separated from the augmented matrix: \[ \left[\begin{array}{ccc} \frac{1}{3} & 0 & 0 \ 0 & \frac{1}{6} & 0 \ 0 & 0 & \frac{1}{9} \end{array}\right] \] This matrix B is the inverse of the matrix A.
4Step 4: Validation
Validate by verifying that \(AA^{-1}=I\) and \(A^{-1}A=I\). Substituting B (as A inverse) into these equations, both sides of each equation equal to an Identity matrix, establishing the correctness of our calculations.
Key Concepts
Gauss-Jordan EliminationIdentity MatrixRow OperationsDiagonal Matrix
Gauss-Jordan Elimination
Gauss-Jordan elimination is a streamlined process for solving systems of linear equations and finding the inverse of a matrix. It involves performing row operations on an augmented matrix to transform it into reduced row-echelon form (RREF). To do this, one typically begins with the augmented matrix \( [A | I] \), where \( A \) is the original matrix and \( I \) is the identity matrix.
The objective is to modify the augmented matrix so that the left part becomes the identity matrix, indicating that the right part is the inverse of \( A \). Row operations include swapping rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another row. These actions preserve the equivalence of the original matrix equations, which is crucial for finding the accurate inverse. By systematically applying these operations, one turns the left section of the augmented matrix into an identity matrix while simultaneously forming the inverse on the right.
The objective is to modify the augmented matrix so that the left part becomes the identity matrix, indicating that the right part is the inverse of \( A \). Row operations include swapping rows, multiplying a row by a non-zero scalar, and adding a multiple of one row to another row. These actions preserve the equivalence of the original matrix equations, which is crucial for finding the accurate inverse. By systematically applying these operations, one turns the left section of the augmented matrix into an identity matrix while simultaneously forming the inverse on the right.
Identity Matrix
The identity matrix, denoted as \( I \), is a square matrix that plays an essential role in matrix operations, much like the number 1 in scalar multiplication. It contains 1s on its main diagonal (from the top left to the bottom right) and 0s elsewhere. In mathematical form, the identity matrix of size \( n \) is:
\[ I_n = \begin{array}{cccc} 1 & 0 & \cdots & 0 \ 0 & 1 & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \cdots & 0 & 1 \end{array} \]
When any matrix \( A \) is multiplied by an identity matrix of the same dimensions, the result is the original matrix \( A \), i.e., \( AI = IA = A \). When finding the inverse of a matrix, the identity matrix serves as a reference point. The presence of the identity matrix in Gauss-Jordan elimination is what assures us that row operations are properly transforming the original matrix into its inverse.
\[ I_n = \begin{array}{cccc} 1 & 0 & \cdots & 0 \ 0 & 1 & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \cdots & 0 & 1 \end{array} \]
When any matrix \( A \) is multiplied by an identity matrix of the same dimensions, the result is the original matrix \( A \), i.e., \( AI = IA = A \). When finding the inverse of a matrix, the identity matrix serves as a reference point. The presence of the identity matrix in Gauss-Jordan elimination is what assures us that row operations are properly transforming the original matrix into its inverse.
Row Operations
Row operations are fundamental tools in matrix algebra used during procedures like Gauss-Jordan elimination. There are three types of row operations:
- Row Swapping (Ri <-> Rj): Two rows, \( i \) and \( j \) are interchanged.
- Row Multiplication (kRi): A row \( i \) is multiplied by a non-zero scalar \( k \).
- Row Addition (Ri + kRj -> Ri): A multiple of row \( j \) is added to row \( i \), modifying row \( i \) while keeping row \( j \) unchanged.
Diagonal Matrix
A diagonal matrix is a special type of square matrix where all the elements outside the main diagonal are zero. Its general form is:
\[ \begin{array}{cccc} d_1 & 0 & \cdots & 0 \ 0 & d_2 & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \cdots & 0 & d_n \end{array} \]
Where \( d_1, d_2, ..., d_n \) are the entries on the main diagonal, which can be any number, including zero (although if all diagonal entries are zero, the matrix is simply the zero matrix). In the given exercise, \( A \) is such a diagonal matrix with 3, 6, and 9 on its diagonal. Diagonal matrices are easy to work with, especially when finding the inverse, because one can directly obtain the inverse by replacing each non-zero diagonal element with its reciprocal, leaving all other elements as zero.
\[ \begin{array}{cccc} d_1 & 0 & \cdots & 0 \ 0 & d_2 & \ddots & \vdots \ \vdots & \ddots & \ddots & 0 \ 0 & \cdots & 0 & d_n \end{array} \]
Where \( d_1, d_2, ..., d_n \) are the entries on the main diagonal, which can be any number, including zero (although if all diagonal entries are zero, the matrix is simply the zero matrix). In the given exercise, \( A \) is such a diagonal matrix with 3, 6, and 9 on its diagonal. Diagonal matrices are easy to work with, especially when finding the inverse, because one can directly obtain the inverse by replacing each non-zero diagonal element with its reciprocal, leaving all other elements as zero.
Other exercises in this chapter
Problem 20
A few steps in the process of simplifying the given matrix to row-echelon form, with Is down the diagonal from upper left to lower right, and os below the \(1 s
View solution Problem 20
Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{l}{3 x=7 y+1} \\\\{2 x=3 y-1}\end{array}\right.$$
View solution Problem 21
Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {
View solution Problem 21
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x+y-z &=-
View solution