Problem 20
Question
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 \(A^{-1} A=I\) $$ A=\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 9 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The inverse of the given matrix \(A^{-1}=\left[\begin{array}{lll} 1/3 & 0 & 0 \ 0 & 1/6 & 0 \ 0 & 0 & 1/9 \end{array}\right]\). This has been verified by showing that both AA^{-1} and A^{-1}A equal the identity matrix, I.
1Step 1: Form the Augmented Matrix [A | I]
Firstly, form the augmented matrix by placing the given matrix, A, on the left and 3x3 identity matrix, I, on the right. So the formed matrix will be: \[ [A | I] = \left[\begin{array}{lll|lll} 3 & 0 & 0 & 1 & 0 & 0 \ 0 & 6 & 0 & 0 & 1 & 0 \ 0 & 0 & 9 & 0 & 0 & 1 \end{array}\right] \]
2Step 2: Use Row Operations to Get [I | B]
Since the matrix A is already a diagonal matrix, apply the operation (1/a_ii)R_i -> R_i for each row i to transform A into the identity matrix. This will give us the required form [I | B]. \[ [I | B] = \left[\begin{array}{lll|lll} 1 & 0 & 0 & 1/3 & 0 & 0 \ 0 & 1 & 0 & 0 & 1/6 & 0 \ 0 & 0 & 1 & 0 & 0 & 1/9 \end{array}\right] \] Here, B is the matrix obtained on the right side.
3Step 3: Conclude A^{-1} from [I | B]
From the form [I | B], the B matrix on the right side is the inverse of A, A^{-1}. So \( A^{-1}= \left[\begin{array}{lll} 1/3 & 0 & 0 \ 0 & 1/6 & 0 \ 0 & 0 & 1/9 \end{array}\right] \)
4Step 4: Verification
Perform matrix multiplication to confirm that AA^{-1} and A^{-1}A both equal I. The results are \[ AA^{-1}=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right] = I \] and \[ A^{-1}A=\left[\begin{array}{lll} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right] = I \] Both are equal to identity matrix, thus verifying that the obtained matrix is indeed the inverse of A.
Key Concepts
Augmented MatrixRow OperationsIdentity MatrixDiagonal Matrix
Augmented Matrix
An augmented matrix is a crucial tool in linear algebra when solving systems of linear equations or finding the inverse of a matrix. It is created by combining two matrices. Specifically, for a given square matrix \(A\), such as a 3x3 matrix, we augment it by attaching the identity matrix of the same size to its right. In the context of finding the inverse of \(A\), we form the matrix \([A | I]\).
- The matrix \(A\) is placed to the left of the identity matrix \(I\).
- It serves as a starting point before applying row operations to transform it into \([I | B]\), where \(B\) will become the inverse.
Row Operations
Row operations are essential for manipulating matrices to find their inverse. There are three elementary row operations you may use:
This will convert \(A\) to the identity matrix, thus giving us \([I | B]\). The beauty of row operations is that they allow you to transition from \([A | I]\) to \([I | B]\) without altering the solutions of the system. This approach effectively finds the inverse matrix through direct and strategic manipulations.
- Swapping two rows.
- Multiplying a row by a non-zero scalar.
- Adding or subtracting the multiples of other rows.
This will convert \(A\) to the identity matrix, thus giving us \([I | B]\). The beauty of row operations is that they allow you to transition from \([A | I]\) to \([I | B]\) without altering the solutions of the system. This approach effectively finds the inverse matrix through direct and strategic manipulations.
Identity Matrix
The identity matrix plays a pivotal role in matrix theory and operations. It is a square matrix with ones on the diagonal and zeroes elsewhere:
\[I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]
When multiplied by any matrix that can conform to its dimensions, the identity matrix leaves the matrix unchanged.
\[I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]
When multiplied by any matrix that can conform to its dimensions, the identity matrix leaves the matrix unchanged.
- This property is similar to multiplying a number by one.
- Essentially, \(AI = A\) and \(IA = A\).
Diagonal Matrix
Diagonal matrices are matrices where all the non-diagonal elements are zero, resembling the structure:
\[D = \begin{bmatrix} d_1 & 0 & 0 \ 0 & d_2 & 0 \ 0 & 0 & d_3 \end{bmatrix}\]
They have a simplified form that makes certain operations like finding inverses or eigenvalues very straightforward.
\[D = \begin{bmatrix} d_1 & 0 & 0 \ 0 & d_2 & 0 \ 0 & 0 & d_3 \end{bmatrix}\]
They have a simplified form that makes certain operations like finding inverses or eigenvalues very straightforward.
- To find the inverse, simply invert each non-zero diagonal element.
- For example, dividing each diagonal by itself transforms \(D\) into an identity matrix.
Other exercises in this chapter
Problem 20
In Exercises \(17-26,\) let $$ A=\left[\begin{array}{rr} -3 & -7 \\ 2 & -9 \\ 5 & 0 \end{array}\right] \text { and } B=\left[\begin{array}{rr} -5 & -1 \\ 0 & 0
View solution Problem 20
Use Cramer's rule to solve each system or to determine that the system is inconsistent or contains dependent equations. $$ \begin{aligned}&3 x=7 y+1\\\&2 x=3 y-
View solution Problem 20
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$\begin{aligned}w+x-y+z &=-2 \\\2 w-x+2 y-z &=7 \
View solution Problem 20
In Exercises \(19-24\), perform each matrix row operation and write the new matrix. \(\left[\begin{array}{rrr|r}3 & -12 & 6 & 9 \\ 1 & -4 & 4 & 0 \\ 2 & 0 & 7 &
View solution