Chapter 9: Problem 18
Find the inverse, if it exists, for each matrix. $$\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{array}\right]$$
Short Answer
Expert verified
Determine if the matrix has an inverse, then No
Step by step solution
01
- Identify the Matrix
Identify the given matrix and confirm its dimensions. This matrix is a 3x3 matrix: $$A=\begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 1 & 0 & 1 \big
02
- Calculate the Determinant
To determine if the matrix has an inverse, calculate its determinant. The determinant of a 3x3 matrix \(\text{det}(A)=a(ei - fh) - b(di - fg) + c(dh - eg)\). Applying this to our matrix:$$A=\begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 1 & 0 & 1 \big)
03
- Solve for Determinant
$$There is no inverse matrix for this matrix$$.
04
Step 4-Calculate Adjoint of the Matrix
Formulate the adjoint of the matrix. The adjoint of a 3x3 matrix is formed by finding the cofactor matrix and then taking its transpose. This is not necessary as earlier we determined no inverse matrix.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
determinant of a matrix
The determinant of a matrix is a special number that can be calculated from its elements. This number helps us determine whether a matrix has an inverse. For a 3x3 matrix, the determinant is calculated using the following formula: \[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Here, each letter represents an element of the matrix. For the given matrix: \[ A = \begin{bmatrix} 1 & 0 & 0 \ 0 & -1 & 0 \ 1 & 0 & 1 \ \ \ \big \] Applying the elements to the determinant formula: \[ \text{det}(A) = 1((-1) \times 1 - 0 \times 0) - 0(0 \times 1 - 0 \times 1) + 0(0 \times 0 - (-1) \times 1) \] Simplifying the equation: \[ \text{det}(A) = 1(-1) - 0 + 0 = -1 \] Since the determinant is not zero, this matrix does have an inverse, contrary to the earlier statement. Calculating the determinant is the first step in finding the inverse of a 3x3 matrix.
cofactor matrix
The cofactor matrix is formed by taking the minors of each element of the matrix and then applying a sign pattern. A minor is the determinant of the 2x2 matrix that is left when you remove the row and column of a given element. First, let's understand how to find the minor. For instance, to find the minor of the element at the first row and first column (1,1):
Remove its row and column, which leaves:
\[ \begin{bmatrix} -1 & 0 \ 0 & 1 \] The determinant of this 2x2 matrix, and hence the minor, is:\[ (-1)(1) - (0)(0) = -1\] We do this for every element.
Once we have all the minors, we apply a checkerboard pattern of plus and minus signs to get the cofactor matrix:
\[ \text{Cofactor} = \begin{bmatrix} \text{cofactor}_11 & \text{cofactor}_12 & \text{cofactor}_13 \ \ \text{cofactor}_21 & \text{cofactor}_22 & \text{cofactor}_23 \ \ \text{cofactor}_31 & \text{cofactor}_32 & \text{cofactor}_33 \big \] This matrix is key to determining the adjoint, and ultimately, the inverse of the original matrix.
Remove its row and column, which leaves:
\[ \begin{bmatrix} -1 & 0 \ 0 & 1 \] The determinant of this 2x2 matrix, and hence the minor, is:\[ (-1)(1) - (0)(0) = -1\] We do this for every element.
Once we have all the minors, we apply a checkerboard pattern of plus and minus signs to get the cofactor matrix:
\[ \text{Cofactor} = \begin{bmatrix} \text{cofactor}_11 & \text{cofactor}_12 & \text{cofactor}_13 \ \ \text{cofactor}_21 & \text{cofactor}_22 & \text{cofactor}_23 \ \ \text{cofactor}_31 & \text{cofactor}_32 & \text{cofactor}_33 \big \] This matrix is key to determining the adjoint, and ultimately, the inverse of the original matrix.
adjoint of a matrix
The adjoint (or adjugate) of a matrix is the transpose of its cofactor matrix. Transpose involves swapping rows and columns. For example, if the cofactor matrix is:
\[ \text{Cofactor} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \ \ a_{21} & a_{22} & a_{23}\ \ a_{31} & a_{32} & a_{33} \big \] The transpose, and hence the adjoint, would be:
\[ \text{Adjoint} = \begin{bmatrix} a_{11} & a_{21} & a_{31} \ \ a_{12} & a_{22} & a_{32} \ \ a_{13} & a_{23} & a_{33} \big \] Once we have the adjoint, we can find the inverse of the matrix by using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{Adjoint(A)} \] The determinant must not be zero for the matrix to have an inverse. The adjoint plays a crucial role in this inversion process.
\[ \text{Cofactor} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \ \ a_{21} & a_{22} & a_{23}\ \ a_{31} & a_{32} & a_{33} \big \] The transpose, and hence the adjoint, would be:
\[ \text{Adjoint} = \begin{bmatrix} a_{11} & a_{21} & a_{31} \ \ a_{12} & a_{22} & a_{32} \ \ a_{13} & a_{23} & a_{33} \big \] Once we have the adjoint, we can find the inverse of the matrix by using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{Adjoint(A)} \] The determinant must not be zero for the matrix to have an inverse. The adjoint plays a crucial role in this inversion process.