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Chapter 6: Problem 23

For each matrix, find \(A^{-1}\) if it exists. $$A=\left[\begin{array}{rrr}1 & 0 & 0 \\\0 & -1 & 0 \\\1 & 0 & 1\end{array}\right]$$

Short Answer

Expert verified
The inverse is \(A^{-1} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\).

Step by step solution

01

Determine if the Matrix is Invertible

To find the inverse of a matrix, the matrix must be square and have a non-zero determinant. We need to calculate the determinant of matrix \(A\).
02

Calculate the Determinant of A

The formula for the determinant \(|A|\) of a 3x3 matrix \( \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \) is \( a(ei - fh) - b(di - fg) + c(dh - eg) \). For matrix \(A\), substitute values: \[|A| = 1((-1)(1) - (0)(0)) - 0(0 - 0) + 0(0 - (-1)) = -1\]. Thus, the determinant is \(-1\).
03

Adjugate and Transpose the Matrix

Since the determinant is non-zero, matrix \(A\) is invertible. The inverse of a 3x3 matrix \(A\) is given by \(A^{-1} = \frac{1}{|A|} \, \text{adj}(A)^T\). First, find the cofactors to form the adjugate matrix and then transpose it.
04

Finding the Cofactor Matrix

Calculate the cofactor for each element of the matrix. Since \(A\) is mostly zeros except for three ones and a minus one, many cofactors will be zero. Calculate the remaining relevant cofactors: - Co-factor for \(1\): \(C_{11} = (-1)^{1+1} \left((-1)(1) - (0)(0)\right) = 1\)- Co-factor for \(-1\): \(C_{22} = (-1)^{2+2} \left(1(1) - 1(0)\right) = 1\) - Co-factor for \(1\) in (3, 3) position: \(C_{33} = (-1)^{3+3} \left(1(-1) - 0(0)\right) = -1 \)
05

Assemble the Adjugate Matrix

The adjugate matrix is the transpose of the cofactor matrix. It will look like: \[\text{adj}(A) = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix}\]
06

Find the Inverse Matrix

Now, using the inverse matrix formula \(A^{-1} = \frac{1}{|A|} \, \text{adj}(A)\), we find: \[A^{-1} = \frac{1}{-1} \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix}\] Thus, \(A^{-1} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant Calculation
Calculating the determinant is one of the first steps in determining if a matrix is invertible. A matrix is invertible if it is square (same number of rows and columns) and its determinant is not zero. For a 3x3 matrix, the determinant can be calculated using the formula:
  • For matrix \(A\) where \(A = \left[ \begin{array}{ccc} a & b & c \ d & e & f \ g & h & i \end{array} \right]\):\[|A| = a(ei - fh) - b(di - fg) + c(dh - eg)\]
In the exercise, applying this formula to matrix \(A = \left[ \begin{array}{ccc} 1 & 0 & 0 \ 0 & -1 & 0 \ 1 & 0 & 1 \end{array} \right]\), the calculated determinant is \(-1\). This non-zero determinant confirms that the matrix is invertible.
Cofactor Expansion
Cofactor expansion is a method of calculating determinants or finding the adjugate of a matrix. It uses specific minors of the matrix. A minor is the determinant of the matrix that remains after removing a particular row and column.
  • Each cofactor \(C_{ij}\) is given by \((-1)^{i+j}\) times the determinant of the minor obtained by deleting the \(i\)-th row and \(j\)-th column.
In this exercise, the only non-zero cofactors are:
  • For the element \(1\) at position (1,1): \[C_{11} = (-1)^{2}( -1 ) = 1 \]
  • For the element \(-1\) at position (2,2): \[C_{22} = (-1)^{4}( 1 ) = 1 \]
  • For the element \(1\) at position (3,3): \[C_{33} = (-1)^{6}( -1 ) = -1 \]
Using these cofactors, we build the cofactor matrix necessary for determining the adjugate matrix.
Adjugate Matrix
The adjugate matrix, sometimes called the adjoint, is a matrix formed by taking the transposition of the cofactor matrix. It is crucial for finding the inverse of a matrix.
  • To construct it, calculate the cofactor for each position in the original matrix.
  • Then, reorganize these cofactors by transposing the matrix of cofactors, which swaps the rows and columns.
For the given exercise, after computing the necessary cofactors:
  • The cofactor matrix is: \[\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix} \]
  • Thus, the adjugate matrix (\(\text{adj}(A)\)) is the same as the cofactor matrix because it is already in its transposed form.
Inverse of a Matrix
The inverse of a matrix is a matrix that, when multiplied by the original matrix, yields the identity matrix. Not all matrices have an inverse, but when they do, the inverse is vital for solving systems of linear equations and other applications.The formula to find the inverse of a matrix \(A\) is:
  • \[A^{-1} = \frac{1}{|A|} \times \text{adj}(A)\]
Where \(|A|\) is the determinant of \(A\), and \(\text{adj}(A)\) is the adjugate of \(A\).
  • In this problem, the determinant \(|A|\) was \(-1\), and the adjugate matrix was \[\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix}\].
  • So, the inverse is calculated as:\[A^{-1} = \frac{1}{-1} \times \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \ 0 & -1 & 0 \ 0 & 0 & 1 \end{bmatrix}\]
This inverse checks out because if you multiply it by the original matrix \(A\), you would get the identity matrix back.

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Most popular questions from this chapter

A manufacturing process requires that oil refineries manufacture at least 2 gallons of gasoline for each gallon of fuel oil. To meet winter demand for fuel oil, at least 3 million gallons a day must be produced. The demand for gasoline is no more than 6.4 million gallons per day. If the price of gasoline is \(\$ 1.90\) per gallon and the price of fuel oil is \(\$ 1.50\) per gallon, how much of each should be produced to maximize revenue?

Solve each system graphically. Give \(x\) - and y-coordinates correct to the nearest hundredth. $$\begin{aligned}&y=\sqrt[3]{x-4}\\\&x^{2}+y^{2}=6\end{aligned}$$

Use the shading capabilities of your graphing calculator to graph each inequality or system of inequalities. $$y \leq x^{3}+x^{2}-4 x-4$$

Let \(A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right], B=\left[\begin{array}{ll}b_{11} & b_{12} \\ b_{21} & b_{22}\end{array}\right],\) and \(C=\left[\begin{array}{ll}c_{11} & c_{12} \\\ c_{21} & c_{22}\end{array}\right]\) where all the elements are real numbers. Use these matrices to show that each statement is true for \(2 \times 2\) matrices. \(A+(B+C)=(A+B)+C\) (associative property)

Farmer Jones raises only pigs and geese. She wants to raise no more than 16 animals, with no more than 12 geese. She spends \(\$ 50\) to raise a pig and \(\$ 20\) to raise a goose. She has \(\$ 500\) available for this purpose. Find the maximum profit she can make if she makes a profit of \(\$ 80\) per goose and \(\$ 40\) per pig. Indicate how many pigs and geese she should raise to achieve this maximum.
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