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Chapter 5: Problem 35

Let \(A=\left[\begin{array}{ll}2 & -5 \\ 1 & -3\end{array}\right] \quad B=\left[\begin{array}{ll}4 & 3 \\ 1 & 1\end{array}\right] \quad C=\left[\begin{array}{rr}2 & 3 \\ -2 & 1\end{array}\right]\) a. Find \(A B C, A^{-1}, B^{-1}\), and \(C^{-1}\). b. Show that \((A B C)^{-1}=C^{-1} B^{-1} A^{-1}\).

Short Answer

Expert verified
In summary, we have calculated the following values: - \(ABC = \begin{bmatrix} 8 & 8 \\ 2 & 3 \end{bmatrix}\) - \(A^{-1} = \begin{bmatrix} -3 & 5 \\ -1 & 2 \end{bmatrix}\) - \(B^{-1} = \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix}\) - \(C^{-1} = \begin{bmatrix} 1 & -3 \\ 2 & 2 \end{bmatrix}\) And proved that \((ABC)^{-1} = C^{-1}B^{-1}A^{-1} = \begin{bmatrix} 8 & 8 \\ -2 & 4 \end{bmatrix}\).

Step by step solution

01

Calculate the product of the matrices ABC

To compute the product ABC, first multiply A by B, then the result by C: \(AB = \begin{bmatrix}2 & -5 \\1 & -3\end{bmatrix}\begin{bmatrix}4 & 3 \\1 & 1\end{bmatrix} = \begin{bmatrix}(2*4 +(-5)*1) & (2*3 + (-5)*1)\\ (1*4 + (-3)*1) & (1*3 + (-3)*1)\end{bmatrix} = \begin{bmatrix}3 & -1\\ 1 & 0\end{bmatrix}\) Now, calculate ABC: \(ABC = \begin{bmatrix}3 & -1\\ 1 & 0\end{bmatrix}\begin{bmatrix}2 & 3\\ -2 & 1\end{bmatrix} = \begin{bmatrix}(3*2 + -1*(-2)) & (3*3 + (-1)*1)\\ (1*2 + 0*(-2)) & (1*3 + 0*1)\end{bmatrix} = \begin{bmatrix}8 & 8\\ 2 & 3\end{bmatrix}\)
02

Calculate the inverse of matrix A

To calculate the inverse of a 2x2 matrix: \(A^{-1} = \frac{1}{(ad - bc)}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\) For matrix A: \(A^{-1} = \frac{1}{2*(-3) -(-5)*1}\begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix}\) \(A^{-1} = \frac{1}{1}\begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix} = \begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix}\)
03

Calculate the inverse of matrix B

For matrix B: \(B^{-1} = \frac{1}{4*1 - 3*1}\begin{bmatrix} 1 & -3\\ -1 & 4\end{bmatrix}\) \(B^{-1} = \frac{1}{1}\begin{bmatrix} 1 & -3\\ -1 & 4\end{bmatrix} = \begin{bmatrix} 1& -3\\ -1 & 4\end{bmatrix}\)
04

Calculate the inverse of matrix C

For matrix C: \(C^{-1} = \frac{1}{2*1 -(-2)*3}\begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix}\) \(C^{-1} = \frac{1}{8}\begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix} = \frac{1}{8}\begin{bmatrix} 8 & -24\\ 16 & 16\end{bmatrix} = \begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix}\)
05

Prove (ABC)^(-1) = C^(-1)B^(-1)A^(-1)

Since we already calculated ABC and the inverses of A, B, and C, we can perform this step computationally. Calculate C^(-1)B^(-1)A^(-1): \(C^{-1}B^{-1}A^{-1} = \begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix}\begin{bmatrix} 1& -3\\ -1 & 4\end{bmatrix}\begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix}\) First multipy C^(-1) and B^(-1): \(\begin{bmatrix} 1 & -3\\ 2 & 2\end{bmatrix}\begin{bmatrix} 1& -3\\ -1 & 4\end{bmatrix} = \begin{bmatrix}(1*1 +(-3)*(-1)) & (1*(-3) + (-3)*4))\\ (2*1 + 2*(-1)) & (2*(-3) + 2*4))\end{bmatrix} = \begin{bmatrix}4 & -15\\ 0 & 2\end{bmatrix}\) Now multiply the result by A^(-1): \(\begin{bmatrix}4 & -15\\ 0 & 2\end{bmatrix}\begin{bmatrix}-3 & 5\\ -1 & 2\end{bmatrix} = \begin{bmatrix}(4*(-3) +(-15)*(-1)) & (4*5 + (-15)*2))\\ (0*(-3) + 2*(-1)) & (0*5 + 2*2))\end{bmatrix} = \begin{bmatrix}8 & 8\\ -2 & 4\end{bmatrix}\) Comparing the computed value for (ABC)^(-1) with C^(-1)B^(-1)A^(-1), we see they are equal: (ABC)^(-1) = C^(-1)B^(-1)A^(-1) = \(\begin{bmatrix}8 & 8\\ -2 & 4\end{bmatrix}\) This proves the desired result.

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Key Concepts

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

Matrix Multiplication
Matrix multiplication is an operation that takes two matrices, A and B, and produces another matrix C. This new matrix is only defined when the number of columns in the first matrix (A) is equal to the number of rows in the second matrix (B). The result matrix will have dimensions based on the rows of the first and columns of the second.
For example, to multiply a 2x3 matrix with a 3x2 matrix, the result will be a 2x2 matrix. Each element of the resulting matrix is calculated by taking the dot product of the corresponding row in the first matrix and column in the second matrix.
  • The entry in the first row and first column of the product (AB) is found by multiplying each element in the first row of A by the corresponding element in the first column of B and summing the products.
  • Similarly, for each subsequent position, these steps are repeated with the respective rows and columns.
This operation is essential for many mathematical computations in various fields like computer graphics, physics, and statistics.
Inverse of a Matrix
The inverse of a matrix is like a partner in multiplication that results in the identity matrix, similar to how dividing a number by itself yields 1. Only certain matrices, known as invertible or non-singular matrices, have inverses. In the context of a 2x2 matrix \( A \), it can be inverted using the formula:
\[ A^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
where the elements of the matrix are \(a, b, c,\) and \(d\). The term \(ad - bc\) is the determinant, which must not be zero for the inverse to exist.
  • The determinant not being zero is crucial, as this implies that the transformations a matrix represents are reversible.
  • If the determinant is zero, the matrix is called singular or non-invertible.
Inverting matrices is a fundamental concept in linear algebra, with applications ranging from solving linear system equations to finding the inverse of transformation operations in graphics.
Determinant of a Matrix
The determinant is a special number that can be calculated from a square matrix. It provides a scalar value which is often used to determine if a matrix is invertible or to find properties about the matrix's linear transformations. For a simple 2x2 matrix, its determinant \(\text{det}(A)\) is calculated as \(ad - bc\) where:
  • \( a \) and \( d \) are the elements on the main diagonal,
  • \( b \) and \( c \) are the elements off the main diagonal.

Here are some important points:
  • The determinant being zero means that the matrix has no inverse and is singular, indicating potential dependencies between its rows and columns.
  • For higher-dimensional matrices, the determinant can be more complex to calculate but serves similar purposes.
Determinants are essential in advanced topics like eigenvalues and eigenvectors, and are widely used in physics to determine orientations and solve differential equations.

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Most popular questions from this chapter

Find the inverse of the matrix, if it exists. Verify your answer. \(\left[\begin{array}{ll}4 & 2 \\ 6 & 3\end{array}\right]\)

The Campus Bookstore's inventory of books is Hardcover: textbooks, 5280 ; fiction, 1680 ; nonfiction, 2320; reference, 1890 Paperback: fiction, 2810; nonfiction, 1490; reference, \(2070 ;\) textbooks, 1940 The College Bookstore's inventory of books is Hardcover: textbooks, \(6340 ;\) fiction, 2220 ; nonfiction, \(1790 ;\) reference, 1980 Paperback: fiction, 3100; nonfiction, 1720; reference, \(2710 ;\) textbooks, 2050 a. Represent Campus's inventory as a matrix \(A\). b. Represent College's inventory as a matrix \(B\). c. The two companies decide to merge, so now write a matrix \(C\) that represents the total inventory of the newly amalgamated company.

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. \(\begin{aligned} x+2 y+z &=b_{1} \\ x+y+z &=b_{2} \\ 3 x+y+z &=b_{3} \\\ \text { where } & \text { (i) } b_{1}=7, b_{2}=4, b_{3}=2 \\ \text { and } & \text { (ii) } b_{1}=5, b_{2}=-3, b_{3}=-1 \end{aligned}\)

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. \(3 x+2 y-z=b_{1}\) \(2 x-3 y+z=b_{2}\) \(x-y-z=b_{3}\) where \(\quad\) (i) \(b_{1}=2, b_{2}=-2, b_{3}=4\) and \(\quad\) (ii) \(b_{1}=8, b_{2}=-3, b_{3}=6\)

Kaitlin and her friend Emma returned to the United States from a tour of four cities: Oslo, Stockholm, Copenhagen, and Saint Petersburg. They now wish to exchange the various foreign currencies that they have accumulated for U.S. dollars. Kaitlin has 82 Norwegian krones, 68 Swedish krones, 62 Danish krones, and 1200 Russian rubles. Emma has 64 Norwegian krones, 74 Swedish krones, 44 Danish krones, and 1600 Russian rubles. Suppose the exchange rates are U.S. \(\$ 0.1651\) for one Norwegian krone, U.S. \(\$ 0.1462\) for one Swedish krone, U.S. \(\$ 0.1811\) for one Danish krone, and U.S. \(\$ 0.0387\) for one Russian ruble. a. Write a \(2 \times 4\) matrix \(A\) giving the values of the various foreign currencies held by Kaitlin and Emma. (Note: The answer is not unique.) b. Write a column matrix \(B\) giving the exchange rate for the various currencies. c. If both Kaitlin and Emma exchange all their foreign currencies for U.S. dollars, how many dollars will each have?
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