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Chapter 2: 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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