- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.16 Exercises
1. $\quad$ If $A = \begin{bmatrix} 1 & -4 & 2 \\ -1 & 4 & -2 \end{bmatrix},$ $B = \begin{bmatrix} 1 & 2 \\ -1 & 3 \\ 5 & -2 \end{bmatrix},$ $C = \begin{bmatrix} 2 & 2 \\ 1 & -1 \\ 1 & -3 \end{bmatrix},$ compute $B + C,$ $AB,$ $BA,$ $AC,$ $CA,$ $A(2B - 3C).$
Solution. $\quad$ In Section 13, we saw that if $A = (a_{ij})$ and $B = (b_{ij})$ are two $m \times n$ matrices, then the sum $A + B$ is defined as \begin{align*} A + B &= (a_{ij} + b_{ij}) \end{align*} In Section 15, we saw that if $A$ is any $m \times p$ matrix and $B$ is any $p \times n$ matrix, namely: \begin{align*} A &= (a_{ij})_{i, j = 1}^{m, p} \qquad \text{and} \qquad B = (b_{ij})_{i, j = 1}^{p, n} \end{align*} Then, the product $AB$ is defined to be the $m \times n$ matrix $C = (c_{ij})$ whose $ij$-entry is given by \begin{align*} c_{ij} &= \sum_{k = 1}^p a_{ik}b_{kj} \end{align*} Additionally, we saw in Theorem 2.17 that matrix multiplication is associative and distributive so long as the resulting products are meaningful. As such, we have the following: \begin{align*} B + C &= \begin{bmatrix} 1 & 2 \\ -1 & 3 \\ 5 & -2 \end{bmatrix} + \begin{bmatrix} 2 & 2 \\ 1 & -1 \\ 1 & -3 \end{bmatrix} \\ &= \begin{bmatrix} 3 & 4 \\ 0 & 2 \\ 6 & -5 \end{bmatrix} \\ \\ AB &= \begin{bmatrix} 1 & -4 & 2 \\ -1 & 4 & -2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & 3 \\ 5 & -2 \end{bmatrix} \\ &= \begin{bmatrix} \sum_{i=1}^{p = 3} a_{1k}b_{k1} & \sum_{i=1}^{p = 3} a_{1k}b_{k2} \\ \sum_{i=1}^{p = 3} a_{2k}b_{k1} & \sum_{i=1}^{p = 3} a_{2k}b_{k2} \end{bmatrix} \\ &= \begin{bmatrix} 15 & -14 \\ -15 & 14 \end{bmatrix} \\ \\ BA &= \begin{bmatrix} 1 & 2 \\ -1 & 3 \\ 5 & -2 \end{bmatrix} \begin{bmatrix} 1 & -4 & 2 \\ -1 & 4 & -2 \end{bmatrix} \\ &= \begin{bmatrix} \sum_{k=1}^{p = 2} b_{1k}a_{k1} & \sum_{k=1}^{p = 2} b_{1k}a_{k2} & \sum_{k=1}^{p = 2} b_{1k}a_{k3} \\ \sum_{k=1}^{p = 2} b_{2k}a_{k1} & \sum_{k=1}^{p = 2} b_{2k}a_{k2} & \sum_{k=1}^{p = 2} b_{2k}a_{k3} \\ \sum_{k=1}^{p = 2} b_{3k}a_{k1} & \sum_{k=1}^{p = 2} b_{3k}a_{k2} & \sum_{k=1}^{p = 2} b_{3k}a_{k3} \end{bmatrix} \\ &= \begin{bmatrix} -1 & 4 & -2 \\ -4 & 16 & -8 \\ 7 & -28 & 14 \end{bmatrix} \\ \\ AC &= \begin{bmatrix} 1 & -4 & 2 \\ -1 & 4 & -2 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ 1 & -1 \\ 1 & -3 \end{bmatrix} \\ &= \begin{bmatrix} \sum_{i=1}^{p = 3} a_{1k}c_{k1} & \sum_{i=1}^{p = 3} a_{1k}c_{k2} \\ \sum_{i=1}^{p = 3} a_{2k}c_{k1} & \sum_{i=1}^{p = 3} a_{2k}c_{k2} \end{bmatrix} \\ &= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ \\ A(2B - 3C) &= 2AB - 3AC \\ &= 2\begin{bmatrix} 15 & -14 \\ -15 & 14 \end{bmatrix} - 3\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 30 & -28 \\ -30 & 28 \end{bmatrix} \quad \blacksquare \end{align*}