- 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
3. $\quad$ In each case find $a,$ $b,$ $c,$ $d$ to satisfy the given equation.
$\quad$ (a) $\quad \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} 1 \\ 9 \\ 6 \\ 5 \end{bmatrix};$
$\quad$ (b) $\quad \begin{bmatrix} a & b & c \\ 1 & 4 & 9 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 6 & 6 \\ 1 & 9 & 8 & 4 \end{bmatrix}.$
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.
(a) $\quad$ If we set $A = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix},$ the product $AB$ is a $4 \times 1$ array given by: \begin{align*} \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} &= \begin{bmatrix} \sum_{k = 1}^{p = 4} a_{1k}b_{k1} \\ \sum_{k = 1}^{p = 4} a_{2k}b_{k1} \\ \sum_{k = 1}^{p = 4} a_{3k}b_{k1} \\ \sum_{k = 1}^{p = 4} a_{4k}b_{k1} \end{bmatrix} \\ &= \begin{bmatrix} 0(a) + 0(b) + 1(c) + 0(d) \\ 1(a) + 0(b) + 0(c) + 0(d) \\ 0(a) + 1(b) + 0(c) + 0(d) \\ 0(a) + 0(b) + 0(c) + 1(d) \end{bmatrix} \\ &= \begin{bmatrix} c \\ a \\ b \\ d \end{bmatrix} = \begin{bmatrix} 1 \\ 9 \\ 6 \\ 5 \end{bmatrix} \end{align*} As such, this relation is satisfied when $a = 9, b = 6, c = 1,$ and $d = 5.$
(b) $\quad$ If we set $A = \begin{bmatrix} a & b & c & d \\ 1 & 4 & 9 & 2\end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix},$ the product $AB$ is a $2 \times 4$ array given by: \begin{align*} \begin{bmatrix} a & b & c & d \\ 1 & 4 & 9 & 2\end{bmatrix} \begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} &= \begin{bmatrix} \sum_{k = 1}^{p = 4} a_{1k}b_{k1} & \sum_{k = 1}^{p = 4} a_{1k}b_{k1} & \sum_{k = 1}^{p = 4} a_{1k}b_{k1} & \sum_{k = 1}^{p = 4} a_{1k}b_{k1} \\ \sum_{k = 1}^{p = 4} a_{1k}b_{k1} & \sum_{k = 1}^{p = 4} a_{1k}b_{k1} & \sum_{k = 1}^{p = 4} a_{1k}b_{k1} & \sum_{k = 1}^{p = 4} a_{1k}b_{k1} \end{bmatrix} \\ &= \begin{bmatrix} a & c & 2a + b + d & b \\ 1 & 9 & 8 & 4 \\ \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 & 6 & 6 \\ 1 & 9 & 8 & 4 \end{bmatrix} \end{align*} As we can see, this relation is satisfied when $a = 1, b = 6, c = 0,$ and $d = -2. \quad \blacksquare$