- 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
2. $\quad$ Let $A = \begin{bmatrix} 0 & 1 \\ 0 & 2 \end{bmatrix}.$ Find all 2 x 2 matrices $B$ such that (a) $AB = O$; (b) $BA = O.$
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.
Let $B$ be a $2 \times 2$ matrix defined as follows:
\begin{align*}
B &= \begin{bmatrix}
b_{11} & b_{12}
\\
b_{21} & b_{22}
\end{bmatrix}
\end{align*}
$\quad$ For $AB$ to be zero, $B$ must be such that
\begin{align*}
\begin{bmatrix}
\sum_{k=1}^{p=2} a_{1k}b_{k1} & \sum_{k=1}^{p=2} a_{1k}b_{k2}
\\
\sum_{k=1}^{p=2} a_{2k}b_{k1} & \sum_{k=1}^{p=2} a_{2k}b_{k2}
\end{bmatrix} &= \begin{bmatrix}
0(b_{11}) + 1(b_{21}) & 0(b_{12}) + 1(b_{22})
\\
0(b_{11}) + 2(b_{21}) & 0(b_{12}) + 2(b_{22})
\end{bmatrix}
\\
&= \begin{bmatrix}
1(b_{21}) & 1(b_{22})
\\
2(b_{21}) & 2(b_{22})
\end{bmatrix}
\\
&= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\end{align*}
As we can see, this is satisfied by $B = \begin{bmatrix}b_{11} & b_{12} \\ 0 & 0\end{bmatrix},$ where $b_{11}$ and $b_{12}$ are arbitrary scalars.
$\quad$ For $BA$ to be zero, $B$ must be such that
\begin{align*}
\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_{2k}b_{k1} & \sum_{k=1}^{p=2} b_{2k}a_{k2}
\end{bmatrix} &= \begin{bmatrix}
b_{11}(0) + b_{12}(0) & b_{11}(1) + b_{12}(2)
\\
b_{21}(0) + b_{22}(0) & b_{21}(1) + b_{22}(2)
\end{bmatrix}
\\
&= \begin{bmatrix}
0 & b_{11}(1) + b_{12}(2)
\\
0 & b_{21}(1) + b_{22}(2)
\end{bmatrix}
\\
&= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\end{align*}
This relation is satisfied when $b_{11} = -2b_{12}$ and $b_{21} = -2b_{22}$ for arbitrary $b_{12}$ and $b_{22}. \quad \blacksquare$