Problems 1-16
1 If $\mathbf{A}$ is a $3\times5$,
$\mathbf{B}$ a $5\times3$, $\mathbf{C}$ a $5\times1$ and
$\mathbf{D}$ a $3\times1$ matrices with all entries being $1$ and
if matrix multiplication is allowed if and and only if number of
columns of the left handside matrix is equal to the number of rows
of the right handside matrix. In other words, only a matrix
multiplication $\mathbf{A}_{n \times k}\mathbf{B}_{k \times m}$ is
allowed, where $n$, $m$ and $k$ numbers of rows and columns.
Therefore; $\mathbf{B} \mathbf{A}$ $=$ $\mathbf{B}_{5 \times 3}$
$\cdot$ $\mathbf{A}_{3 \times 5}$ is allowed, $\mathbf{A}
\mathbf{B}$ $=$ $\mathbf{A}_{3 \times 5}$ $\cdot$ $\mathbf{B}_{5
\times 3}$ is allowed, $\mathbf{A} \mathbf{B} \mathbf{D}$ $=$
$\mathbf{A}_{3 \times 5}$ $\cdot$ $\mathbf{B}_{5 \times 3}$ $\cdot$
$\mathbf{D}_{3 \times 1}$ is allowed, $\mathbf{D} \mathbf{C}$ $=$
$\mathbf{D}_{3 \times 1}$ $\cdot$ $\mathbf{C}_{5 \times 1}$ is not
allowed because operand dimensions don’t match, and $\mathbf{A}
(\mathbf{B} + \mathbf{C})$ $=$ $\mathbf{A}_{3 \times 5}$ $\cdot$
$(\mathbf{B}_{5 \times 3}$ $+$ $\mathbf{C}_{5 \times 1})$ is not
allowed because dimensions of $\mathbf{B}$ and $\mathbf{C}$ don’t
match.
2 Given some matrices $\mathbf{A}$ and
$\mathbf{B}$ (a) the second column of product $\mathbf{A}
\mathbf{B}$ is $\mathbf{Ab}_2$, where $\mathbf{b}_2$ is third
column of $\mathbf{B}$, (b) the first row of $\mathbf{AB}$ is
$\mathbf{a}_{1m}\mathbf{B}$, (c) the entry in row $3$, column
$5$ of $\mathbf{AB}$ is $\mathbf{a}_{3m} \cdot \mathbf{b}_{n5}$ and
(d) the entry in row $1$, column $1$ of $\mathbf{CDE}$ $=$
$\mathbf{E}_{1k} \cdot \mathbf{d}_{kk} \cdot \mathbf{e}_{k1}$.
3 If
$$
\mathbf{A} = \begin{bmatrix}
1 & 5 \\
2 & 3
\end{bmatrix}
, \quad
\mathbf{B} = \begin{bmatrix}
0 & 2 \\
0 & 1
\end{bmatrix}
, \quad
\mathbf{C} = \begin{bmatrix}
3 & 1 \\
0 & 0
\end{bmatrix}
,
$$
$$
\begin{array}{c}
\mathbf{A} = \begin{bmatrix}
1 & 5 \\
2 & 3
\end{bmatrix},
\\[1.0em]
\mathbf{B} = \begin{bmatrix}
0 & 2 \\
0 & 1
\end{bmatrix},
\\[1.0em]
\mathbf{C} = \begin{bmatrix}
3 & 1 \\
0 & 0
\end{bmatrix},
\end{array}
$$
then
$$
\mathbf{AB} = \begin{bmatrix}
1(0) + 5(0) & 2(1) + 1(5) \\
2(0) + 3(0) & 2(2) + 1(3)
\end{bmatrix}
=
\begin{bmatrix}
0 & 7 \\
0 & 7
\end{bmatrix}
$$
and
$$
\mathbf{AC} = \begin{bmatrix}
1(3) + 5(0) & 1(1) + 1(0) \\
2(3) + 3(0) & 2(1) + 1(0)
\end{bmatrix}
=
\begin{bmatrix}
3 & 1 \\
6 & 2
\end{bmatrix}
$$
so
$$
\begin{array}{r c l}
\mathbf{AB} + \mathbf{AC} & = & \begin{bmatrix}
0 + 3 & 7 + 1 \\
0 + 6 & 7 + 2
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
3 & 8 \\
6 & 9
\end{bmatrix}
\end{array}
$$
and
$$
\begin{array}{r c l}
\mathbf{A}(\mathbf{B} + \mathbf{C})
& = &
\mathbf{A}\begin{bmatrix}
0 + 3 & 2 + 1 \\
0 + 0 & 1 + 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1 & 5 \\
2 & 3
\end{bmatrix}
\begin{bmatrix}
3 & 3 \\
0 & 1
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1(3) + 5(0) & 1(3) + 5(1) \\
2(3) + 3(0) & 2(3) + 3(1)
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
3 & 8 \\
6 & 9
\end{bmatrix},
\end{array}
$$
$$
\begin{array}{r c l}
& &
\mathbf{A}(\mathbf{B} + \mathbf{C})
\\[0.5em] & = &
\mathbf{A}\begin{bmatrix}
0 + 3 & 2 + 1 \\
0 + 0 & 1 + 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1 & 5 \\
2 & 3
\end{bmatrix}
\begin{bmatrix}
3 & 3 \\
0 & 1
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1(3) + 5(0) & 1(3) + 5(1) \\
2(3) + 3(0) & 2(3) + 3(1)
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
3 & 8 \\
6 & 9
\end{bmatrix},
\end{array}
$$
which means that for the given matrices $\mathbf{A}$, $\mathbf{B}$
and $\mathbf{C}$, $\mathbf{AB} + \mathbf{AC}$ $=$
$\mathbf{A}(\mathbf{B} + \mathbf{C})$.
4 Continuing with the previous
exercises matrices
$$
\begin{array}{r c l}
\mathbf{A}(\mathbf{BC})
& = &
\begin{bmatrix}
1 & 5 \\
2 & 3
\end{bmatrix}
\begin{bmatrix}
0(3) + 2(0) & 1(0) + 0(2) \\
0(3) + 1(0) & 1(0) + 0(1)
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1 & 5 \\
2 & 3
\end{bmatrix}
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
0(3) + 7(0) & 0(1) + 7(0) \\
0(3) + 7(0) & 0(1) + 7(0) \\
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
0 & 7 \\
0 & 7
\end{bmatrix}
\begin{bmatrix}
3 & 1 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
(\mathbf{AB})\mathbf{C}.
\end{array}
$$
$$
\begin{array}{r c l}
& &
\mathbf{A}(\mathbf{BC})
\\[0.5em] & = & \begin{bmatrix}
1 & 5 \\
2 & 3
\end{bmatrix}
\begin{bmatrix}
0(3) + 2(0) & 1(0) + 0(2) \\
0(3) + 1(0) & 1(0) + 0(1)
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1 & 5 \\
2 & 3
\end{bmatrix}
\begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix} 0 & 0 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
0(3) + 7(0) & 0(1) + 7(0) \\
0(3) + 7(0) & 0(1) + 7(0) \\
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
0 & 7 \\
0 & 7
\end{bmatrix}
\begin{bmatrix}
3 & 1 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
(\mathbf{AB})\mathbf{C}.
\end{array}
$$
5 When
$$
\mathbf{A} = \begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
$$
and when
$$
\mathbf{A} = \begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix},
$$
the product
$$
\begin{array}{r c l}
\mathbf{A}^3
& = &
(\mathbf{A}^2)\mathbf{A}
\\[0.5em] & = &
(\mathbf{AA})\mathbf{A}
\\[0.5em] & = &
\left(
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\right)
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\\[0.5em] & = &
\left(
\begin{bmatrix}
1(1) + b(0) & 1(b) + b(1) \\
0(1) + 1(0) & 0(b) + 1(1) \\
\end{bmatrix}
\right)
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1 & 2b \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1(1) + 2b(0) & 1(b) + 2b(1) \\
0(1) + 1(0) & 0(b) + 1(1) \\
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1 & 3b \\
0 & 1 \\
\end{bmatrix}
\end{array}
$$
$$
\begin{array}{r c l}
& &
\mathbf{A}^3
\\[0.25em] & = &
(\mathbf{A}^2)\mathbf{A}
\\[0.5em] & = &
(\mathbf{AA})\mathbf{A}
\\[0.5em] & = &
\left(
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\right)
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\\[0.5em] & = &
\left(
\begin{bmatrix}
1(1) + b(0) \hspace{-0.25em} & 1(b) + b(1) \\
0(1) + 1(0) \hspace{-0.25em} & 0(b) + 1(1) \\
\end{bmatrix}
\right)
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1 & 2b \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & b \\
0 & 1
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1(1) + 2b(0) & 1(b) + 2b(1) \\
0(1) + 1(0) & 0(b) + 1(1) \\
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
1 & 3b \\
0 & 1 \\
\end{bmatrix}
\end{array}
$$
so the pattern is
$$
\mathbf{A}^n = \begin{bmatrix}
1 & nb \\
0 & 1 \\
\end{bmatrix}
$$
and
$$
\begin{array}{r c l}
\mathbf{A}^3
& = &
(\mathbf{A}^2)\mathbf{A}
\\[0.5em] & = &
(\mathbf{AA})\mathbf{A}
\\[0.5em] & = &
\left(
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\right)
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\left(
\begin{bmatrix}
2(2) + 2(0) & 2(2) + 2(0) \\
0(2) + 0(0) & 0(2) + 0(0) \\
\end{bmatrix}
\right)
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
4 & 4 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
4(2) + 4(0) & 4(2) + 4(0) \\
0(2) + 0(0) & 0(2) + 0(0) \\
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
8 & 8 \\
0 & 0 \\
\end{bmatrix}.
\end{array}
$$
$$
\begin{array}{r c l}
& &
\mathbf{A}^3
\\[0.5em] & = &
(\mathbf{A}^2)\mathbf{A}
\\[0.5em] & = &
(\mathbf{AA})\mathbf{A}
\\[0.5em] & = &
\left(
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\right)
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\left(
\begin{bmatrix}
2(2) + 2(0) & \hspace{-0.35em} 2(2) + 2(0) \\
0(2) + 0(0) & \hspace{-0.35em} 0(2) + 0(0) \\
\end{bmatrix}
\right)
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
4 & 4 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
2 & 2 \\
0 & 0
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
4(2) + 4(0) & 4(2) + 4(0) \\
0(2) + 0(0) & 0(2) + 0(0) \\
\end{bmatrix}
\\[0.5em] & = &
\begin{bmatrix}
8 & 8 \\
0 & 0 \\
\end{bmatrix}.
\end{array}
$$
so
$$
\mathbf{A}^n = \begin{bmatrix}
2^n & 2^n \\
0 & 0
\end{bmatrix}.
$$