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
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
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
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
so the pattern is
$$ \mathbf{A}^n = \begin{bmatrix} 1 & nb \\ 0 & 1 \\ \end{bmatrix} $$and
so
$$ \mathbf{A}^n = \begin{bmatrix} 2^n & 2^n \\ 0 & 0 \end{bmatrix}. $$6 When
$$ \mathbf{A} = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix} \quad \mathbf{B} = \begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix}, $$then
but
The terms of the previous equation are
$$ \begin{array}{r c l} \mathbf{A}^2 & = & \begin{bmatrix} 1(1) + 0(2) & 2(1) + 0(2) \\ 1(0) + 0(0) & 2(0) + 0(0) \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}, \\[0.0em] & & \\[0.0em] 2\mathbf{AB} & = & 2 \begin{bmatrix} 1(1) + 3(2) & 0(1) + 0(2) \\ 1(0) + 3(0) & 0(0) + 0(0) \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} 2(7) & 2(0) \\ 2(0) & 0(0) \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} 14 & 0 \\ 0 & 0 \end{bmatrix}, \\[0.0em] & & \\[0.0em] \mathbf{B}^2 & = & \begin{bmatrix} 1(1) + 3(0) & 0(1) + 0(0) \\ 1(3) + 3(0) & 0(3) + 0(0) \end{bmatrix} \\[0.5em] & = & \begin{bmatrix} 1 & 0 \\ 3 & 0 \end{bmatrix}, \end{array} $$which adds to
$$ \begin{array}{r c l} \begin{bmatrix} 1 + 14 + 1 & 2 + 0 + 0 \\ 0 + 0 + 3 & 0 + 0 + 0 \end{bmatrix} & = & \begin{bmatrix} 16 & 2 \\ 3 & 0 \end{bmatrix}, \end{array} $$so $(\mathbf{A} + \mathbf{B})^2$ $\neq$ $\mathbf{A}^2$ $+$ $2\mathbf{AB}$ $+$ $\mathbf{B}^2$ for the given matrices. The correct rule for $(\mathbf{A}$ $+$ $\mathbf{B})$$(\mathbf{A}$ $+$ $\mathbf{B})$ $=$ $\mathbf{AA}$ $+$ $\mathbf{AB}$ $+$ $\mathbf{BA}$ $+$ $\mathbf{BB}$ $=$ $\mathbf{A}^2$ $+$ $\mathbf{AB}$ $+$ $\mathbf{BA}$ $+$ $\mathbf{B}^2$, and that’s it, because matrix multiplication operation does not (always) commute.
7 (a) When
$$ \mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} $$and
$$ \mathbf{B} = \begin{bmatrix} b & 0 & b \\ b & 1 & b \\ b & 2 & b \\ \end{bmatrix}, $$then
$$ \mathbf{AB} = \begin{bmatrix} a_{11}b + a_{12}b + a_{13}b & 0a_{11} + 1a_{12} + 2a_{13} & a_{11}b + a_{12}b + a_{13}b \\ a_{21}b + a_{22}b + a_{23}b & 0a_{21} + 1a_{22} + 2a_{23} & a_{21}b + a_{22}b + a_{23}b \\ a_{31}b + a_{32}b + a_{33}b & 0a_{31} + 1a_{32} + 2a_{33} & a_{31}b + a_{32}b + a_{33}b \\ \end{bmatrix}, $$from where it can be seen that columns $(\mathbf{AB})_{n1}$ $=$ $(\mathbf{AB})_{n3}$.
(b) If $\mathbf{B}$, i.e. rows $1$ and $3$ of $\mathbf{B}$ are same, is rotated $90$ degrees counterclockwise and $\mathbf{A}$ remains the same, then
$$ \mathbf{AB} = \begin{bmatrix} a_{11}b + a_{12}0 + a_{13}b & a_{11}b + 1a_{12} + a_{13}b & a_{11}b + a_{12}2 + a_{13}b \\ a_{21}b + a_{22}0 + a_{23}b & a_{21}b + 1a_{22} + a_{23}b & a_{21}b + a_{22}2 + a_{23}b \\ a_{31}b + a_{32}0 + a_{33}b & a_{31}b + 1a_{32} + a_{33}b & a_{31}b + a_{32}2 + a_{33}b \\ \end{bmatrix}, $$from where it can be seen that columns $(\mathbf{AB})_{n1}$ $\neq$ $(\mathbf{AB})_{n3}$.
(c) By the extension of the previous matrix product, the columns $1$ and $3$ of $\mathbf{ABC}$ are note same either.
(d) If $\mathbf{A}$ is as above and
$$ \mathbf{B} = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \\ \end{bmatrix} $$then
$$ \begin{array}{r c l} (\mathbf{AB})^2 \hspace{-0.8em} & = & \hspace{-0.8em} \begin{bmatrix} a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} & \hspace{-0.8em} a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32} & \hspace{-0.8em} a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33} \hspace{-0.8em} \\ a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} & \hspace{-0.8em} a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32} & \hspace{-0.8em} a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33} \hspace{-0.8em} \\ a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31} & \hspace{-0.8em} a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32} & \hspace{-0.8em} a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33} \hspace{-0.8em} \\ \end{bmatrix}^2. \end{array} $$Computing the square as $(\mathbf{AB})(\mathbf{ab})_{i}$, where tha latter term is a column of $(\mathbf{AB})$ and $i$ $=$ $1,2,3$ shows that:
$$ \begin{array}{r c l} (\mathbf{AB})(\mathbf{ab})_1 & = & \left[ \begin{array}{l} b_{11}(a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31}) \\ + b_{21}(a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32}) \\ + b_{31} (a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33}) \\[0.5em] b_{11}(a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31}) \\ + b_{21}(a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32}) \\ + b_{31} (a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33}) \\[0.5em] b_{11}(a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31}) \\ + b_{21}(a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32}) \\ + b_{31} (a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33}) \\[0.5em] \end{array} \right], \\[1em] (\mathbf{AB})(\mathbf{ab})_2 & = & \left[ \begin{array}{l} b_{12}(a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31}) \\ + b_{22}(a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32}) \\ + b_{32} (a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33}) \\[0.5em] b_{12}(a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31}) \\ + b_{22}(a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32}) \\ + b_{32} (a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33}) \\[0.5em] b_{12}(a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31}) \\ + b_{22}(a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32}) \\ + b_{32} (a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33}) \\[0.5em] \end{array} \right], \\[1em] (\mathbf{AB})(\mathbf{ab})_3 & = & \left[ \begin{array}{l} b_{13}(a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31}) \\ + b_{23}(a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32}) \\ + b_{33} (a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33}) \\[0.5em] b_{13}(a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31}) \\ + b_{23}(a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32}) \\ + b_{33} (a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33}) \\[0.5em] b_{13}(a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31}) \\ + b_{23}(a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32}) \\ + b_{33} (a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33}) \\[0.5em] \end{array} \right]. \end{array} $$and omputing the squares and the product shows that:
$$ \begin{array}{l c l} \mathbf{A}^2 \hspace{-0.5em} & = & \hspace{-0.5em} \begin{bmatrix} a_{11}a_{11} + a_{21}a_{12} + a_{31}a_{13} & a_{12}a_{11} + a_{22}a_{12} + a_{32}a_{13} & a_{13}a_{11} + a_{23}a_{12} + a_{33}a_{13} \hspace{-1.0em} \\ a_{11}a_{21} + a_{21}a_{22} + a_{31}a_{23} & a_{12}a_{21} + a_{22}a_{22} + a_{32}a_{23} & a_{13}a_{21} + a_{23}a_{22} + a_{33}a_{23} \hspace{-1.0em} \\ a_{11}a_{31} + a_{21}a_{32} + a_{31}a_{33} & a_{12}a_{31} + a_{22}a_{32} + a_{32}a_{33} & a_{13}a_{31} + a_{23}a_{32} + a_{33}a_{33} \hspace{-1.0em} \\ \end{bmatrix}, \\[0.5em] \mathbf{B}^2 \hspace{-0.5em} & = & \hspace{-0.5em} \begin{bmatrix} b_{11}b_{11} + b_{21}b_{12} + b_{31}b_{13} & b_{12}b_{11} + b_{22}b_{12} + b_{32}b_{13} & b_{13}b_{11} + b_{23}b_{12} + b_{33}b_{13} \hspace{-1.0em} \\ b_{11}b_{21} + b_{21}b_{22} + b_{31}b_{23} & b_{12}b_{21} + b_{22}b_{22} + b_{32}b_{23} & b_{13}b_{21} + b_{23}b_{22} + b_{33}b_{23} \hspace{-1.0em} \\ b_{11}b_{31} + b_{21}b_{32} + b_{31}b_{33} & b_{12}b_{31} + b_{22}b_{32} + b_{32}b_{33} & b_{13}b_{31} + b_{23}b_{32} + b_{33}b_{33} \hspace{-1.0em} \\ \end{bmatrix}. \end{array} $$From these results, it can be seen that the first element of $\mathbf{A}^2\mathbf{B}^2$ shows that it’s not equal to the first element of $(\mathbf{AB})(\mathbf{ab})_1$, so $\mathbf{AB}^2$ $\neq$ $\mathbf{A}^2\mathbf{B}^2$.