# Section 5.7

Recall the generalized matrix

$$
A=\left[\begin{matrix}a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\
a_{31} & a_{32} & a_{33} & \cdots & a_{3n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn}
\end{matrix}\right]
$$

The number of **rows** a matrix has by the number of **columns** a matrix has will be called the **dimension** of the matrix. For example, $A$ is a $m\times n$ matrix.

:::{prf:definition} $A=B$
:label: matrixEquality

Two matrices are equal if they have the same **dimension** and if corresponding elements, position by position, are equal.

:::

$$
\left[\begin{matrix}
    2 & 1\\
    3 & -5
\end{matrix}\right]=
\left[\begin{matrix}
    2 & 1\\
    3 & -5
\end{matrix}\right]
$$

but

$$
\left[\begin{matrix}2 & 1\\
3 & -5
\end{matrix}\right]\ne\left[\begin{matrix}1 & 2\\
-5 & 3
\end{matrix}\right]
$$

::::{prf:example}
:label: equalMatrix1

Find the values of the variables for which each statement is true,
if possible.

1. $\left[\begin{matrix}2 & 1\\
p & q
\end{matrix}\right]=\left[\begin{matrix}x & y\\
-1 & 0
\end{matrix}\right]$
2. $\left[\begin{matrix}x\\
y
\end{matrix}\right]=\left[\begin{matrix}1\\
4\\
0
\end{matrix}\right]$

:::{dropdown} Solution:

For (1) By definition we have

$$
\left[\begin{matrix}2 & 1\\
p & q
\end{matrix}\right]=\left[\begin{matrix}x & y\\
-1 & 0
\end{matrix}\right]\to\begin{matrix}2=x & 1=y\\
p=-1 & q=0
\end{matrix}
$$

That is,  $x=2$, $y=1$, $p=-1$, and $q=0$.

---

For (2), by definition since the dimensions of the two matrices do not match the two matrices cannot equal.

:::
::::

:::{prf:definition} Addition of Matrices
:label: addMatrix

To add matrices of the same dimension, add corresponding elements. **Only matrices of the same dimension can be added.

:::

::::{prf:example}
:label: addMatrixExam1

Add

$$
\left[\begin{matrix}5 & -6\\
8 & 9
\end{matrix}\right]+\left[\begin{matrix}-4 & 6\\
8 & -3
\end{matrix}\right]
$$

:::{dropdown} Solution:

\begin{align*}
\left[\begin{matrix}5 & -6\\
8 & 9
\end{matrix}\right]+\left[\begin{matrix}-4 & 6\\
8 & -3
\end{matrix}\right] & =\left[\begin{matrix}5+(-4) & -6+6\\
8+8 & 9+(-3)
\end{matrix}\right]\\
 & =\left[\begin{matrix}1 & 0\\
16 & 6
\end{matrix}\right]
\end{align*}

:::
::::

:::{prf:definition} Scalar and Scalar Multiplication
:label: scalarSalarMult

In the context of matrices, a real number is called a **scalar** to distinguish it from a matrix.

The product of a scalar $k$ and a matrix $X$ is the matrix $kX$, each of whose elements is $k$ times the corresponding element of $X$.
:::

For example

$$
k\left[\begin{matrix}a & b\\
c & d
\end{matrix}\right]=\left[\begin{matrix}ka & kb\\
kc & kd
\end{matrix}\right]
$$

Also,

$$
(-1)\left[\begin{matrix}a & b\\
c & d
\end{matrix}\right]=-\left[\begin{matrix}a & b\\
c & d
\end{matrix}\right]=\left[\begin{matrix}-a & -b\\
-c & -d
\end{matrix}\right]
$$

:::{prf:definition} Properties of Scalar Multiplication
:label: scalarMultiplicationProperty

Let $A$ and $B$ be matrices of the same dimension, and let $c$
and $d$ be scalars. Then these properties hold true.

1. $(c+d)A=cA+dA$
2. $(cA)d=(cd)A$
3. $c(A+B)=cA+cB$
4. $(cd)A=c(dA)$

:::

:::{prf:example} Subtraction of Matrices
:label: subMatrix

If $A$ and $B$ are two matrices of the same dimension, then the
following holds true

$$
A-B=A+(-B).
$$

:::

## Multiplication

For multiplication the process is not intuitive. That is,

$$
\left[\begin{matrix}a & b\\
c & d
\end{matrix}\right]\cdot\left[\begin{matrix}w & x\\
y & z
\end{matrix}\right]\ne\left[\begin{matrix}aw & bx\\
cy & dz
\end{matrix}\right]
$$

Instead, multiplication between two matrices would be performed in
the following way

\begin{align*}
\left[\begin{matrix}3 & 5\end{matrix}\right]\cdot\left[\begin{matrix}-2\\
4
\end{matrix}\right] & =\left[\begin{matrix}(3)(-2)+(5)(4)\end{matrix}\right]\\
 & =\left[\begin{matrix}14\end{matrix}\right]
\end{align*}

If $A=\left[\begin{matrix}3 & 5\end{matrix}\right]$ (which is $1\times2$) and $B=\left[\begin{matrix}-2\\ 4\end{matrix}\right]$ (which is $2\times1$) we see that in order to perform the calculation the number of columns of the first matrix **must** match the number of rows of the second matrix.

:::{prf:definition} Matrix Multiplication
:label: multMatrix

The number of columns of an $m\times n$ matrix $A$ is the same as the number of rows of and $n\times p$ matrix $B$ (i.e., both $n$). The element $c_{ij}$ of the product matrix $C=AB$ is found as follows

$$
c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+\cdots+a_{in}b_{nj}
$$

Matrix $AB$ will be an $m\times p$ matrix.

:::

:::{prf:theorem}
:label: noncommutativeMatrixMult

In general, if $A$ and $B$ are matrices, then 

$$
AB\ne BA.
$$

That is, Matrix Multiplication is not commutative.
:::

:::{prf:theorem} Matrix Multiplication Properties
:label: matrixMultProp

If $A$, $B$, and $C$ are matrices such that all the following products
and sums exist, then these properties hold true

1. $(AB)C=A(BC)$
2. $A(B+C)=AB+AC$
3. $(B+C)A=BA+CA$ (pay attention to the order of the multiplication since matrix multiplication is not commutative.)

:::

:::{prf:example}
:label: matrixMultPropExam1

Suppose $A$ is a $3\times 2$ matrix, while $B$ is a $2\times 4$
matrix.

The sum $A+B$ is impossible since the dimensions of the two matrices do not equal.

The product of $A\cdot B$ does exist since the number of columns for matrix $A$ (2) matches the number of rows for matrix $B$ (2). The resulting matrix (after the product) will be a $3\times 4$ matrix.

The product of $B\cdot A$ does not exists since the number of columns for matrix $B$ (4) does not equal the number of rows for matrix $A$ (3).
:::

::::{prf:example}
:label: matrixMultExam1

Let $A=\left[\begin{matrix}1 & -3\\
7 & 2
\end{matrix}\right]$ and $B=\left[\begin{matrix}1 & 0 & -1 & 2\\
3 & 1 & 4 & -1
\end{matrix}\right]$. 

Find each product, if possible.

Multiply $A\cdot B$.

:::{dropdown} Solution:

Since $A$ has dimension $2\times n=2\times2$ where $n=2$ and $B$ has dimension $n\times4$ where $n=2$ the two matrices can be multiplied. The resulting matrix should be a $2\times4$ matrix.

\begin{align*}
\left[\begin{matrix}1 & -3\\
7 & 2
\end{matrix}\right]\cdot\left[\begin{matrix}1 & 0 & -1 & 2\\
3 & 1 & 4 & -1
\end{matrix}\right] & =\left[\begin{matrix}1*1+(-3)*3 & 1*(0)+(-3)*1 & 1*(-1)+(-3)*4 & 1*(2)+(-3)*(-1)\\
7*(1)+2*(3) & 7*(0)+2*(1) & 7*(-1)+2*(4) & 7*(2)+2*(-1)
\end{matrix}\right]\\
 & =\left[\begin{matrix}-8 & -3 & -13 & 5\\
13 & 2 & 1 & 12
\end{matrix}\right]
\end{align*}

:::

Multiply $B\cdot A$.

:::{dropdown} Solution:

Since the number of columns and rows do not match the product is not possible.

:::
::::

::::{prf:example}
:label: matrixMultExam2

Let $A=\left[\begin{matrix}2 & -5\\
-6 & 1
\end{matrix}\right]$ and $B=\left[\begin{matrix}3 & -4\\
2 & 3
\end{matrix}\right]$.

Multiply $A\cdot B$.

:::{dropdown} Solution:

The product $A\cdot B$ is possible since the number of columns of the first matrix matches the number of rows of the second. The resulting matrix will be a $2\times2$ matrix.

\begin{align*}
\left[\begin{matrix}2 & -5\\
-6 & 1
\end{matrix}\right]\cdot\left[\begin{matrix}3 & -4\\
2 & 3
\end{matrix}\right] & =\left[\begin{matrix}(2)(3)+(-5)(2) & (2)(-4)+(-5)(3)\\
(-6)(3)+(1)(2) & (-6)(-4)+(1)(3)
\end{matrix}\right]\\
 & =\left[\begin{matrix}-4 & -23\\
-16 & 27
\end{matrix}\right]
\end{align*}

:::

Multiply $B\cdot A$

:::{dropdown} Solution:

The product of $B\cdot A$ is possible since the number of columns of the first matrix matches the number of rows of the second. The resulting matrix will be a $2\times2$ matrix.

\begin{align*}
\left[\begin{matrix}3 & -4\\
2 & 3
\end{matrix}\right]\cdot\left[\begin{matrix}2 & -5\\
-6 & 1
\end{matrix}\right] & =\left[\begin{matrix}(3)(2)+(-4)(-6) & (3)(-5)+(-4)(1)\\
(2)(2)+(3)(-6) & (2)(-5)+(3)(1)
\end{matrix}\right]\\
 & =\left[\begin{matrix}30 & -19\\
-14 & -7
\end{matrix}\right]
\end{align*}

:::
::::

Notice in the last example $A\cdot B\ne B\cdot A$.