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1.3 Matrices and Matrix Operations: Class Notes

Definitions

#6

Use the following matrices to compute the indicated expression if it is defined.

$A=\left[\begin{array}{cc}3& 0\\ -1& 2\\ 1& 1\end{array}\right]$, $B=\left[\begin{array}{cc}4& -1\\ 0& 2\end{array}\right]$, $C=\left[\begin{array}{ccc}1& 4& 2\\ 3& 1& 5\end{array}\right]$, $D=\left[\begin{array}{ccc}1& 5& 2\\ -1& 0& 1\\ 3& 2& 4\end{array}\right]$, $E=\left[\begin{array}{ccc}6& 1& 3\\ -1& 1& 2\\ 4& 1& 3\end{array}\right]$

a. $(2D^T-E)A$
b. $(4B)C+2B$
c. $(-AC{)}^T+5D^T$
d. ${(B{A}^T-2C)}^T$
e. $B^T(C{C}^T-A^{T}A)$
f. $D^T{E}^T-(ED{)}^T$

Definition

Solution to #6

a. The rows of $D$ become the columns of $D^T$ (row 1 is shown in blue, row 2 in green, row 3 in pink):

$$D^T=\left[\begin{array}{ccc}1& -1& 3\\ 5& 0& 2\\ 2& 1& 4\end{array}\right]$$

$2D^T$ is scalar multiplication: multiply every entry by 2.

$$2D^T=\left[\begin{array}{ccc}2& -2& 6\\ 10& 0& 4\\ 4& 2& 8\end{array}\right]$$

Subtraction (and addition) are performed using corresponding entries:

$$2D^T-E=\left[\begin{array}{ccc}2& -2& 6\\ 10& 0& 4\\ 4& 2& 8\end{array}\right]-\left[\begin{array}{ccc}6& 1& 3\\ -1& 1& 2\\ 4& 1& 3\end{array}\right]=\left[\begin{array}{ccc}-4& -3& 3\\ 11& -1& 2\\ 0& 1& 5\end{array}\right]$$

To decide whether a matrix product is defined, use the sizes of the matrices being multiplied. $(2D^T-E)$ is a $3\times 3$ matrix and $A$ is a $3\times 2$ matrix. The inner sizes (3 and 3) have to match, and the outer sizes give the size of the result, $3\times 2$.

$$\underset{3\times 3}{(2D^T-E)}\underset{3\times 2}{A}$$

Each entry of the product is a row of the first matrix times a column of the second: multiply corresponding entries and add. The entry in row $i$, column $j$ of the product is row $i$ times column $j$.

$$(2D^T-E)A=\left[\begin{array}{ccc}-4& -3& 3\\ 11& -1& 2\\ 0& 1& 5\end{array}\right]\left[\begin{array}{cc}3& 0\\ -1& 2\\ 1& 1\end{array}\right]$$

$$=\left[\begin{array}{cc}-12+3+3& 0-6+3\\ 33+1+2& 0-2+2\\ 0-1+5& 0+2+5\end{array}\right]=\left[\begin{array}{cc}-6& -3\\ 36& 0\\ 4& 7\end{array}\right]$$

b. $4B$ is $2\times 2$ and $C$ is $2\times 3$, so the product $(4B)C$ exists and is $2\times 3$. But $2B$ is $2\times 2$. Since addition and subtraction are performed on corresponding entries, they can be done only with matrices of the same size. Thus, this sum does not exist.

Relabel the matrices from #6a: let

$$A=\left[\begin{array}{ccc}-4& -3& 3\\ 11& -1& 2\\ 0& 1& 5\end{array}\right],B=\left[\begin{array}{cc}3& 0\\ -1& 2\\ 1& 1\end{array}\right]$$

We found the product $AB$ using the row-column rule.

The row vectors of $A$ (or row matrices) are

$${\overrightarrow{r}}_1=\left[\begin{array}{ccc}-4& -3& 3\end{array}\right],{\overrightarrow{r}}_2=\left[\begin{array}{ccc}11& -1& 2\end{array}\right],{\overrightarrow{r}}_3=\left[\begin{array}{ccc}0& 1& 5\end{array}\right]$$

The column vectors of $B$ (or column matrices) are

$${\overrightarrow{c}}_1=\left[\begin{array}{c}3\\ -1\\ 1\end{array}\right],{\overrightarrow{c}}_2=\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]$$

$AB$ can also be found using the column method:

$$AB=\left[\begin{array}{cc}\hfill {\displaystyle A{\overrightarrow{c}}_1}\hfill & \hfill {\displaystyle A{\overrightarrow{c}}_2}\hfill \end{array}\right]$$

This is a partitioned matrix. Its first column is

$$\underset{3\times 33\times 1}{A{\overrightarrow{c}}_1}=\left[\begin{array}{ccc}-4& -3& 3\\ 11& -1& 2\\ 0& 1& 5\end{array}\right]\left[\begin{array}{c}3\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}-12+3+3\\ 33+1+2\\ 0-1+5\end{array}\right]=\left[\begin{array}{c}-6\\ 36\\ 4\end{array}\right](3\times 1)$$

and $A{\overrightarrow{c}}_2$ likewise gives column 2 of $AB$.

We can also use the row method:

$$AB=\left[\begin{array}{c}{\overrightarrow{r}}_{1}B\\ {\overrightarrow{r}}_{2}B\\ {\overrightarrow{r}}_{3}B\end{array}\right]$$

For instance,

$${\overrightarrow{r}}_{1}B=\left[\begin{array}{ccc}-4& -3& 3\end{array}\right]\left[\begin{array}{cc}3& 0\\ -1& 2\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}-12+3+3& 0-6+3\end{array}\right]=\left[\begin{array}{cc}-6& -3\end{array}\right]$$

which is the first row of $AB$.

#17

Use the column-row expansion of $AB$ to express this product as a sum of matrix products.

$A=\left[\begin{array}{cc}4& -3\\ 2& -1\end{array}\right]$, $B=\left[\begin{array}{ccc}0& 1& 2\\ -2& 3& 1\end{array}\right]$

Definition

Solution to #17

Multiply column 1 of $A$ by row 1 of $B$, and column 2 of $A$ by row 2 of $B$:

$$AB=\left[\begin{array}{c}4\\ 2\end{array}\right]\left[\begin{array}{ccc}0& 1& 2\end{array}\right]+\left[\begin{array}{c}-3\\ -1\end{array}\right]\left[\begin{array}{ccc}-2& 3& 1\end{array}\right]$$

Each term is $(2\times 1)(1\times 3)$: the inner sizes match, so each product is $2\times 3$.

$$=\left[\begin{array}{ccc}0& 4& 8\\ 0& 2& 4\end{array}\right]+\left[\begin{array}{ccc}6& -9& -3\\ 2& -3& -1\end{array}\right]=\left[\begin{array}{ccc}6& -5& 5\\ 2& -1& 3\end{array}\right]$$

The linear system

$$\begin{array}{cc}\hfill {\displaystyle a_{11}{x}_1+a_{12}{x}_2+\dots +a_{1n}{x}_n}& {\displaystyle =b_1}\hfill \\ \hfill {\displaystyle a_{21}{x}_1+a_{22}{x}_2+\dots +a_{2n}{x}_n}& {\displaystyle =b_2}\hfill \\ \hfill {\displaystyle ⋮}& {\displaystyle ⋮}\hfill \\ \hfill {\displaystyle a_{m1}{x}_1+a_{m2}{x}_2+\dots +a_{mn}{x}_n}& {\displaystyle =b_m}\hfill \end{array}$$

can be expressed using matrix multiplication.

Definition

The coefficients $a_{ij}$ of the system are exactly the entries of $A$. This is a matrix equation.

#13

a. Express the matrix equation as a system of linear equations.

$$\left[\begin{array}{ccc}5& 6& -7\\ -1& -2& 3\\ 0& 4& -1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}2\\ 0\\ 3\end{array}\right]$$

Solution

Multiply out the left-hand side; each row of the coefficient matrix times the column of unknowns gives one entry:

$$\left[\begin{array}{c}5x_1+6x_2-7x_3\\ -x_1-2x_2+3x_3\\ 4x_2-x_3\end{array}\right]=\left[\begin{array}{c}2\\ 0\\ 3\end{array}\right]$$

Two matrices are equal exactly when their corresponding entries are equal (see the definition below), so each pair of matching entries gives one equation:

Definition

Definition

In #13, we have

$$A=\left[\begin{array}{ccc}5& 6& -7\\ -1& -2& 3\\ 0& 4& -1\end{array}\right],\overrightarrow{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right],\overrightarrow{b}=\left[\begin{array}{c}2\\ 0\\ 3\end{array}\right]$$

Note that the left-hand side splits by columns into a linear combination of the columns of $A$, with the unknowns as coefficients:

$$\left[\begin{array}{c}5x_1+6x_2-7x_3\\ -x_1-2x_2+3x_3\\ 4x_2-x_3\end{array}\right]=\left[\begin{array}{c}5x_1\\ -x_1\\ 0\end{array}\right]+\left[\begin{array}{c}6x_2\\ -2x_2\\ 4x_2\end{array}\right]+\left[\begin{array}{c}-7x_3\\ 3x_3\\ -x_3\end{array}\right]$$

$$=\left[\begin{array}{c}5\\ -1\\ 0\end{array}\right]x_1+\left[\begin{array}{c}6\\ -2\\ 4\end{array}\right]x_2+\left[\begin{array}{c}-7\\ 3\\ -1\end{array}\right]x_3$$

#22

Example 6 of section 1.2 presents the linear system

$$\begin{array}{ccccccc}\hfill {\displaystyle x_1}& \hfill {\displaystyle +3x_2}& \hfill {\displaystyle -2x_3}& \hfill {\displaystyle }& \hfill {\displaystyle +2x_5}& \hfill {\displaystyle }& {\displaystyle =0}\hfill \\ \hfill {\displaystyle 2x_1}& \hfill {\displaystyle +6x_2}& \hfill {\displaystyle -5x_3}& \hfill {\displaystyle -2x_4}& \hfill {\displaystyle +4x_5}& \hfill {\displaystyle -3x_6}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle }& \hfill {\displaystyle }& \hfill {\displaystyle 5x_3}& \hfill {\displaystyle +10x_4}& \hfill {\displaystyle }& \hfill {\displaystyle +15x_6}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle 2x_1}& \hfill {\displaystyle +6x_2}& \hfill {\displaystyle }& \hfill {\displaystyle +8x_4}& \hfill {\displaystyle +4x_5}& \hfill {\displaystyle +18x_6}& {\displaystyle =0}\hfill \end{array}$$

and the reduced row echelon form of its associated augmented matrix as

$$\left[\begin{array}{ccccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Express the solution as a linear combination of column vectors that contain only numerical entries.

Solution

The columns go with $x_1,\dots ,x_6$. The leading 1s are in the $x_1$-, $x_3$-, and $x_6$-columns, so solve each nonzero row for its leading variable:

$$\begin{array}{cc}\hfill {\displaystyle x_1}& {\displaystyle =-3x_2-4x_4-2x_5}\hfill \\ \hfill {\displaystyle x_3}& {\displaystyle =-2x_4}\hfill \\ \hfill {\displaystyle x_6}& {\displaystyle =0}\hfill \end{array}$$

The free variables are $x_2$, $x_4$, and $x_5$. Let $r=x_2$, $s=x_4$, $t=x_5$. Then

$$\overrightarrow{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\end{array}\right]=\left[\begin{array}{c}-3r-4s-2t\\ r\\ -2s\\ s\\ t\\ 0\end{array}\right]$$

Theorem 1.3.1

See the left-hand side of #13 split into columns, just before #22, and the answer to #22.

Definition

Definition