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1.6 More on Linear Systems and Invertible Matrices: Class Notes

Theorem 1.6.1

Proof

Since we know a system can have no solution or one solution, we’ll establish that if there is more than one solution, there are infinitely many.

If $A\overrightarrow{x}=\overrightarrow{b}$ has solutions ${\overrightarrow{x}}_1$ and ${\overrightarrow{x}}_2$, then

$$A{\overrightarrow{x}}_2-A{\overrightarrow{x}}_1=\overrightarrow{b}-\overrightarrow{b}=\overrightarrow{0}.$$

Let ${\overrightarrow{x}}_0={\overrightarrow{x}}_2-{\overrightarrow{x}}_1$. Then $A{\overrightarrow{x}}_0=A({\overrightarrow{x}}_2-{\overrightarrow{x}}_1)=\overrightarrow{0}$. If $k$ is any scalar, then

$$A({\overrightarrow{x}}_1+k{\overrightarrow{x}}_0)=A{\overrightarrow{x}}_1+A(k{\overrightarrow{x}}_0)=A{\overrightarrow{x}}_1+kA{\overrightarrow{x}}_0=\overrightarrow{b}+\overrightarrow{0}=\overrightarrow{b}.$$

So ${\overrightarrow{x}}_1+k{\overrightarrow{x}}_0$ is a solution for every $k$. Infinitely many choices for $k$ give infinitely many solutions.

Theorem 1.6.2

There is exactly one solution because $A^{-1}$ is unique.

#4

Solve the system by inverting the coefficient matrix and using Theorem 1.6.2.

$$\begin{array}{cc}\hfill {\displaystyle 5x_1+3x_2+2x_3}& {\displaystyle =4}\hfill \\ \hfill {\displaystyle 3x_1+3x_2+2x_3}& {\displaystyle =2}\hfill \\ \hfill {\displaystyle x_2+x_3}& {\displaystyle =5}\hfill \end{array}$$

Solution

Use the inversion algorithm on the coefficient matrix $A$:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_2\to 3R_1-5R_2$:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -6}\hfill & \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_3\to R_2+6R_3$:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -6}\hfill & \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 6}\hfill \end{array}\right]$$

$R_1\to R_1-R_3$ and $R_2\to R_2+2R_3$:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle -6}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -6}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 9}\hfill & \hfill {\displaystyle -15}\hfill & \hfill {\displaystyle 12}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 6}\hfill \end{array}\right]$$

$R_1\to 2R_1+R_2$:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 10}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -6}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 9}\hfill & \hfill {\displaystyle -15}\hfill & \hfill {\displaystyle 12}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 6}\hfill \end{array}\right]$$

Then $R_1\to \frac{1}{10}{R}_1$, $R_2\to -\frac{1}{6}{R}_2$, $R_3\to \frac{1}{2}{R}_3$:

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

By Theorem 1.6.2, $A\overrightarrow{x}=\overrightarrow{b}\Rightarrow \overrightarrow{x}=A^{-1}\overrightarrow{b}$:

$$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{ccc}1/2& -1/2& 0\\ -3/2& 5/2& -2\\ 3/2& -5/2& 3\end{array}\right]\left[\begin{array}{c}4\\ 2\\ 5\end{array}\right]=\left[\begin{array}{c}1\\ -11\\ 16\end{array}\right]$$

#11

Solve the linear systems. Using the given values for the $b$’s solve the systems together by reducing an appropriate augmented matrix to reduced row echelon form.

$$\begin{array}{cc}\hfill {\displaystyle 4x_1-7x_2}& {\displaystyle =b_1}\hfill \\ \hfill {\displaystyle x_1+2x_2}& {\displaystyle =b_2}\hfill \end{array}$$

i. $b_1=0,b_2=1$
ii. $b_1=-4,b_2=6$
iii. $b_1=-1,b_2=3$
iv. $b_1=-5,b_2=1$

Solution

Reduce the augmented matrix with general $b_1$ and $b_2$, then substitute each pair of values.

$$\left[\begin{array}{ccc}\hfill {\displaystyle 4}\hfill & \hfill {\displaystyle -7}\hfill & \hfill {\displaystyle b_1}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle b_2}\hfill \end{array}\right]$$

$R_2\to R_1-4R_2$:

$$\left[\begin{array}{ccc}\hfill {\displaystyle 4}\hfill & \hfill {\displaystyle -7}\hfill & \hfill {\displaystyle b_1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -15}\hfill & \hfill {\displaystyle b_1-4b_2}\hfill \end{array}\right]$$

$R_1\to 15R_1-7R_2$:

$${\displaystyle \left[\begin{array}{ccc}\hfill {\displaystyle 60}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 8b_1+28b_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -15}\hfill & \hfill {\displaystyle b_1-4b_2}\hfill \end{array}\right]\to \left[\begin{array}{ccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle \frac{2}{15}{b}_1+\frac{7}{15}{b}_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -\frac{1}{15}{b}_1+\frac{4}{15}{b}_2}\hfill \end{array}\right]}$$

For example, in ii, $(b_1,b_2)=(-4,6)$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle x_1}& {\displaystyle =-\frac{8}{15}+\frac{42}{15}=\frac{34}{15}}\hfill \\ \hfill {\displaystyle x_2}& {\displaystyle =\frac{4}{15}+\frac{24}{15}=\frac{28}{15}}\hfill \end{array}}$$

Tip: We could also put all four right-hand sides into one augmented matrix and row reduce it: $\left[\begin{array}{cccccc}\hfill {\displaystyle 4}\hfill & \hfill {\displaystyle -7}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle -5}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$.

#17

Determine conditions on the $b_i$’s, if any, in order to guarantee that the linear system is consistent.

$$\begin{array}{cc}\hfill {\displaystyle x_1-x_2+3x_3+2x_4}& {\displaystyle =b_1}\hfill \\ \hfill {\displaystyle -2x_1+x_2+5x_3+x_4}& {\displaystyle =b_2}\hfill \\ \hfill {\displaystyle -3x_1+2x_2+2x_3-x_4}& {\displaystyle =b_3}\hfill \\ \hfill {\displaystyle 4x_1-3x_2+x_3+3x_4}& {\displaystyle =b_4}\hfill \end{array}$$

Solution

$$\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle b_1}\hfill \\ \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle b_2}\hfill \\ \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle b_3}\hfill \\ \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle b_4}\hfill \end{array}\right]\to \left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -8}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle -b_1-b_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -11}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle -2b_1-b_2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle b_1-b_2+b_3}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -2b_1+b_2+b_4}\hfill \end{array}\right]$$

The last two rows have no variables, so this is consistent if

$$\begin{array}{cc}\hfill {\displaystyle b_1-b_2+b_3}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle -2b_1+b_2+b_4}& {\displaystyle =0.}\hfill \end{array}$$

This becomes a system in the $b_i$’s:

$$\left[\begin{array}{cccc}1& -1& 1& 0\\ -2& 1& 0& 1\end{array}\right]\to \left[\begin{array}{cccc}1& -1& 1& 0\\ 0& 1& -2& -1\end{array}\right]\to \left[\begin{array}{cccc}1& 0& -1& -1\\ 0& 1& -2& -1\end{array}\right]$$

so $b_1-b_3-b_4=0$ and $b_2-2b_3-b_4=0$.

Theorem 1.6.4 Equivalent Statements (extends Theorem 1.5.3)

Theorem 1.6.5

Proof

Let ${\overrightarrow{x}}_0$ be a solution of $B\overrightarrow{x}=\overrightarrow{0}$. Then, by associativity of matrix multiplication,

$$(AB){\overrightarrow{x}}_0=A(B{\overrightarrow{x}}_0)=A\overrightarrow{0}=\overrightarrow{0}.$$

Since $AB$ is invertible, $(AB)\overrightarrow{x}=\overrightarrow{0}$ has only the trivial solution (Theorem 1.6.4 (a) and (b)), so ${\overrightarrow{x}}_0=\overrightarrow{0}$. Then $B\overrightarrow{x}=\overrightarrow{0}$ has only the trivial solution. Thus $B$ is invertible ($B^{-1}$ exists), and

$$A=A(B{B}^{-1})=(AB)B^{-1}.$$

Since $AB$ and $B^{-1}$ are invertible, their product is invertible, so $A$ is invertible.

Theorem 1.4.6 together with Theorem 1.6.5 give

$$A\text{ and }B\text{ are invertible}\iff AB\text{ is invertible},$$

with $\Rightarrow$ from Theorem 1.4.6 and $\Leftarrow$ from Theorem 1.6.5.