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1.5 Elementary Matrices and a Method for Finding A⁻¹: Class Notes

Definition

Definition

Elementary matrices are useful for developing theory.

Theorem 1.5.1 Row Operations by Matrix Multiplication

#6

An elementary matrix $E$ and a matrix $A$ are given. Identify the row operation corresponding to $E$ and verify that the product $EA$ results from applying the row operation to $A$.

a. $E=\left[\begin{array}{cc}-6& 0\\ 0& 1\end{array}\right]$, $A=\left[\begin{array}{cccc}-1& -2& 5& -1\\ 3& -6& -6& -6\end{array}\right]$

b. $E=\left[\begin{array}{ccc}1& 0& 0\\ -4& 1& 0\\ 0& 0& 1\end{array}\right]$, $A=\left[\begin{array}{ccccc}2& -1& 0& -4& -4\\ 1& -3& -1& 5& 3\\ 2& 0& 1& 3& -1\end{array}\right]$

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

Solution

a. $E$ comes from $I_2$ by multiplying row 1 by $-6$, so the row operation is $R_1\to -6R_1$.

$$EA=\left[\begin{array}{cc}-6& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cccc}-1& -2& 5& -1\\ 3& -6& -6& -6\end{array}\right]=\left[\begin{array}{cccc}6& 12& -30& 6\\ 3& -6& -6& -6\end{array}\right]$$

This is $R_1\to -6R_1$ applied to $A$.

b. $E$ comes from $I_3$ by adding $-4$ times row 1 to row 2, so the row operation is $R_2\to -4R_1+R_2$.

$$EA=\left[\begin{array}{ccc}1& 0& 0\\ -4& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccccc}2& -1& 0& -4& -4\\ 1& -3& -1& 5& 3\\ 2& 0& 1& 3& -1\end{array}\right]=\left[\begin{array}{ccccc}2& -1& 0& -4& -4\\ -7& 1& -1& 21& 19\\ 2& 0& 1& 3& -1\end{array}\right]$$

Applying $R_2\to -4R_1+R_2$ to $A$ directly:

so row 2 becomes $\left[\begin{array}{ccccc}-7& 1& -1& 21& 19\end{array}\right]$, the same matrix as $EA$.

Proof of Theorem 1.5.1

Case I: $E$ results from multiplying the $i$th row of $I$ by $k\ne 0$.

Let ${\overrightarrow{e}}_1=\left[\begin{array}{cccc}1& 0& \cdots & 0\end{array}\right]$, ${\overrightarrow{e}}_2=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\end{array}\right]$, …, ${\overrightarrow{e}}_m=\left[\begin{array}{cccc}0& 0& \cdots & 1\end{array}\right]$, and $k{\overrightarrow{e}}_i=\left[\begin{array}{cccccc}0& 0& \cdots & k& \cdots & 0\end{array}\right]$. These are the rows of $E$. Each row of $EA$ is the corresponding row of $E$ times $A$, and ${\overrightarrow{e}}_{j}A$ is row $j$ of $A$:

$$EA=\left[\begin{array}{c}{\overrightarrow{e}}_{1}A\\ {\overrightarrow{e}}_{2}A\\ ⋮\\ k{\overrightarrow{e}}_{i}A\\ ⋮\\ {\overrightarrow{e}}_{m}A\end{array}\right],$$

which results in multiplying the $i$th row of $A$ by $k$.

#7

Use the following matrices and find an elementary matrix $E$ that satisfies the stated equation.

$A=\left[\begin{array}{ccc}3& 4& 1\\ 2& -7& -1\\ 8& 1& 5\end{array}\right]$, $B=\left[\begin{array}{ccc}8& 1& 5\\ 2& -7& -1\\ 3& 4& 1\end{array}\right]$, $C=\left[\begin{array}{ccc}3& 4& 1\\ 2& -7& -1\\ 2& -7& 3\end{array}\right]$, $D=\left[\begin{array}{ccc}8& 1& 5\\ -6& 21& 3\\ 3& 4& 1\end{array}\right]$, $F=\left[\begin{array}{ccc}8& 1& 5\\ 8& 1& 1\\ 3& 4& 1\end{array}\right]$

c. $EA=C$

Solution

Rows 1 and 2 of $A$ and $C$ are the same. Row 3 of $C$ results from adding $-2$ times row 1 of $A$ to row 3 ($R_3\to R_3-2R_1$). Performing that operation on $I_3$ gives

$$E=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -2& 0& 1\end{array}\right]$$

Check:

$$EA=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -2& 0& 1\end{array}\right]\left[\begin{array}{ccc}3& 4& 1\\ 2& -7& -1\\ 8& 1& 5\end{array}\right]=\left[\begin{array}{ccc}3& 4& 1\\ 2& -7& -1\\ 2& -7& 3\end{array}\right]=C✓$$

Theorem 1.5.2

Proof

Let $E$ be the elementary matrix that results from performing a single row operation on $I$, and let $E\prime$ be the matrix that results from performing the inverse operation on $I$. Then by Theorem 1.5.1,

$$EE\prime =I\text{and}E\prime E=I\Rightarrow E\prime =E^{-1}.$$

Theorem 1.5.3 Equivalent Statements

Proof

To prove that statements are equivalent, we can show that $(a)\Rightarrow (b)\Rightarrow (c)\Rightarrow (d)\Rightarrow (a)$.

$(a)\Rightarrow (b)$: $A$ invertible $\Rightarrow$ there exists $A^{-1}$ such that $A^{-1}A=I$.

$$A\overrightarrow{x}=\overrightarrow{0}\Rightarrow A^{-1}A\overrightarrow{x}=A^{-1}\overrightarrow{0}\Rightarrow \overrightarrow{x}=\overrightarrow{0}.$$

If $\overrightarrow{x}\prime$ is any other solution, then $A\overrightarrow{x}=\overrightarrow{0}$ and $A\overrightarrow{x}\prime =\overrightarrow{0}$, so

$$A\overrightarrow{x}-A\overrightarrow{x}\prime =\overrightarrow{0}\Rightarrow \overrightarrow{x}-\overrightarrow{x}\prime =A^{-1}\overrightarrow{0}=\overrightarrow{0}\Rightarrow \overrightarrow{x}=\overrightarrow{x}\prime .$$

So $A\overrightarrow{x}=\overrightarrow{0}$ has only the trivial solution.

$(b)\Rightarrow (c)$: $A\overrightarrow{x}=\overrightarrow{0}$ has only the trivial solution, so the system contains no free variables. By Theorem 1.2.1, the reduced row echelon form has $n$ nonzero rows. By Theorem 1.4.3, the reduced row echelon form is then $I_n$.

$(c)\Rightarrow (d)$: The reduced row echelon form of $A$ is $I_n$, so $I_n$ can be obtained from $A$ by a series of row operations. By Theorem 1.5.1 we can express this as

$$I_n=E_r\cdots E_2{E}_{1}A.$$

Then, by Theorems 1.4.6 and 1.5.2,

$$A=E_1^{-1}{E}_2^{-1}\cdots E_r^{-1}{I}_n,$$

so $A$ is equal to a product of elementary matrices.

$(d)\Rightarrow (a)$: $A$ is a product of elementary matrices, and elementary matrices are invertible (Theorem 1.5.2). So $A$ is a product of invertible matrices, and so is invertible (Theorem 1.4.6).

#15

Use the inversion algorithm to find the inverse of the matrix (if the inverse exists).

$$\left[\begin{array}{ccc}2& 6& 6\\ 2& 7& 6\\ 2& 7& 7\end{array}\right]$$

Solution

The inversion algorithm is based on the $(c)\Rightarrow (d)$ proof above. Since $I_n=E_r\cdots E_2{E}_{1}A$, multiplying on the right by $A^{-1}$ gives

$$A^{-1}=E_r\cdots E_2{E}_1{I}_n.$$

So the elementary row operations that reduce $A$ to $I_n$ will yield $A^{-1}$ when applied to $I_n$.

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 7}\hfill & \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 7}\hfill & \hfill {\displaystyle 7}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$ clear the first column below the leading 2:

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

$R_3\to R_3-R_2$:

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

$R_1\to R_1-6R_3$:

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

$R_1\to R_1-6R_2$:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 7}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -6}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_1\to \frac{1}{2}{R}_1$:

$$\left[\begin{array}{cccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 7/2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -3}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

In general,

$$\left[\begin{array}{cc}\hfill {\displaystyle A}\hfill & \hfill {\displaystyle I_n}\hfill \end{array}\right]\stackrel{\text{elementary row operations}}{\to }\left[\begin{array}{cc}\hfill {\displaystyle I_n}\hfill & \hfill {\displaystyle A^{-1}}\hfill \end{array}\right].$$

#18

$$\left[\begin{array}{cccc}0& 0& 2& 0\\ 1& 0& 0& 1\\ 0& -1& 3& 0\\ 2& 1& 5& -3\end{array}\right]$$

Solution

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

$R_1\leftrightarrow R_2$:

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

$R_4\to R_4-2R_1$:

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

$R_4\to R_4+R_3$, then $R_2\leftrightarrow R_3$:

$$\left[\begin{array}{cccccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle -5}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_4\to R_4-4R_3$, $R_2\to -R_2$:

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

$R_1\to R_1+\frac{1}{5}{R}_4$:

$R_2\to R_2+\frac{3}{2}{R}_3$:

Then $R_3\to \frac{1}{2}{R}_3$ and $R_4\to -\frac{1}{5}{R}_4$:

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

#20

Find the inverse of each of the following $4\times 4$ matrices, where $k_1$, $k_2$, $k_3$, $k_4$, $k$ are all nonzero.

a. $\left[\begin{array}{cccc}0& 0& 0& k_1\\ 0& 0& k_2& 0\\ 0& k_3& 0& 0\\ k_4& 0& 0& 0\end{array}\right]$  b. $\left[\begin{array}{cccc}k& 0& 0& 0\\ 1& k& 0& 0\\ 0& 1& k& 0\\ 0& 0& 1& k\end{array}\right]$

Solution

a.

$$\left[\begin{array}{cccccccc}\hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k_1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k_2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k_3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle k_4}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_1\leftrightarrow R_4$ and $R_2\leftrightarrow R_3$:

$$\left[\begin{array}{cccccccc}\hfill {\displaystyle k_4}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k_3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k_2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k_1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Multiply each row by the reciprocal of its diagonal entry ($\frac{1}{k_4}{R}_1$, $\frac{1}{k_3}{R}_2$, $\frac{1}{k_2}{R}_3$, $\frac{1}{k_1}{R}_4$):

b. Clear the 1 below each diagonal $k$, working down the matrix.

$$\left[\begin{array}{cccccccc}\hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_2\to k{R}_2-R_1$:

$$\left[\begin{array}{cccccccc}\hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k^2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_3\to k^2{R}_3-R_2$:

$$\left[\begin{array}{cccccccc}\hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k^2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k^3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -k}\hfill & \hfill {\displaystyle k^2}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

and so on: $R_4\to k^3{R}_4-R_3$:

$$\left[\begin{array}{cccccccc}\hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k^2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k^3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -k}\hfill & \hfill {\displaystyle k^2}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle k^4}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle k}\hfill & \hfill {\displaystyle -k^2}\hfill & \hfill {\displaystyle k^3}\hfill \end{array}\right]$$

Finally divide the rows by $k$, $k^2$, $k^3$, $k^4$:

#27

Show that the matrices $A$ and $B$ are row equivalent by finding a sequence of elementary row operations that produces $B$ from $A$, and then use that result to find a matrix $C$ such that $CA=B$.

$A=\left[\begin{array}{ccc}1& 2& 3\\ 1& 4& 1\\ 2& 1& 9\end{array}\right]$, $B=\left[\begin{array}{ccc}1& 0& 5\\ 0& 2& -2\\ 1& 1& 4\end{array}\right]$

Solution

To obtain $B$ from $A$, keep track of each row operation and its corresponding elementary matrix (the same operation performed on $I_3$).

$R_2\to R_2-R_1$:

$$\left[\begin{array}{ccc}1& 2& 3\\ 0& 2& -2\\ 2& 1& 9\end{array}\right],E_1=\left[\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 0& 0& 1\end{array}\right]$$

$R_1\to R_1-R_2$:

$$\left[\begin{array}{ccc}1& 0& 5\\ 0& 2& -2\\ 2& 1& 9\end{array}\right],E_2=\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$

$R_3\to R_3-R_1$:

$$\left[\begin{array}{ccc}1& 0& 5\\ 0& 2& -2\\ 1& 1& 4\end{array}\right]=B,E_3=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -1& 0& 1\end{array}\right]$$

So $A$ and $B$ are row equivalent, and

$$E_3{E}_2{E}_{1}A=B\Rightarrow C=E_3{E}_2{E}_1$$

$$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -1& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& 0\\ -1& 1& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ -1& 1& 0\\ 0& 0& 1\end{array}\right]$$

Tip: Alternative approach: apply the same row operations to $\left[\begin{array}{cc}\hfill {\displaystyle A}\hfill & \hfill {\displaystyle I}\hfill \end{array}\right]$. Reducing it to $\left[\begin{array}{cc}\hfill {\displaystyle B}\hfill & \hfill {\displaystyle C}\hfill \end{array}\right]$ gives $C$ directly.

#23

Express the matrix and its inverse as products of elementary matrices.

$$A=\left[\begin{array}{cc}-3& 1\\ 2& 2\end{array}\right]$$

Solution

Reduce $A$ to $I_2$, keeping track of the operations, the corresponding elementary matrix, and its inverse.

$R_2\to R_2+\frac{2}{3}{R}_1$:

$$\left[\begin{array}{cc}-3& 1\\ 0& 8/3\end{array}\right]E_1=\left[\begin{array}{cc}1& 0\\ 2/3& 1\end{array}\right]E_1^{-1}=\left[\begin{array}{cc}1& 0\\ -2/3& 1\end{array}\right]$$

$R_1\to R_1-\frac{3}{8}{R}_2$:

$$\left[\begin{array}{cc}-3& 0\\ 0& 8/3\end{array}\right]E_2=\left[\begin{array}{cc}1& -3/8\\ 0& 1\end{array}\right]E_2^{-1}=\left[\begin{array}{cc}1& 3/8\\ 0& 1\end{array}\right]$$

$R_1\to -\frac{1}{3}{R}_1$:

$$\left[\begin{array}{cc}1& 0\\ 0& 8/3\end{array}\right]E_3=\left[\begin{array}{cc}-1/3& 0\\ 0& 1\end{array}\right]E_3^{-1}=\left[\begin{array}{cc}-3& 0\\ 0& 1\end{array}\right]$$

$R_2\to \frac{3}{8}{R}_2$:

$$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]E_4=\left[\begin{array}{cc}1& 0\\ 0& 3/8\end{array}\right]E_4^{-1}=\left[\begin{array}{cc}1& 0\\ 0& 8/3\end{array}\right]$$

So $E_4{E}_3{E}_2{E}_{1}A=I_2$, and therefore

#33

Prove that if $B$ is obtained from $A$ by performing a sequence of elementary row operations, then there is a second sequence of elementary row operations, which when applied to $B$ recovers $A$.

Proof

Let $E_1,E_2,\dots ,E_r$ be elementary matrices such that when the associated row operations are applied to $A$, the result is $B$. That is,

$$B=E_r\cdots E_2{E}_{1}A.$$

Since each $E_i$ is invertible by Theorem 1.5.2 (and its inverse is also an elementary matrix), we can multiply on the left to produce

$$E_1^{-1}{E}_2^{-1}\cdots E_r^{-1}B=A.$$

So $A$ is obtained from $B$ using the row operations represented by the $E_i^{-1}$.