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1.2 Gaussian Elimination: Class Notes

#7

Solve the system by Gaussian elimination.

$$\begin{array}{cc}\hfill {\displaystyle x-y+2z-w}& {\displaystyle =-1}\hfill \\ \hfill {\displaystyle 2x+y-2z-2w}& {\displaystyle =-2}\hfill \\ \hfill {\displaystyle -x+2y-4z+w}& {\displaystyle =1}\hfill \\ \hfill {\displaystyle 3x-3w}& {\displaystyle =-3}\hfill \end{array}$$

Solution

Gaussian elimination works with the augmented matrix of the system:

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

Tip: Watch the difference between $0$ (zero) and $\varnothing$, the empty set: $\varnothing =\{\}$.

Step 1. Row 1 already starts with a leading 1. Make the entries below it in column 1 zero, using $R_2\to R_2-2R_1$, $R_3\to R_3+R_1$, and $R_4\to R_4-3R_1$:

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

$$\left[\begin{array}{ccccc}-1& 2& -4& 1& 1\end{array}\right]+\left[\begin{array}{ccccc}1& -1& 2& -1& -1\end{array}\right]=\left[\begin{array}{ccccc}0& 1& -2& 0& 0\end{array}\right]$$

$$\left[\begin{array}{ccccc}3& 0& 0& -3& -3\end{array}\right]+\left[\begin{array}{ccccc}-3& 3& -6& 3& 3\end{array}\right]=\left[\begin{array}{ccccc}0& 3& -6& 0& 0\end{array}\right]$$

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

Step 2. Now work on column 2, below row 1. Row 3 already has a 1 there, so swap it into row 2 with $R_2\leftrightarrow R_3$:

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

Then clear the 3s below that leading 1 with $R_3\to R_3-3R_2$ and $R_4\to R_4-3R_2$:

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

This is in row echelon form, reached by Gaussian elimination. The system is now

$$\begin{array}{cc}\hfill {\displaystyle (1)x-y+2z-w}& {\displaystyle =-1}\hfill \\ \hfill {\displaystyle (2)y-2z}& {\displaystyle =0}\hfill \end{array}$$

Solve each equation for its first variable:

$$\begin{array}{cc}\hfill {\displaystyle (1)x}& {\displaystyle =y-2z+w-1}\hfill \\ \hfill {\displaystyle (2)y}& {\displaystyle =2z}\hfill \end{array}$$

Substituting $y=2z$ into equation (1) gives $x=2z-2z+w-1$, so

$$x=w-1,y=2z$$

The solution is $(x,y,z,w)=(w-1,2z,z,w)$. Let $s=z$ and $t=w$. Then $x=t-1$, $y=2s$, $z=s$, $w=t$:

Definitions

Back to #7: the row echelon form was

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

The $-1$ above the leading 1 in column 2 breaks property 4. Clear it with $R_1\to R_1+R_2$:

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

The columns go with $x$, $y$, $z$, $w$, and the constants:

$$\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle -1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]⟶\begin{array}{cc}\hfill {\displaystyle x-w}& {\displaystyle =-1}\hfill \\ \hfill {\displaystyle y-2z}& {\displaystyle =0}\hfill \end{array}$$

This is reduced row echelon form, reached by Gauss–Jordan elimination. It gives $x=w-1$ and $y=2z$ directly, without substituting. Letting $s=z$ and $t=w$ gives the same result as before.

In #7 the leading 1s are in the $x$- and $y$-columns, so $x$ and $y$ are leading variables. The remaining variables, $z$ and $w$, are free variables.

Definitions

Every variable in a system is either leading or free.

Definition

Theorem 1.2.1 Free Variable Theorem for Homogeneous Systems

The $r$ nonzero rows correspond to the leading 1s, so $r$ counts the leading variables, and the other $n-r$ variables are free.

Theorem 1.2.2

For example, the system below has 3 unknowns but only 2 equations:

$$\begin{array}{cc}\hfill {\displaystyle 2x+3y-5z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle 4x+8y-7z}& {\displaystyle =0}\hfill \end{array}$$

Proof

The number of equations in a system is the number of rows in its augmented matrix. The number of leading 1s is no more than the number of rows. Thus, such a system has at least 1 free variable. A homogeneous system is always consistent, since it has the trivial solution $x_1=x_2=\cdots =x_n=0$. A consistent system with a free variable has infinitely many solutions.

#28

What condition, if any, must $a$, $b$, and $c$ satisfy for the linear system to be consistent?

$$\begin{array}{cc}\hfill {\displaystyle x+3y+z}& {\displaystyle =a}\hfill \\ \hfill {\displaystyle -x-2y+z}& {\displaystyle =b}\hfill \\ \hfill {\displaystyle 3x+7y-z}& {\displaystyle =c}\hfill \end{array}$$

Solution

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle a}\hfill \\ \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle b}\hfill \\ \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 7}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle c}\hfill \end{array}\right]$$

Step 1. Clear column 1 below the leading 1 with $R_2\to R_2+R_1$ and $R_3\to R_3-3R_1$:

$$\left[\begin{array}{cccc}-1& -2& 1& b\end{array}\right]+\left[\begin{array}{cccc}1& 3& 1& a\end{array}\right]=\left[\begin{array}{cccc}0& 1& 2& a+b\end{array}\right]$$

$$\left[\begin{array}{cccc}3& 7& -1& c\end{array}\right]+\left[\begin{array}{cccc}-3& -9& -3& -3a\end{array}\right]=\left[\begin{array}{cccc}0& -2& -4& -3a+c\end{array}\right]$$

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle a}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle a+b}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle -3a+c}\hfill \end{array}\right]$$

Step 2. Clear column 2 below the leading 1 with $R_3\to R_3+2R_2$:

$$\left[\begin{array}{cccc}0& -2& -4& -3a+c\end{array}\right]+\left[\begin{array}{cccc}0& 2& 4& 2a+2b\end{array}\right]=\left[\begin{array}{cccc}0& 0& 0& -a+2b+c\end{array}\right]$$

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle a}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle a+b}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -a+2b+c}\hfill \end{array}\right]$$

Row 3 is the equation $0=-a+2b+c$. The system has a solution exactly when this is true.

Solve the given linear system by any method.

#16

$$\begin{array}{cc}\hfill {\displaystyle 2x-y-3z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle -x+2y-3z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle x+y+4z}& {\displaystyle =0}\hfill \end{array}$$

Solution

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

Tip: For a homogeneous system, writing the column of zeros is optional: row operations never change it. The work below leaves it off.

Step 1. Clear column 1 below row 1 with $R_2\to R_1+2R_2$ and $R_3\to R_3+R_2$ (both use the rows as they are now):

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

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

Then ${\displaystyle R_2\to \frac{1}{3}{R}_2}$:

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

Step 2. Clear the 3 below the leading 1 in column 2 with $R_3\to R_3-3R_2$:

$$\left[\begin{array}{ccc}0& 3& 1\end{array}\right]+\left[\begin{array}{ccc}0& -3& 9\end{array}\right]=\left[\begin{array}{ccc}0& 0& 10\end{array}\right]$$

Then ${\displaystyle R_3\to \frac{1}{10}{R}_3}$:

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

Step 3. Clear column 3 above the leading 1 with $R_2\to R_2+3R_3$ and $R_1\to R_1+3R_3$:

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

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

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

Step 4. Clear the $-1$ in column 2 with $R_1\to R_1+R_2$, then ${\displaystyle R_1\to \frac{1}{2}{R}_1}$:

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

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]⟶\begin{array}{c}{\displaystyle x=0}\hfill \\ {\displaystyle y=0}\hfill \\ {\displaystyle z=0}\hfill \end{array}$$

This is the trivial solution.

#19

$$\begin{array}{cc}\hfill {\displaystyle 2x+2y+4z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle w-y-3z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle 2w+3x+y+z}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle -2w+x+3y-2z}& {\displaystyle =0}\hfill \end{array}$$

Solution

Divide the first equation by 2 and use $x+y+2z=0$ instead. With the columns in the order $w$, $x$, $y$, $z$ (leaving off the column of zeros), the matrix is

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

Row 1 starts with 0, so swap it with row 2: $R_1\leftrightarrow R_2$.

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

Column 1. Use $R_3\to R_3-2R_1$ and $R_4\to R_4+R_3$ (adding row 3 as it is now, before it changes):

$$\left[\begin{array}{cccc}2& 3& 1& 1\end{array}\right]+\left[\begin{array}{cccc}-2& 0& 2& 6\end{array}\right]=\left[\begin{array}{cccc}0& 3& 3& 7\end{array}\right]$$

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

$$\left[\begin{array}{cccc}1& 0& -1& -3\\ 0& 1& 1& 2\\ 0& 3& 3& 7\\ 0& 4& 4& -1\end{array}\right]$$

Column 2. Use $R_3\to R_3-3R_2$ and $R_4\to R_4-4R_2$:

$$\left[\begin{array}{cccc}0& 3& 3& 7\end{array}\right]+\left[\begin{array}{cccc}0& -3& -3& -6\end{array}\right]=\left[\begin{array}{cccc}0& 0& 0& 1\end{array}\right]$$

$$\left[\begin{array}{cccc}0& 4& 4& -1\end{array}\right]+\left[\begin{array}{cccc}0& -4& -4& -8\end{array}\right]=\left[\begin{array}{cccc}0& 0& 0& -9\end{array}\right]$$

Then $R_4\to R_4+9R_3$ clears row 4, giving row echelon form:

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

Clear the column above the leading 1 in column 4 with $R_1\to R_1+3R_3$ and $R_2\to R_2-2R_3$ to reach reduced row echelon form (rref). With the column of zeros put back:

$$\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]⟶\begin{array}{cc}\hfill {\displaystyle w-y}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle x+y}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle z}& {\displaystyle =0}\hfill \end{array}$$

The leading 1s are in the $w$-, $x$-, and $z$-columns, so $w$, $x$, and $z$ are leading variables, and $y$ is a free variable.

Let $t=y$. Then $w=t$, $x=-t$, and $z=0$.

#23

The augmented matrix for a linear system is given in which the asterisk represents an unspecified real number. Determine whether the system is consistent, and if so whether the solution is unique. Answer “inconclusive” if there is not enough information to make a decision.

a. $\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill \end{array}\right]$  b. $\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$

c. $\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$  d. $\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle *}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle *}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle *}\hfill \end{array}\right]$

Solution

a. Every variable has a leading 1, so there are no free variables, and no row is a contradiction.

b. Row 3 says $0=0$, which is always true. The third column has no leading 1, so there is a free variable.

c. Row 3 says $0=1$, a contradiction.

d. Call the variables $x$, $y$, $z$. Row 3 says $z$ equals its last entry. If the unspecified entry in row 2 is not zero, row 2 forces $z=0$. Then the system is consistent if the last entry of row 3 is 0, and inconsistent if it is not 0. Both can happen:

If rows 2 and 3 end in

$$\left[\begin{array}{cccc}\hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 5}\hfill \end{array}\right]⟶\left[\begin{array}{cccc}0& 0& 2& 0\end{array}\right]+\left[\begin{array}{cccc}0& 0& -2& -10\end{array}\right]=\left[\begin{array}{cccc}0& 0& 0& -10\end{array}\right]$$

then subtracting twice the second of these rows from the first gives $0=-10$: no solution. But if they are

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

subtracting gives $0=0$, and the system is consistent.