College Algebra, Chapter 7, 7.4, Section 7.4, Problem 48

Solve the system $\left\{\begin{equation}
\begin{aligned}

x + y =& 1
\\
y + z =& 2
\\
z + w =& 3
\\
w - x =& 4

\end{aligned}
\end{equation} \right.$ using Cramer's Rule.

For this system we have

$\displaystyle |D| = \left| \begin{array}{cccc}
1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
-1 & 0 & 0 & 1
\end{array} \right| $

If we add row 1 to row 4, we get

$\displaystyle \left| \begin{array}{cccc}
1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 \\
0 & 1 & 0 & 1
\end{array} \right| $

So,

$|D| = 1 \left| \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1
\end{array} \right| - 1 \left| \begin{array}{ccc}
0 & 1 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array} \right|$

Now, we add -1 times row 1 to row 3 in the first matrix, we have


$
\begin{equation}
\begin{aligned}

|D| =& 1 \left| \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 1 \\
0 & -1 & 1
\end{array} \right| - 1 \left| \begin{array}{ccc}
0 & 1 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array} \right| \qquad \text{Expand}
\\
\\
|D| =& (1)(1) \left| \begin{array}{cc}
1 & 1 \\
-1 & 1
\end{array} \right| - (1)(1) \left| \begin{array}{cc}
0 & 1 \\
0 & 1
\end{array} \right| - (1)(-1) \left| \begin{array}{cc}
0 & 1 \\
0 & 1
\end{array} \right| + (1)(1) \left| \begin{array}{cc}
0 & 0 \\
0 & 1
\end{array} \right|
\\
\\
|D| =& 1 \cdot 1 - 1 \cdot (-1)
\\
\\
|D| =& 2

\end{aligned}
\end{equation}
$


For $D_x$,

$\displaystyle |D_x| = \left| \begin{array}{cccc}
1 & 1 & 0 & 0 \\
2 & 1 & 1 & 0 \\
3 & 0 & 1 & 1 \\
4 & 0 & 0 & 1
\end{array} \right|$

If we add -1 times row 1 to row 2, we get

$\displaystyle \left| \begin{array}{cccc}
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
3 & 0 & 1 & 1 \\
4 & 0 & 0 & 1
\end{array} \right|$

So,

$\displaystyle |D_x| = 1 \left| \begin{array}{ccc}
0 & 1 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array} \right| - 1 \left| \begin{array}{ccc}
1 & 1 & 0 \\
3 & 1 & 1 \\
4 & 0 & 1
\end{array} \right|$

Now, applying -1 times row 1 to row 2 in the second matrix and since the first matrix is equal to zero, this gives us


$
\begin{equation}
\begin{aligned}

|D_x| =& -1 \left| \begin{array}{ccc}
1 & 1 & 0 \\
2 & 0 & 1 \\
4 & 0 & 1
\end{array} \right|
\qquad \text{Expand}
\\
\\
=& -1(1) \left| \begin{array}{cc}
0 & 1 \\
0 & 1
\end{array} \right| - (1)(-1) \left| \begin{array}{cc}
2 & 1 \\
4 & 1
\end{array} \right|
\\
\\
=& 2 \cdot 1 - 4 \cdot 1
\\
\\
=& -2

\end{aligned}
\end{equation}
$


For $D_y$

$\displaystyle |D_y| = \left| \begin{array}{cccc}
1 & 1 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 3 & 1 & 1 \\
-1 & 4 & 0 & 1
\end{array} \right|$

If we add row 1 to row 4, we get

$\displaystyle \left| \begin{array}{cccc}
1 & 1 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 3 & 1 & 1 \\
0 & 5 & 0 & 1
\end{array} \right|$

So,

$\displaystyle |D_y| = 1 \left| \begin{array}{ccc}
2 & 1 & 0 \\
3 & 1 & 1 \\
5 & 0 & 1
\end{array} \right| - 1 \left| \begin{array}{ccc}
0 & 1 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array} \right|$

Now, since the second matrix is equal to zero. We add -1 times row 1 to row 2 in the first matrix. We have,


$
\begin{equation}
\begin{aligned}

|D_y| =& 1 \left| \begin{array}{ccc}
2 & 1 & 0 \\
1 & 0 & 1 \\
5 & 0 & 1
\end{array} \right| \qquad \text{Expand}
\\
\\
=& 1 (2) \left| \begin{array}{cc}
0 & 1 \\
0 & 1
\end{array} \right| - 1 \left| \begin{array}{cc}
1 & 1 \\
5 & 1
\end{array} \right|
\\
\\
=& -1 (1 \cdot 1 - 1 \cdot 5)
\\
\\
=& 4

\end{aligned}
\end{equation}
$


For $D_z$,

$\displaystyle |D_z| = \left| \begin{array}{cccc}
1 & 1 & 1 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 3 & 1 \\
-1 & 0 & 4 & 1
\end{array} \right|$

If we add row 1 to row 4, we get

$\displaystyle \left| \begin{array}{cccc}
1 & 1 & 1 & 0 \\
0 & 1 & 2 & 0 \\
0 & 0 & 3 & 1 \\
0 & 1 & 5 & 1
\end{array} \right|$

So,

$\displaystyle |D_z| = 1 \left| \begin{array}{ccc}
1 & 2 & 0 \\
0 & 3 & 1 \\
1 & 5 & 1
\end{array} \right| - 1\left| \begin{array}{ccc}
0 & 2 & 0 \\
0 & 3 & 1 \\
0 & 5 & 1
\end{array} \right| + 1 \left| \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 1 & 1
\end{array} \right|$

Since the second and third matrix are equal to zero, we can disregard it. Then if we add -1 times row 1 to row 3 in the first matrix. We have


$
\begin{equation}
\begin{aligned}

|D_z| =& 1 \left| \begin{array}{ccc}
1 & 2 & 0 \\
0 & 3 & 1 \\
0 & 3 & 1
\end{array} \right| \qquad \text{Expand}
\\
\\
=& 1 (1) \left| \begin{array}{cc}
3 & 1 \\
3 & 1
\end{array} \right| - 1(2) \left| \begin{array}{cc}
0 & 1 \\
0 & 1
\end{array} \right|
\\
\\
=& 3 \cdot 1 - 3 \cdot 1
\\
\\
=& 0


\end{aligned}
\end{equation}
$


For $D_w$

$\displaystyle |D_w| = \left| \begin{array}{cccc}
1 & 1 & 0 & 1 \\
0 & 1 & 1 & 2 \\
0 & 0 & 1 & 3 \\
-1 & 0 & 0 & 4
\end{array} \right|$

If we add row 1 to row 4, we get

$\displaystyle \left| \begin{array}{cccc}
1 & 1 & 0 & 1 \\
0 & 1 & 1 & 2 \\
0 & 0 & 1 & 3 \\
0 & 1 & 0 & 5
\end{array} \right|$

So,

$\displaystyle |D_w| = 1 \left| \begin{array}{ccc}
1 & 1 & 2 \\
0 & 1 & 3 \\
1 & 0 & 5
\end{array} \right| - 1 \left| \begin{array}{ccc}
0 & 1 & 2 \\
0 & 1 & 3 \\
0 & 0 & 5
\end{array} \right| - 1 \left| \begin{array}{ccc}
0 & 1 & 1 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array} \right|$

Since the second and third matrix are both equal to zero. Then we can disregard it. If we add -1 times row 1 to row 3, we have


$
\begin{equation}
\begin{aligned}

|D_w| =& 1 \left| \begin{array}{ccc}
1 & 1 & 2 \\
0 & 1 & 3 \\
0 & -1 & 3
\end{array} \right| \qquad \text{Expand}
\\
\\
=& 1(1) \left| \begin{array}{cc}
1 & 3 \\
-1 & 3
\end{array} \right| - 1(1) \left| \begin{array}{cc}
0 & 3 \\
0 & 3
\end{array} \right| + 2 \left| \begin{array}{cc}
0 & 1 \\
0 & 0
\end{array} \right|
\\
\\
=& 1 \cdot 3 - 3 \cdot (-1)
\\
\\
=& 6

\end{aligned}
\end{equation}
$


The solution is


$
\begin{equation}
\begin{aligned}

x =& \frac{|D_x|}{|D|} = \frac{-2}{2} = -1
\\
\\
y =& \frac{|D_y|}{|D|} = \frac{4}{2} = 2
\\
\\
z =& \frac{|D_z|}{|D|} = \frac{0}{2} = 0
\\
\\
w =& \frac{|D_w|}{|D|} = \frac{6}{2} = 3

\end{aligned}
\end{equation}
$