### 3 Do matrices always have an LU decomposition?

No. Sometimes it is impossible to write a matrix in the form

#### 3.1 Why not?

An invertible matrix $A$ has an $LU$ decomposition provided that all its leading submatrices have non-zero determinants. The ${k}^{\mathrm{\text{th}}}$ leading submatrix of $A$ is denoted ${A}_{k}$ and is the $k×k$ matrix found by looking only at the top $k$ rows and leftmost $k$ columns. For example if

$\phantom{\rule{2em}{0ex}}A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 8\hfill & \hfill 14\hfill \\ \hfill 2\hfill & \hfill 6\hfill & \hfill 13\hfill \end{array}\right]$

$\phantom{\rule{2em}{0ex}}{A}_{1}=1,\phantom{\rule{2em}{0ex}}{A}_{2}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 8\hfill \end{array}\right],\phantom{\rule{2em}{0ex}}{A}_{3}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 8\hfill & \hfill 14\hfill \\ \hfill 2\hfill & \hfill 6\hfill & \hfill 13\hfill \end{array}\right].$

The fact that this matrix $A$ has an $LU$ decomposition can be guaranteed in advance because none of these determinants is zero:

$\begin{array}{rcll}& & \left|{A}_{1}\right|=1,\phantom{\rule{2em}{0ex}}& \text{}\\ & & \left|{A}_{2}\right|=\left(1×8\right)-\left(2×3\right)=2,& \text{}\\ & & |{A}_{3}|=\left|\begin{array}{cc}\hfill 8\hfill & \hfill 14\hfill \\ \hfill 6\hfill & \hfill 13\hfill \end{array}\right|-2\left|\begin{array}{cc}\hfill 3\hfill & \hfill 14\hfill \\ \hfill 2\hfill & \hfill 13\hfill \end{array}\right|+4\left|\begin{array}{cc}\hfill 3\hfill & \hfill 8\hfill \\ \hfill 2\hfill & \hfill 6\hfill \end{array}\right|=20-\left(2×11\right)+\left(4×2\right)=6\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

(where the $3×3$ determinant was found by expanding along the top row).

##### Example 7

Show that $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$ does not have an $LU$ decomposition.

##### Solution

The second leading submatrix has determinant equal to

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 4\hfill \end{array}\right|=\left(1×4\right)-\left(2×2\right)=0$

which means that an $LU$ decomposition is not possible in this case.

Which, if any, of these matrices have an $LU$ decomposition?

1. $A=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ ,
2. $A=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right]$ ,
3. $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 7\hfill \\ \hfill -2\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -2\hfill \end{array}\right]$ .

$|{A}_{1}|=3$ and $|{A}_{2}|=|A|=3$ . Neither of these is zero, so $A$ does have an $LU$ decomposition.

$\left|{A}_{1}\right|=0$ so $A$ does not have an $LU$ decomposition.

$|{A}_{1}|=1$ , $|{A}_{2}|=6-6=0$ , so $A$ does not have an $LU$ decomposition.

#### 3.2 Can we get around this problem?

Yes. It is always possible to re-order the rows of an invertible matrix so that all of the submatrices have non-zero determinants.

##### Example 8

Reorder the rows of $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$ so that the reordered matrix has an $LU$ decomposition.

##### Solution

Swapping the first and second rows does not help us since the second leading submatrix will still have a zero determinant. Let us swap the second and third rows and consider

$\phantom{\rule{2em}{0ex}}B=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 5\hfill \end{array}\right]$

$\phantom{\rule{2em}{0ex}}{B}_{1}=1,\phantom{\rule{1em}{0ex}}{B}_{2}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 3\hfill \end{array}\right],\phantom{\rule{1em}{0ex}}{B}_{3}=B.$

Now $|{B}_{1}|=1$ , $|{B}_{2}|=3×1-2×1=1$ and (expanding along the first row)

$\phantom{\rule{2em}{0ex}}\left|{B}_{3}\right|=1\left(15-16\right)-2\left(5-8\right)+3\left(4-6\right)=-1+6-6=-1.$

All three of these determinants are non-zero and we conclude that $B$ does have an $LU$ decomposition.

Reorder the rows of $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 7\hfill \\ \hfill -2\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -2\hfill \end{array}\right]$ so that the reordered matrix has an $LU$ decomposition.

Let us swap the second and third rows and consider

$\phantom{\rule{2em}{0ex}}B=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -3\hfill & \hfill 7\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill -2\hfill \\ \hfill -2\hfill & \hfill 6\hfill & \hfill 1\hfill \end{array}\right]$

$\phantom{\rule{2em}{0ex}}{B}_{1}=1,\phantom{\rule{1em}{0ex}}{B}_{2}=\left[\begin{array}{cc}\hfill 1\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 3\hfill \end{array}\right],\phantom{\rule{1em}{0ex}}{B}_{3}=B$

which have determinants $1$ , $3$ and $45$ respectively. All of these are non-zero and we conclude that $B$ does indeed have an $LU$ decomposition.

##### Exercises
1. Calculate $LU$ decompositions for each of these matrices
1. $A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill -6\hfill \end{array}\right]$
2. $A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill -2\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill -11\hfill \end{array}\right]$
3. $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 8\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 11\hfill & \hfill 4\hfill \end{array}\right]$
2. Check each answer in Question 1, by multiplying out $LU$ to show that the product equals $A$ .
3. Using the answers obtained in Question 1, solve the following systems of equations.
1. $\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill -6\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \end{array}\right]$
2. $\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill -4\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill -2\hfill \\ \hfill 6\hfill & \hfill 3\hfill & \hfill -11\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \\ \hfill {x}_{3}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 4\hfill \\ \hfill 0\hfill \\ \hfill 11\hfill \end{array}\right]$
3. $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 8\hfill & \hfill 5\hfill \\ \hfill 1\hfill & \hfill 11\hfill & \hfill 4\hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\hfill \\ \hfill {x}_{2}\hfill \\ \hfill {x}_{3}\hfill \end{array}\right]=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill 3\hfill \\ \hfill 0\hfill \end{array}\right]$
4. Consider $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 6\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 12\hfill & \hfill 5\hfill \\ \hfill -1\hfill & \hfill -3\hfill & \hfill -1\hfill \end{array}\right]$
1. Show that $A$ does not have an $LU$ decomposition.
2. Re-order the rows of $A$ and find an $LU$ decomposition of the new matrix.
3. Hence solve $\begin{array}{rcll}{x}_{1}+6{x}_{2}+2{x}_{3}& =& 9& \text{}\\ 2{x}_{1}+12{x}_{2}+5{x}_{3}& =& -4\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ -{x}_{1}-3{x}_{2}-{x}_{3}& =& 17& \text{}\end{array}$
1. We let

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill -4& \hfill -6\end{array}\right]=LU=\left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill {L}_{21}& \hfill 1\end{array}\right]\left[\begin{array}{cc}\hfill {U}_{11}& \hfill {U}_{12}\\ \hfill 0& \hfill {U}_{22}\end{array}\right]=\left[\begin{array}{cc}\hfill {U}_{11}& \hfill {U}_{12}\\ \hfill {L}_{21}{U}_{11}& \hfill {L}_{21}{U}_{12}+{U}_{22}\end{array}\right].$

Comparing the left-hand and right-hand sides row by row gives us that ${U}_{11}=2$ , ${U}_{12}=1$ , ${L}_{21}{U}_{11}=-4$ which implies that ${L}_{21}=-2$ and, finally, ${L}_{21}{U}_{12}+{U}_{22}=-6$ from which we see that ${U}_{22}=-4$ . Hence

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill -4& \hfill -6\end{array}\right]=\left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill -2& \hfill 1\end{array}\right]\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill 0& \hfill -4\end{array}\right]$

is an $LU$ decomposition of the given matrix.

2. We let

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 2& \hfill 1& \hfill -4\\ \hfill 2& \hfill 2& \hfill -2\\ \hfill 6& \hfill 3& \hfill -11\end{array}\right]=LU=\left[\begin{array}{ccc}\hfill {U}_{11}& \hfill {U}_{12}& \hfill {U}_{13}\\ \hfill {L}_{21}{U}_{11}& \hfill {L}_{21}{U}_{12}+{U}_{22}& \hfill {L}_{21}{U}_{13}+{U}_{23}\\ \hfill {L}_{31}{U}_{11}& \hfill {L}_{31}{U}_{12}+{L}_{32}{U}_{22}& \hfill {L}_{31}{U}_{13}+{L}_{32}{U}_{23}+{U}_{33}\end{array}\right].$

Looking at the top row we see that ${U}_{11}=2$ , ${U}_{12}=1$ and ${U}_{13}=-4$ . Now, from the second row, ${L}_{21}=1$ , ${U}_{22}=1$ and ${U}_{23}=2$ . The last three unknowns come from the bottom row: ${L}_{31}=3$ , ${L}_{32}=0$ and ${U}_{33}=1$ . Hence

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 2& \hfill 1& \hfill -4\\ \hfill 2& \hfill 2& \hfill -2\\ \hfill 6& \hfill 3& \hfill -11\\ \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 3& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 2& \hfill 1& \hfill -4\\ \hfill 0& \hfill 1& \hfill 2\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill \end{array}\right]$

is an $LU$ decomposition of the given matrix.

3. We let

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 2\\ \hfill 2& \hfill 8& \hfill 5\\ \hfill 1& \hfill 11& \hfill 4\end{array}\right]=LU=\left[\begin{array}{ccc}\hfill {U}_{11}& \hfill {U}_{12}& \hfill {U}_{13}\\ \hfill {L}_{21}{U}_{11}& \hfill {L}_{21}{U}_{12}+{U}_{22}& \hfill {L}_{21}{U}_{13}+{U}_{23}\\ \hfill {L}_{31}{U}_{11}& \hfill {L}_{31}{U}_{12}+{L}_{32}{U}_{22}& \hfill {L}_{31}{U}_{13}+{L}_{32}{U}_{23}+{U}_{33}\end{array}\right].$

Looking at the top row we see that ${U}_{11}=1$ , ${U}_{12}=3$ and ${U}_{13}=2$ . Now, from the second row, ${L}_{21}=2$ , ${U}_{22}=2$ and ${U}_{23}=1$ . The last three unknowns come from the bottom row: ${L}_{31}=1$ , ${L}_{32}=4$ and ${U}_{33}=-2$ . Hence

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 2\\ \hfill 2& \hfill 8& \hfill 5\\ \hfill 1& \hfill 11& \hfill 4\\ \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 2& \hfill 1& \hfill 0\\ \hfill 1& \hfill 4& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 2\\ \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill 0& \hfill -2\\ \hfill \end{array}\right]$

is an $LU$ decomposition of the given matrix.

1. Direct multiplication provides the necessary check.
1. We begin by solving

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill -2& \hfill 1\end{array}\right]\left[\begin{array}{c}\hfill {y}_{1}\\ \hfill {y}_{2}\end{array}\right]=\left[\begin{array}{c}\hfill 1\\ \hfill 2\end{array}\right]$

Clearly ${y}_{1}=1$ and therefore ${y}_{2}=4$ . The values ${y}_{1}$ and ${y}_{2}$ appear on the right-hand side of the second system we need to solve:

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill 0& \hfill -4\end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\\ \hfill {x}_{2}\end{array}\right]=\left[\begin{array}{c}\hfill 1\\ \hfill 4\end{array}\right]$

The second equation implies that ${x}_{2}=-1$ and therefore, from the first equation, ${x}_{1}=1$ .

2. We begin by solving the system

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 3& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill {y}_{1}\\ \hfill {y}_{2}\\ \hfill {y}_{3}\end{array}\right]=\left[\begin{array}{c}\hfill 4\\ \hfill 0\\ \hfill 11\end{array}\right].$

Starting with the top equation we see that ${y}_{1}=4$ . The second equation then implies that ${y}_{2}=-4$ and then, from the third equation, ${y}_{3}=-1$ . These values now appear on the right-hand side of the second system

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 2& \hfill 1& \hfill -4\\ \hfill 0& \hfill 1& \hfill 2\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\\ \hfill {x}_{2}\\ \hfill {x}_{3}\end{array}\right]=\left[\begin{array}{c}\hfill 4\\ \hfill -4\\ \hfill -1\end{array}\right].$

The bottom equation shows us that ${x}_{3}=-1$ . Moving up to the middle equation we obtain ${x}_{2}=-2$ . The top equation yields ${x}_{1}=1$ .

3. We begin by solving the system

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill 2& \hfill 1& \hfill 0\\ \hfill 1& \hfill 4& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill {y}_{1}\\ \hfill {y}_{2}\\ \hfill {y}_{3}\end{array}\right]=\left[\begin{array}{c}\hfill 2\\ \hfill 3\\ \hfill 0\end{array}\right].$

Starting with the top equation we see that ${y}_{1}=2$ . The second equation then implies that ${y}_{2}=-1$ and then, from the third equation, ${y}_{3}=2$ . These values now appear on the right-hand side of the second system

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill 3& \hfill 2\\ \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill 0& \hfill -2\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\\ \hfill {x}_{2}\\ \hfill {x}_{3}\end{array}\right]=\left[\begin{array}{c}\hfill 2\\ \hfill -1\\ \hfill 2\end{array}\right].$

The bottom equation shows us that ${x}_{3}=-1$ . Moving up to the middle equation we obtain ${x}_{2}=0$ . The top equation yields ${x}_{1}=4$ .

1. The second leading submatrix has determinant $1×12-6×2=0$ and this implies that $A$ has no $LU$ decomposition.
2. Swapping the second and third rows gives $\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill -1& \hfill -3& \hfill -1\\ \hfill 2& \hfill 12& \hfill 5\end{array}\right].$ We let

$\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill -1& \hfill -3& \hfill -1\\ \hfill 2& \hfill 12& \hfill 5\end{array}\right]=LU=\left[\begin{array}{ccc}\hfill {U}_{11}& \hfill {U}_{12}& \hfill {U}_{13}\\ \hfill {L}_{21}{U}_{11}& \hfill {L}_{21}{U}_{12}+{U}_{22}& \hfill {L}_{21}{U}_{13}+{U}_{23}\\ \hfill {L}_{31}{U}_{11}& \hfill {L}_{31}{U}_{12}+{L}_{32}{U}_{22}& \hfill {L}_{31}{U}_{13}+{L}_{32}{U}_{23}+{U}_{33}\end{array}\right].$

Looking at the top row we see that ${U}_{11}=1$ , ${U}_{12}=6$ and ${U}_{13}=2$ . Now, from the second row, ${L}_{21}=-1$ , ${U}_{22}=3$ and ${U}_{23}=1$ . The last three unknowns come from the bottom row: ${L}_{31}=2$ , ${L}_{32}=0$ and ${U}_{33}=1$ . Hence

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill -1& \hfill -3& \hfill -1\\ \hfill 2& \hfill 12& \hfill 5\\ \hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill 0& \hfill 3& \hfill 1\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill \end{array}\right]$

is an $LU$ decomposition of the given matrix.

3. We begin by solving the system

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill {y}_{1}\\ \hfill {y}_{2}\\ \hfill {y}_{3}\end{array}\right]=\left[\begin{array}{c}\hfill 9\\ \hfill 17\\ \hfill -4\end{array}\right].$

(Note that the second and third rows of the right-hand side vector have been swapped too.) Starting with the top equation we see that ${y}_{1}=9$ . The second equation then implies that ${y}_{2}=26$ and then, from the third equation, ${y}_{3}=-22$ . These values now appear on the right-hand side of the second system

$\phantom{\rule{2em}{0ex}}\left[\begin{array}{ccc}\hfill 1& \hfill 6& \hfill 2\\ \hfill 0& \hfill 3& \hfill 1\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill \end{array}\right]\left[\begin{array}{c}\hfill {x}_{1}\\ \hfill {x}_{2}\\ \hfill {x}_{3}\end{array}\right]=\left[\begin{array}{c}\hfill 9\\ \hfill 26\\ \hfill -22\end{array}\right].$

The bottom equation shows us that ${x}_{3}=-22$ . Moving up to the middle equation we obtain ${x}_{2}=16$ . The top equation yields ${x}_{1}=-43$ .