Answer
See the answers
Work Step by Step
\[
\left[\begin{array}{cccc}
-5 & 10 & -5 & 4 \\
8 & 3 & -4 & 7 \\
4 & -9 & 5 & -3 \\
-3 & -2 & 5 & 4
\end{array}\right]
\]
Given matrix
\[
\left.\begin{array}{cccc}
-5 & 10 & -5 & 4 \\
0 & 19 & -12 & 67 / 5 \\
0 & -1 & 1 & 1 / 5 \\
0 & -8 & 8 & 8 / 5
\end{array}\right]
\]
$R_{3}=R_{3}+\frac{4}{5} R_{1}$
$R_{4}=R_{4}-\frac{3}{5} R_{1}$
\[
\left[\begin{array}{cccc}
-5 & 10 & -5 & 4 \\
0 & 19 & -12 & 67 / 5 \\
0 & 0 & 7 / 19 & 86 / 95 \\
0 & 0 & 56 / 19 & 688 / 95
\end{array}\right]
\]
$R_{3}=R_{3}+\frac{1}{19} R_{2}$
$R_{4}=R_{4}+\frac{8}{19} R_{2}$
\begin{aligned}
&\left[\begin{array}{cccc}
-5 & 10 & -5 & 4 \\
0 & 19 & -12 & 67 / 5 \\
0 & 0 & 7 / 19 & 86 / 95 \\
0 & 0 & 0 & 0
\end{array}\right]\\
&R_{4}=R_{4}-8 R_{3}\\
&\left[\begin{array}{cccc}
-5 & 10 & 0 & 114 / 7 \\
0 & 19 & 0 & 1501 / 35 \\
0 & 0 & 7 / 19 & 86 / 95 \\
0 & 0 & 0 & 0
\end{array}\right]\\
&R_{1}=R_{1}+\frac{228}{7} R_{3}\\
&R_{2}=R_{2}+\frac{95}{7} R_{3}\\
&\left[\begin{array}{cccc}
-5 & 0 & 0 & -44 / 7 \\
0 & 19 & 0 & 1501 / 35 \\
0 & 0 & 7 / 19 & 86 / 95 \\
0 & 0 & 0 & 0
\end{array}\right]\\
&R_{1}=R_{1}-\frac{10}{19} R_{3}\\
&\left[\begin{array}{cccc}
1 & 0 & 0 & 44 / 35 \\
0 & 1 & 0 & 79 / 35 \\
0 & 0 & 1 & 86 / 35 \\
0 & 0 & 0 & 0
\end{array}\right]
\end{aligned}
Simplifying Matrix by taking common out
\[
\left[\begin{array}{cccc}
1 & 0 & 0 & 1.2571 \\
0 & 1 & 0 & 2.2571 \\
0 & 0 & 1 & 2.4571 \\
0 & 0 & 0 & 0
\end{array}\right]
\]
There is no pivot in the fourth column of the standard matrix A, so the equation $A x=0$ has a nontrivial solution. By Theorem $11,$ the transformation $\mathrm{T}$ is not one-to-one.
