dublattm

2022-01-22

Solve the given system of equations, or else show that there is no solution.
a) ${x}_{1}+2{x}_{2}-{x}_{3}=1$
$2{x}_{1}+{x}_{2}+{x}_{3}=1$
${x}_{1}-{x}_{2}+2{x}_{3}=1$
b) ${x}_{1}+2{x}_{2}-{x}_{3}=-2$
$-2{x}_{1}-4{x}_{2}+2{x}_{3}=4$
$2{x}_{1}+4{x}_{2}-2{x}_{3}=-4$

### Answer & Explanation

coolbananas03ok

Step 1
Consider the following system:
${x}_{1}+2{x}_{2}-{x}_{3}=1$
$2{x}_{1}+{x}_{2}+{x}_{3}=1$
${x}_{1}-{x}_{2}+2{x}_{3}=1$
Solve the system
Write augmented matrix
$\left[\begin{array}{cccc}1& 2& -1& 1\\ 2& 1& 1& 1\\ 1& -1& 2& 1\end{array}\right]$
${\stackrel{\to }{{R}_{3}-{R}_{1}}}^{{R}_{2}-2{R}_{1}}$
$\left[\begin{array}{cccc}1& 2& -1& 1\\ 0& -3& 3& -1\\ 0& -3& 3& 0\end{array}\right]$
$\underset{\to }{{R}_{3}-{R}_{2}}$
$\left[\begin{array}{cccc}1& 2& -1& 1\\ 0& -3& 3& -1\\ 0& 0& 0& 1\end{array}\right]$
From the last row, we got $0=1$ which is absqrt.
Hence the system has no solution
Step 2
Consider the following syste:
${x}_{1}+2{x}_{2}-{x}_{3}=-2$
$-2{x}_{1}-4{x}_{2}+2{x}_{3}=4$
$2{x}_{1}+4{x}_{2}-2{x}_{3}=-4$
Solve the system
Write the augmented matrix
$\left[\begin{array}{cccc}1& 2& -1& -2\\ -2& -4& 2& 4\\ 2& 4& -2& -4\end{array}\right]$
$\begin{array}{c}{R}_{2}+2{R}_{1}\\ \to \\ {R}_{3}-2{R}_{1}\end{array}$
$\left[\begin{array}{cccc}1& 2& -1& -2\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$
From this, we get ${x}_{1}+2{x}_{2}-{x}_{3}=-2$
Here, are free variables.
So choose , where s, t are any parameters.
Then ${x}_{1}=-2-2s+t$
Hence the solution to the given system is
$\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)=\left(\begin{array}{c}-2-2s+t\\ s\\ t\end{array}\right)$, where s, t are any parameters

ocretz56

Step 1
Part A
The coefficient matrix is :
$A=\left[\begin{array}{ccc}1& 2& -1\\ 2& 1& 1\\ 1& -1& 2\end{array}\right]$
Now
$det\left(A\right)=\left[\begin{array}{ccc}1& 2& -1\\ 2& 1& 1\\ 1& -1& 2\end{array}\right]=1×\left(2+1\right)-2×\left(4-1\right)-1×\left(-2-1\right)$
$⇒det\left(A\right)=3-6+3=0$
Since $det\left(A\right)=0$, the matrix is not invertible and thus, has no solution.
Step 2
Part B:
The coefficient matrix is:
$A=\left[\begin{array}{ccc}1& 2& -1\\ -2& -4& 2\\ 2& 4& -2\end{array}\right]$
Now
$det\left(A\right)==\left[\begin{array}{ccc}1& 2& -1\\ -2& -4& 2\\ 2& 4& -2\end{array}\right]=1×\left(8-8\right)-2×\left(4-4\right)-1×\left(-8+8\right)$
$⇒det\left(A\right)=0+0+0=0$
Since $det\left(A\right)=0$, the matrix is not invertible and thus, has no solution.

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