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I just found this website on the Internet and you are really my last chance to help me with one task. For $ a \in \mathbb{R}$, let the following linear equation system be given: $$ \begin{array}{rcrcrcc} (a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a-2)x_{3} & = & 1\\ (a+1)(a-3)x_{1} & + & (a^2-6a+9)x_{2} & + & 3x_{3} & = & a-3\\ (a+1)x_{1} & + & (-a^2+6a-9)x_{2} & + & (a+1)x_{3} & = & 1\\ \end{array} $$ Now I should bring that in the Matrix form $\mathbf{A}x=b$, which isn't a problem. $$ \left(\begin{array}{ccc} (a+1) & (-a^2+6a-9) & (a-2) \\ (a+1)(a-3) & (a^2-6a+9) & 3 \\ (a+1) & (-a^2+6a-9) & (a+1) \end{array}\right) \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} 1 \\ a-3 \\ 1 \end{pmatrix} $$

But now I should bring it to the advanced Coefficient matrix in row form. Can someone explain to me how to do it? Thank you very much!

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  • $\begingroup$ Thank you, I saw that, i just need a hint how to bring it to the coefficient matrix $\endgroup$
    – Alice
    Jan 31 '19 at 8:36
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This is a situation where Cramer's Rule applies quite nicely.

First, note that $$ \det(A) = 3 \, {\left(a + 1\right)} {\left(a - 2\right)} {\left(a - 3\right)}^{2} $$ So, $A$ is invertible as long as $a\notin\{-1, 2, 3\}$.

Next, define $A_i$ as the matrix obtained by replacing the $i$th column of $A$ with the vector $\vec{b}=\left\langle1,\,a - 3,\,1\right\rangle$. This gives $$ \begin{align*} \det(A_1) &= 3 \, {\left(a - 2\right)} {\left(a - 3\right)}^{2} & \det(A_2) &= 0 & \det(A_3) &= 0 \end{align*} $$ By Cramer's Rule, we have $$ \begin{align*} x_1 &= \frac{\det(A_1)}{\det(A)} = \frac{1}{a + 1} & x_2 &= \frac{\det(A_2)}{\det(A)} = 0 & x_3 &= \frac{\det(A_3)}{\det(A)} = 0 \end{align*} $$ Of course, we should also account for what happens if $a\in\{-1, 2, 3\}$.

For $a=-1$, we can row-reduce the system to obtain $$ \operatorname{rref}\left[\begin{array}{rrr|r} 0 & -16 & -3 & 1 \\ 0 & 16 & 3 & -4 \\ 0 & -16 & 0 & 1 \end{array}\right] =\left[\begin{array}{rrr|r} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] $$ For $a=2$, we have $$ \operatorname{rref}\left[\begin{array}{rrr|r} 3 & -1 & 0 & 1 \\ -3 & 1 & 3 & -1 \\ 3 & -1 & 3 & 1 \end{array}\right] =\left[\begin{array}{rrr|r} 1 & -\frac{1}{3} & 0 & \frac{1}{3} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ For $a=3$, we have $$ \operatorname{rref}\left[\begin{array}{rrr|r} 4 & 0 & 1 & 1 \\ 0 & 0 & 3 & 0 \\ 4 & 0 & 4 & 1 \end{array}\right] =\left[\begin{array}{rrr|r} 1 & 0 & 0 & \frac{1}{4} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$

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Problem statement

Solve the linear system $\mathbf{A}x=b$: $$ \left[\begin{array}{crc} a+1 & -(a-3)^2 & a-2 \\ (a-3) (a+1) & (a-3)^2 & 3 \\ a+1 & -(a-3)^2 & a+1 \end{array}\right] \left[\begin{array}{c} x_{1} \\ x_{2} \\x_{3} \end{array}\right] = \left[\begin{array}{c} 1 \\ a-3 \\ 1 \end{array}\right]. $$

Gaussian Elimination

Two reduction steps are used resulting in $$ \mathbf{E}_{2}\mathbf{E}_{1}\mathbf{A}x=\mathbf{E}_{2}\mathbf{E}_{1}b $$ where $$ \mathbf{E}_{1} = \left[ \begin{array}{ccc} \frac{1}{a+1} & 0 & 0 \\ 0 & \frac{1}{a+1} & 0 \\ 0 & 0 & \frac{1}{a+1} \\ \end{array} \right], \qquad \mathbf{E}_{2} = \left[ \begin{array}{rcc} 1 & 0 & 0 \\ -1 & \frac{1}{a-3} & 0 \\ -1 & 0 & 1 \\ \end{array} \right]. $$

Solution via back substitution

The final linear system is $$ \left[ \begin{array}{ccc} 1 & -\frac{(a-3)^2}{a+1} & \frac{a-2}{a+1} \\ 0 & \frac{a^2-5 a+6}{a+1} & \frac{a^2-5 a+3}{-a^2+2 a+3} \\ 0 & 0 & \frac{3}{a+1} \\ \end{array} \right] \left[\begin{array}{c} x_{1} \\ x_{2} \\x_{3} \end{array}\right] = \left[\begin{array}{c} \frac{1}{a+1} \\ 0 \\ 0 \end{array}\right]. $$ Solution vis back substitution produces $$ \left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] = \left[\begin{array}{c} \frac{1}{a+1} \\ 0 \\ 0 \end{array}\right], \qquad a\ne1. $$

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