# How to bring a Matrix to an advanced Coefficient matrix in row form?

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!

• Thank you, I saw that, i just need a hint how to bring it to the coefficient matrix Jan 31 '19 at 8:36

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 $$\operatorname{rref}\left[\begin{array}{rrr|r} 4 & 0 & 1 & 1 \\ 0 & 0 & 3 & 0 \\ 4 & 0 & 4 & 1 \end{array}\right] =\left[\begin{array}{rrr|r} 1 & 0 & 0 & \frac{1}{4} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

## 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 $$\left[\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] = \left[\begin{array}{c} \frac{1}{a+1} \\ 0 \\ 0 \end{array}\right], \qquad a\ne1.$$