\( \begin{bmatrix} {\color{blue}0.05} & 0.10833 & 0.00833 & 0 & 0 \\ {\color{red}0.10833} & 0.5 & 0.21666 & 0.00833 & 0 \\ {\color{blue}0.00833} & 0.21666 & 0.55 & 0.21666 & 0.00833 \\ {\color{blue}0} & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ {\color{blue}0} & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix}1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} \)
We start with the row pivoting operation. In the first column (marked in blue and red) we are looking for the highest value per module. It is an expression \( A_{2,1}=0.10833 \) (marked in red). So we swap the first line with the second one
\( \begin{bmatrix} {\color{red}0.10833} & 0.5 & 0.21666 & 0.00833 & 0 \\ {\color{blue}0.05} & 0.10833 & 0.00833 & 0 & 0 \\ {\color{blue}0.00833} & 0.21666 & 0.55 & 0.21666 & 0.00833 \\ {\color{blue}0} & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ {\color{blue}0} & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix}1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} \)
Now, we are performing the step of the "normal" Gaussian elimination algorithm. We scale the first row to get one on the diagonal
\( 1^{st}=1^{st}A_{1,1}=1^{st}\frac{1}{0.10833}=1^{st}9.231053 \)
\( \begin{bmatrix} 0.10833{\color{red}*9.23105} & 0.5{\color{red}*9.23105} & 0.21666{\color{red}*9.23105} & 0.00833{\color{red}*9.23105} & 0{\color{red}*9.23105} \\ 0.05 & 0.10833 & 0.00833 & 0 & 0 \\ 0.00833 & 0.21666 & 0.55 & 0.21666 & 0.00833 \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 1{\color{red}*9.23105} \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} {\color{red}1} & {\color{red}4,61553} & {\color{red}2.00006} & {\color{red}0,07693} & {\color{red}0} \\ 0.05 & 0.10833 & 0.00833 & 0 & 0 \\ 0.00833 & 0.21666 & 0.55 & 0.21666 & 0.00833 \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix}{\color{red}9.2311} \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} \)
Then we subtract the first row from the second to get zero in position \( A_{2,1} \) \( 2^{nd}=2^{nd}-1^{st}A_{2,1}=2^{nd}-1^{st}0.05 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ {\color{red}0} & 0.10833{\color{red}-4.61553*0.05} & 0.00833{\color{red}-2*0.05} & 0{\color{red}-0.07693*0.05} & 0 \\ 0.00833 & 0.21666 & 0.55 & 0.21666 & 0.00833 \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix}9.23105 \\ 1{\color{red}-9.23105*0.05} \\ 1 \\ 1 \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ {\color{red}0} & {\color{red}-0.12244} & {\color{red}-0.09167} & {\color{red}-0,00385} & 0 \\ 0.00833 & 0.21666 & 0.55 & 0.21666 & 0.00833 \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix}\begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ {\color{red}0.53844} \\ 1 \\ 1 \\ 1 \end{bmatrix} \)
We now subtract the first row from the third to get zero in position \( A_{3,1} \)
\( 3^{rd}=3^{rd}-1^{st}A_{3,1}=3^{rd}-1^{st}0.00833 \) \( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & -0.12244 & -0.09167 & -0,00385 & 0 \\ {\color{red}0} & 0.21666{\color{red}-4.61552*0.00833} & 0.55{\color{red}-2*0.00833} & 0.21666 {\color{red}-0.07689*0.00833} & 0.00833{\color{red}-0} \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 0.53844 \\ 1{\color{red}-9.23105*0.00833} \\ 1 \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & -0.12244 & -0.09167 & -0,00385& 0 \\ {\color{red}0} & {\color{red}0.17821} & {\color{red}0.53334} & {\color{red}0.21603} & {\color{red}0.00833} \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 0.53844 \\ {\color{red}0.92310} \\ 1 \\ 1 \end{bmatrix} \)
We are now doing the row pivoting operation again.
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & {\color{blue}-0.12244} & -0.09167 & -0,00385 & 0 \\ 0 & {\color{red}0.17821} & 0.53334 & 0.21603 & 0.00833 \\ 0 & {\color{blue}0.00833} & 0.21666 & 0.5 & 0.10833 \\ 0 & {\color{blue}0} & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix}9.23105 \\ 0.53844 \\ 0.92310 \\ 1 \\ 1 \end{bmatrix} \)
In the second column from the diagonal down (marked in blue and red) we are looking for the highest value per module. It is an expression A(3,1)=0.17821 (marked in red) because \( |0.17821|>|-0.12244|=0.12244 \). So we swap the second row with the third one.
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & {\color{red}0.17821} & 0.53334 & 0.21603 & 0.00833 \\ 0 & {\color{blue}-0.12244} & -0.09167 & -0,00385 & 0 \\ 0 & {\color{blue}0.00833} & 0.21666 & 0.5 & 0.10833 \\ 0 & {\color{blue}0} & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 0.92310 \\ 0.53844 \\ 1 \\ 1 \end{bmatrix} \)
Now, we scale the new second row to get one on the diagonal \( 2^{nd}=2^{nd}\frac{1}{A_{2,2}}=2^{nd}\frac{1}{0.17821}=2^{nd}5.61135 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 0.17821{\color{red}*5.61135} & 0.53334{\color{red}*5.61135} & 0.21603{\color{red}*5.61135} & 0.00833{\color{red}*5.61135} \\ 0 & -0.12244 & -0.09167 & -0,00385 & 0 \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 0.92310{\color{red}*5.61135} \\ 0.53844 \\ 1 \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & {\color{red}1} & {\color{red}2.99275} & {\color{red}1.21222} & {\color{red}0.046742} \\ 0 & -0.12244 & -0.09167 & -0,00385 & 0 \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ {\color{red}5.17983} \\ 0.53844 \\ 1 \\ 1 \end{bmatrix} \)
We now subtract the second row from the third scaled to get zero in position
\( A_{3,2}=-0.12244 \)
\( 3^{rd}=3^{rd}-2^{nd}A_{3,2}=3^{rd}-2^{nd}(-0.12244) \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & {\color{red}0} & -0.09167{\color{red}+2.99275*0.12244} & -0,00385{\color{red}+1.21222*0.12244} & 0{\color{red}+0.046742*0.12244} \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 0.53844{\color{red}+5.17983*0.12244} \\ 1 \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & {\color{red}0} & {\color{red}0.27479} & {\color{red}0.14459} & {\color{red}0.00573} \\ 0 & 0.00833 & 0.21666 & 0.5 & 0.10833 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.1800 \\ {\color{red}1.1727} \\ 1 \\ 1 \end{bmatrix} \)
Likewise, we subtract the second row from the fourth
\( 4^{th}=4^{th}-2^{nd}A_{(4,2}=4^{th}-2^{nd}0.00833 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 0.27476 & 0.14457 & 0.005723 \\ 0 & {\color{red}0} & 0.21666{\color{red}-2.99275*0.00833} & 0.5{\color{red}-1.21210*0.00833} & 0.10833{\color{red}-0.046742*0.00833} \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 1.17265 \\ 1{\color{red}-5.17983*0.00833} \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 0.27476 & 0.14457 & 0.005723 \\ 0 & {\color{red}0} & {\color{red}0.19173} & {\color{red}0.48990} & {\color{red}0.10794} \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix}9.23105 \\ 5.17999 \\ 1.17270 \\ {\color{red}0.95684} \\ 1 \end{bmatrix} \)
We are left with a 3 by 3 sub matrix in the lower right corner, for which we continue Gaussian elimination with row pivoting. We first look for the maximum (per module) value in the third column from the diagonal down (marked in red and blue). It is the value from the diagonal \( A_{3,3} \) marked in red, so we don't need to swap the rows. \( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & {\color{red}0.27476} & 0.14457 & 0.005723 \\ 0 & 0 & {\color{blue}0.19173} & 0.48990 & 0.10794 \\ 0 & 0 & {\color{blue}0.00833} & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 1.17265 \\ 0.95685 \\ 1 \end{bmatrix} \)
We now scale row three to get one on the diagonal
\( 3^{rd}=3^{rd}\frac{1}{A_{3,3}}=3^{rd}\frac{1}{0.27476}=3^{rd}3.63953 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 0.27476{\color{red}*3.63953} & 0.14457{\color{red}*3.63953} & 0.005723{\color{red}*3.63953} \\ 0 & 0 & 0.19173 & 0.48990 & 0.10794 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 1.17265{\color{red}*3.63953} \\ 0.95685 \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & {\color{red}1} & {\color{red}0.52617} & {\color{red}0.02082} \\ 0 & 0 & 0.19173 & 0.48990 & 0.10794 \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17999 \\ {\color{red}4.26763} \\ 0.95683 \\ 1 \end{bmatrix} \)
We will now use the third row to generate the zeros in the third column below the diagonal
\( 4^{th}=4^{th}-3^{rd}A_{4,3}=4^{th}-3^{rd}0.19173 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & {\color{red}0} & 0.48990{\color{red}-0.52618*0.19173} & 0.10794{\color{red}-0.02083*0.19173} \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ 0.95685{\color{red}-4.26789*0.19173} \\ 1 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & {\color{red}0} & {\color{red}0.38901} & {\color{red}0.10395} \\ 0 & 0 & 0.00833 & 0.10833 & 0.05\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ {\color{red}0.13860} \\ 1 \end{bmatrix} \)
\( 5^{th}=5^{th}-3^{rd}A_{5,3}=5^{th}-3^{rd}0.00833 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & 0 & 0.38901 & 0.10395 \\ 0 & 0 & {\color{red}0} & 0.10833{\color{red}-0.52618*0.00833} & 0.05{\color{red}-0.02082*0.00833}\\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ 0.13860 \\ 1{\color{red}-4.26789*0.00833} \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & 0 & 0.38901 & 0.10395 \\ 0 & 0 & 0 & {\color{red}0.10395} & {\color{red}0.04983}\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ 0.13860 \\ {\color{red}0.96445} \end{bmatrix} \)
Then we scale the fourth row so that it has a one on the diagonal
\( 4^{th}=4^{th}\frac{1}{A_{4,4}}=4^{th}\frac{1}{0.38901}=4^{th}2.5706 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & 0 & 0.38365{\color{red}*2.5706} & 0.10395{\color{red}*2.5706} \\ 0 & 0 & 0 & 0.10395 & 0.04982\\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ {\color{red}0.13860*2.5706} \\ 0.91110 \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & 0 & {\color{red}1} & {\color{red}0.26721} \\ 0 & 0 & 0 & 0.10395 & 0.04983 \\ \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ {\color{red}0.35629} \\ 0.96444 \end{bmatrix} \)We will now use the fourth row to generate a zero in the fourth column below the diagonal \( 5^{th}=5^{th}-4^{rd}A_{5,4}=5^{th}-4^{rd}0.10395 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & 0 & 1 & 0.26721 \\ 0 & 0 & 0 & {\color{red}0} & 0.04983{\color{red}-0.26721*0.10395} \\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ 0.35629\\ 0.96444{\color{red}-0.35629*0.10395} \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & 0 & 1 & 0.26721 \\ 0 & 0 & 0 & {\color{red}0} & {\color{red}0.022053} \\ \end{bmatrix} \nonumber\\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ 0.35629 \\ {\color{red}0.92740} \end{bmatrix} \)
Finally, we scale the fifth row so that it has a one on the diagonal
\( 5^{th}=5^{th}\frac{1}{A_{5,5}}=5^{th}\frac{1}{0.022053}=5^{th}45.351 \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & 0 & 1 & 0.26721 \\ 0 & 0 & 0 & {\color{red}0} & {\color{red}1} \\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ 0.35629\\ 0.92740{\color{red}*45.351} \end{bmatrix} \)
\( \begin{bmatrix} 1 & 4.61553 & 2 & 0.07693 & 0 \\ 0 & 1 & 2.99287 & 1.21226 & 0.04676 \\ 0 & 0 & 1 & 0.52617 & 0.02082 \\ 0 & 0 & 0 & 1 & 0.26721 \\ 0 & 0 & 0 & {\color{red}0} & {\color{red}1} \\ \end{bmatrix} \\ \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \\ u_{5} \\ \end{bmatrix} = \begin{bmatrix} 9.23105 \\ 5.17983 \\ 4.26789 \\ 0.35629\\ {\color{red}42.059} \end{bmatrix} \)
Writes \( 1u_1+4.61553u_2+2u_3+0.07693u_4=9.23105 \\ 1u_2+2.99287u_3+1.21226u_4+0.04676u_5=5.17983 \\ 1u_3+0.52617u_4+0.02082u_5=4.26789 \\ 1u_4+0.26721u_5=0.35629 \\ 1u_5=42.059 \)
And we solve from the last equation
\( u_5={\color{blue}42.059} \\ u_4=0.35629-0.26721u_5 =0.35629-0.26721*{\color{blue}42.059}={\color{red}-10.88229} \\ u_3=4.26789-0.52617u_4-0.02082u_5 =4.26789-0.52617{\color{red}-10.88229}-0.02082{\color{blue}42.059}={\color{green}9.11815} \\ u_2=5.17983-2.99287u_3-1.21226u_4-0.04676u_5= \\ =5.17983-2.99287{\color{green}9.11815}-1.21226{\color{red}-10.88229}-0.04676{\color{blue}42.059} = {\color{cyan}-10,88412} \\ u_1=9.23105-4.61553u_2-2u_3-0.07693u_4 = 42.0679 \)
This time our solution is correct.