Respuesta :
If the coefficient matrix has a pivot in each column, it means that it is shaped like this:
[tex]A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right][/tex]
So, the correspondant system
[tex]Ax = b[/tex]
will look like this:
[tex]\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right][/tex]
This turn into the following system of equations:
[tex]\begin{cases}a_{1,1}x_1+a_{1,2}x_2+a_{1,3}x_3+a_{1,4}x_4=b_1\\a_{2,2}x_2+a_{2,3}x_3+a_{2,4}x_4=b_2\\a_{3,3}x_3+a_{3,4}x_4=b_3\\a_{4,4}x_4=b_4\end{cases}[/tex]
The last equation is solvable for [tex]x_4[/tex]: we easily have
[tex]x_4=\dfrac{b_4}{a_{4,4}}[/tex]
Once the value for [tex]x_4[/tex] is known, we can solve the third equation for [tex]x_3[/tex]:
[tex]x_3 = \dfrac{b_3-a_{3,4}x_4}{a_{3,3}}[/tex]
(recall that [tex]x_4[/tex] is now known)
The pattern should be clear: you can use the last equation to solve for [tex]x_4[/tex]. Once it is known, the third equation involves the only variable [tex]x_3[/tex]. Once
