Question

Suppose the coefficient matrix of a linear system of four equations in four variables has a pivot in each column. Explain why the system has a unique solution. What must be true of a linear system for it to have a unique​ solution? Select all that apply.

Answer 1

If the coefficient matrix has a pivot in each column, it means that it is shaped like this:

$A=\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]$

So, the correspondant system

$Ax = b$

will look like this:

$\left[\begin{array}{cccc}a_{1,1}&a_{1,2}&a_{1,3}&a_{1,4}\\0&a_{2,2}&a_{2,3}&a_{2,4}\\0&0&a_{3,3}&a_{3,4}\\0&0&0&a_{4,4}\end{array}\right]\cdot \left[\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right] = \left[\begin{array}{c}b_1\\b_2\\b_3\\b_4\end{array}\right]$

This turn into the following system of equations:

$\begin{cases}a_{1,1}x_1+a_{1,2}x_2+a_{1,3}x_3+a_{1,4}x_4=b_1\\a_{2,2}x_2+a_{2,3}x_3+a_{2,4}x_4=b_2\\a_{3,3}x_3+a_{3,4}x_4=b_3\\a_{4,4}x_4=b_4\end{cases}$

The last equation is solvable for $x_4$: we easily have

$x_4=\dfrac{b_4}{a_{4,4}}$

Once the value for $x_4$ is known, we can solve the third equation for $x_3$:

$x_3 = \dfrac{b_3-a_{3,4}x_4}{a_{3,3}}$

(recall that $x_4$ is now known)

The pattern should be clear: you can use the last equation to solve for $x_4$. Once it is known, the third equation involves the only variable $x_3$. Once