# Matrices: help with homework

I need to prove that $$\vec{x}$$ is a solution of $$A\vec{x}=\vec{b}$$:

$$\begin{vmatrix} 2&-7&-3\\ -4&1&5\\ 1&3&-1\\ \end{vmatrix} \cdot \begin{vmatrix} 5\\ -1\\ 7\\ \end{vmatrix} = \begin{vmatrix} -4\\ 14\\ -5\\ \end{vmatrix}$$

I've done scalar multiplication and have gotten the right answer: $$\begin{vmatrix} 2\cdot5+(-7)\cdot(-1)+(-3)\cdot7\\ (-4)\cdot5+1\cdot(-1)+5\cdot7\\ 1\cdot5+3\cdot(-1)+(-1)\cdot7\\ \end{vmatrix} = \begin{vmatrix} -4\\ 14\\ -5\\ \end{vmatrix}$$ My question is how would I go the other way, with only $$A$$ and $$\vec{b}$$ get $$\vec{x}$$?

• You can use Gaussian elimination for this. – st.math Mar 23 at 10:20
• Multiply both sides by the inverse of A? – John_dydx Mar 23 at 10:20
• Solve the linear equations. – amitava Mar 23 at 10:20
• Take a look at brilliant.org/wiki/gaussian-elimination – SinTan1729 Mar 23 at 10:21
• Three excellent (if somewhat brief) answers. All posted as comments. – Arthur Mar 23 at 10:21

You can reduce the augmented matrix to the Row Reduced Echelon Form (RREF). The RREF will have a row of zeros which means there is a free variable. So the system has infinitely many solutions. You can pick an "appropriate" value of the free variable and show that (5,-1,7) is one of the solutions. This process of solving the system is called Gauss-Jordan elimination.

Since it is a homework question, I cannot give out well-written complete solution. However, you can find many examples on Gauss-Jordan elimination if you google it.