# How To Solve A Matrix Problem $\require\left[ \right]$ To convert it into the final form we will start in the upper left corner and work in a counter-clockwise direction until the first two columns appear as they should be.

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In this part we won’t put in as much explanation for each step.

We will mark the next number that we need to change in red as we did in the previous part.

Before we get into the method we first need to get some definitions out of the way.

An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable. Here is the system of equations that we looked at in the previous section.

This is mostly dependent on the instructor and/or textbook being used.

Next, we need to discuss elementary row operations.This means that we need to change the red three into a zero.This will almost always require us to use third row operation.The second row is the constants from the second equation with the same placement and likewise for the third row.The dashed line represents where the equal sign was in the original system of equations and is not always included.However, the only way to change the -2 into a zero that we had to have as well was to also change the 1 in the lower right corner as well. Sometimes it will happen and trying to keep both ones will only cause problems. $\require\left[ \right]\begin\ \to \end\left[ \right]\begin\ \to \end\left[ \right]$ The solution to this system is then $$x = 2$$ and $$y = 1$$.Let’s first write down the augmented matrix for this system.Let’s start with a system of two equations and two unknowns.$\beginax by & = p\ cx dy & = q\end$ We first write down the augmented matrix for this system, $\left[ \right]$ and use elementary row operations to convert it into the following augmented matrix.If we add -3 times row 1 onto row 2 we can convert that 3 into a 0. $\require\left[ \right]\begin\ \to \end\left[ \right]$ Next, we need to get a 1 into the lower right corner of the first two columns. This is usually accomplished with the second row operation.If we divide the second row by -11 we will get the 1 in that spot that we need.

## Comments How To Solve A Matrix Problem

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