Solving Systems of Linear Equations Using Matrices

The matrix method of solving systems of linear equations can be very handy in a number of situations. Although it''s not the only way to solve linear equations, it provides us with many advantages that are difficult to replicate elsewhere. But how exactly does this method work, and when should we use it? Let''s find out:

How does the matrix method work?

The matrix method allows us to rewrite systems of linear equations in an easier way, reducing complex algebraic operations to simple arithmetic. When you really break it down, it is simply a variant of the elimination method.

Recall that a matrix is a group of numbers. When we place numbers inside a matrix, they become "elements" of that matrix. The advantage of this system is that it lets us do away with variables and focus only on the coefficients. As long as we know that a certain column is associated with a certain variable, there''s no need to rewrite these variables in our operations.

How to solve systems of linear equations with matrices

So how exactly do you use matrices to solve linear equations? Let''s walk through the process, starting with our system of linear equations:

$\{\begin{array}{r}3x+4y=5\\ 2x-y=7\end{array}$

These linear equations are ready to be turned into matrices because they are already in standard form $Ax+By=C$ . If they were not in standard form, we would need to turn them into standard expressions first. This is because when we work with matrices, everything must line up in the same columns. Let''s see what this system looks like when we turn it into a matrix:

$\left[\begin{array}{cc}3& 4\\ 2& -1\end{array}\right|\begin{array}{c}5\\ 7\end{array}]$

The goal here is to turn the matrix on the left side into an identity matrix. Recall that an identity matrix is essentially the matrix equivalent of the value "1." If we can turn the left side into an identity matrix, it means that we are multiplying the right side by 1. This means that the right side must equal our x and y value -- which is exactly what we need to solve this system of linear equations.

In order to turn the left side of the matrix into an identity matrix, we need the main diagonal terms to equal $1$ . All other values must equal $0$ . Let''s start with row 1, column 1. We know that in order to make this value a "0," we need to add $4$ . We can do this by multiplying row 2 by four and adding it to row 1. Note that this will leave us with an $11$ in the top left corner (because $2\times 4+3=11$ ). Also, note that this gives us a value of $33$ at the top of column 3 (the constant matrix) because $7\times 4+5=33$ .

$\left[\begin{array}{cc}11& 0\\ 2& -1\end{array}\right|\begin{array}{c}33\\ 7\end{array}]$

Now let''s focus on the top left corner. In order to form an identity matrix, we need to make this value a "1." We can do this by dividing row 1 by 11. This leaves us with the following:

$\left[\begin{array}{cc}1& 0\\ 2& -1\end{array}\right|\begin{array}{c}3\\ 7\end{array}]$

Great! Our identity matrix is starting to take shape. Let''s move on to the bottom left corner. We need a zero for this element, so we''re going to multiply row 1 by (-2) and add it to row 2. This leaves us with the following:

$\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right|\begin{array}{c}3\\ 1\end{array}]$

Our last operation involves the bottom right corner. We need a value of "1" for this element, so we''re going to multiply row 2 by (-1). Here is the final result:

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|\begin{array}{c}3\\ -1\end{array}]$

This is essentially telling us that This is essentially telling us that $3\times 1=x$ and $-1\times 1=y$ . Remember that the entire identity matrix is equivalent to the value of "1."

In other words:

$\begin{array}{c}x=3\\ y=-1\end{array}$

We have just solved our system of linear equations using the matrix method.

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Pair your student with a tutor who understands how to solve systems of linear equations with matrices

Using matrices to solve systems of linear equations can be very handy, especially as students progress further into more advanced mathematics. However, it can easily catch students off guard, since there are a few very notable differences between matrix operations and normal arithmetic or algebra. Taking the time to revisit key concepts and practice skills during 1-on-1 tutoring sessions allows students to pursue greater confidence. Students can ask as many questions as they like during these tutoring sessions, and they can progress at a manageable yet engaging pace that matches their ability level. Speak with our Educational Directors today to learn more, and rest assured: Varsity Tutors will pair your student with a suitable tutor.