Lecture 3: Systems of Equations

Instructions

• This section covers the concepts listed below. You can click the button to go directly to that topic.
• After each video, there are notes for the material in the video.
• At the end of the page, there are exercises covering the material in the section.
• When you have finished the material below, you can go to the next section or return to the main page
Introduction to Systems of Equations

Examples

1. Solve the equation:\begin{align*}
\begin{pmatrix}
1 & 1 & 0\\
1 & 0 & 1\\
1 & -1 & -1
\end{pmatrix}
\begin{pmatrix}
c_1\\c_2\\c_3
\end{pmatrix}
=\begin{pmatrix}2\\5\\2\end{pmatrix}.
\end{align*}
We can write this as:
\begin{align*}
c_1 + c_2 + 0c_3 = 2\\
c_1 + 0c_2 + c_3 = 5\\
c_1 - c_2 - c_3 = 2.
\end{align*}

We can systematically reduce this to a simpler set of equations in the following way. First, subtract the first equation from the second and third equations to get:
\begin{align*}
c_1 + c_2 + 0c_3 = 2\\
0c_1 - c_2 + c_3 = 3\\
0c_1 - 2c_2 - c_3 = 0.
\end{align*}
Now multiply the second and third equations by $$-1$$:
\begin{align*}
c_1 + c_2 + 0c_3 = 2\\
0c_1 + c_2 - c_3 = -3\\
0c_1 + 2c_2 + c_3 = 0.
\end{align*}
Subtract $$2$$ times the second equation from the third:
\begin{align*}
c_1 + c_2 + 0c_3 = 2\\
0c_1 + c_2 - c_3 = -3\\
0c_1 + 0c_2 + 3c_3 = 6.
\end{align*}
This shows that $$c_3 = 2$$. Then the second equation becomes:
\begin{align*}
c_2 - 2 = -3
\end{align*}
and so $$c_2 = -1$$. These values can be plugged into the first equation to get $$c_1 = 3$$.

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Linear Algebra for Math 308: L3V2

2. Can you find two real numbers $$x_1$$ and $$x_2$$ such that the following equations are true? If you can find such real
numbers, are they unique?
1. \begin{align*}
x_1\begin{pmatrix}1\\0\\1\end{pmatrix} + x_2\begin{pmatrix}1\\2\\0\end{pmatrix}
= \begin{pmatrix}3\\4\\1\end{pmatrix}.
\end{align*}
2. \begin{align*}
x_1\begin{pmatrix}1\\0\\1\end{pmatrix} + x_2\begin{pmatrix}1\\2\\0\end{pmatrix}
= \begin{pmatrix}3\\5\\1\end{pmatrix}.
\end{align*}

1. Observe that this can be written as a system of equations:
\begin{align*}
1x_1 + 1x_2 &= 3\\
0x_1 + 2x_2 &= 4\\
1x_1 + 0x_2 &= 1.
\end{align*}
Subtract the first equation from the third and then multiply the new equation 3 by $$-1$$:
\begin{align*}
1x_1 + 1x_2 &= 3\\
0x_1 + 2x_2 &= 4\\
0x_1 + x_2 &= 2.
\end{align*}
Now subtract two times equation 3 from equation 2, and subtract equation 3 from equation 1:
\begin{align*}
1x_1 + 0x_2 &= 3\\
0x_1 + 0x_2 &= 0\\
0x_1 + x_2 &= 2.
\end{align*}
So $$x+1 = 3$$ and $$x_2 = 2$$. This could have been solve just by looking at the second and third equations, but I wanted to illustrate the systematic approach to solving systems.
2. Observe that this can be written as a system of equations:
\begin{align*}
1x_1 + 1x_2 &= 3\\
0x_1 + 2x_2 &= 5\\
1x_1 + 0x_2 &= 1.
\end{align*}
Subtract the first equation from the third and then multiply the new equation 3 by $$-1$$:
\begin{align*}
1x_1 + 1x_2 &= 3\\
0x_1 + 2x_2 &= 5\\
0x_1 + x_2 &= 2.
\end{align*}
Now subtract two times equation 3 from equation 2, and subtract equation 3 from equation 1:
\begin{align*}
1x_1 + 0x_2 &= 3\\
0x_1 + 0x_2 &= 1\\
0x_1 + x_2 &= 2.
\end{align*}
The second equation says:
\begin{align*}
0x_1 + 0x_2 &= 1.
\end{align*}
That is, $$0 = 1$$; there are no values of $$x_1$$ and $$x_2$$ that make this equation true, so there are no solutions to this equation. This also could have been solve just by looking at the second and third equations, but I wanted to illustrate the systematic approach to solving systems.

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Linear Algebra for Math 308: L3V3

Gauss-Jordan Elimination

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Video Notes

Definition. A pivot is the first non–zero entry in each row.

For example, the pivots in the matrix below are $$1$$, $$17$$, and $$2$$
\begin{align*}
\begin{pmatrix}
1 & 6 & 0 & -1\\
0 & 0 & 17 & 5\\
0 & 0 & 2  & 4\\
\end{pmatrix}
\end{align*}
Definition. A matrix is in echelon form if:
1. All non–zero rows (rows with at least one non–zero element) are above any rows of all zeros.
2. The pivots are all strictly to the right of the pivot of the row above it.
Definition. A matrix is in reduced row echelon form if:
1. The matrix is in row echelon form.
2. All pivots are $$1$$.
3. Each column containing a pivot has zeros everywhere else.
Definition. The three elementary row operations are:
1. Multiply a row by a non–zero number.
2. Swap two rows.
3. Add a multiple of a row to another row.

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Video Notes

The algorithm to get a matrix to row echelon form is the following:
1. Make row swaps so that all non–zero rows are above all zero rows.
2. Make row swaps so that all pivots are to the right of the pivot of the row above it or under the pivot of the row above it.
3. For each pivot, divide the row the pivot is on by the pivot (so that the pivot is now $$1$$. Then do row operations to make the entries under the pivot zero. (That is, add multiples of the row containing the pivot to rows underneath that row to make the entries under the pivot zero). Do this until the matrix is in row echelon form.
The algorithm to get a matrix to reduced row echelon form is the following:
1. Get the matrix to row echelon form.
2. For each pivot, do row operations to make the entries above the pivot zero. (That is, add multiples of the row containing the pivot to rows above that row to make the entries above the pivot zero).
Solving Systems of Equations

Example

Solve the system:
\begin{align*}
x_1 + x_2 + 2x_3 &= 1\\
2x_1 + x_2 + x_3 &= 2\\
x_1 + x_2 + x_3  &= 3
\end{align*}

As an augmented matrix this is:
\begin{align*}
\left(\begin{array}{ccc|c}
1 & 1 & 2 & 1\\
2 & 1 & 1 & 2\\
1 & 1 & 1 & 3
\end{array}\right)
\end{align*}
If we subtract the first equation from the third, and replace the third equation with this new equation (or subtract the first row from the third, and replace the third row with this new row), we get (here and below, we will put the system side--by--side with the augmented matrix):

System:  \begin{align*} x_1 + x_2 + 2x_3 & = 1\\ 2x_1 + x_2 + x_3 & = 2\\ 0 + 0 - x_3 & = 2 \end{align*}    Augmented Matrix:  $$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 1\\ 2 & 1 & 1 & 2\\ 0 & 0 & -1 & 2 \end{array}\right)$$

Now, we should subtract $$2$$ times the first equation from the second equation and replace the second equation with this new equation (or subtract $$2$$ times for first row from the second and replace the second row with this new row):

System:  \begin{align*} x_1 + x_2 + 2x_3 & = 1\\ 0 - x_2 - 3x_3 & = 0\\ 0 + 0 - x_3 & = 2 \end{align*}   Augmented Matrix:  $$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 1\\ 0 & -1 & -3 & 0\\ 0 & 0 & -1 & 2 \end{array}\right)$$

Now, we multiply the second and third equations by $$-1$$ (or multiply the second and third rows by $$-1$$):

System:  \begin{align*} x_1 + x_2 + 2x_3 & = 1\\ 0 + x_2 + 3x_3 & = 0\\ 0 + 0 + x_3 & = -2 \end{align*}    Augmented Matrix:  $$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 1\\ 0 & 1 & 3 & 0\\ 0 & 0 & 1 & -2 \end{array}\right)$$

At this point, we can solve the system. The last equation says that $$x_3 = -2$$. We can plug this into the second equation to get $$x_2 - 3(-2) = 0$$ and so $$x_2 = -6$$. Plugging these values in to the first equation gives $$1 = x_1 + 6 + 2(-2) = x_1 + 2$$ and so $$x_1 = -1$$.

For illustrative purposes, we will carry out this procedure a little further because not all systems can be transformed to such a simple system. Next, we will subtract the second equation from the first and replace the first equation with this new onw (or subtract the second row from the first and replace the first row with this new one):

System:  \begin{align*} x_1 + 0 - x_3 & = 1\\ 0 + x_2 + 3x_3 & = 0\\ 0 + 0 + x_3 & = -2 \end{align*}    Augmented Matrix:  $$\left(\begin{array}{ccc|c} 1 & 0 & -1 & 1\\ 0 & 1 & 3 & 0\\ 0 & 0 & 1 & -2 \end{array}\right)$$

Next add the third equation to the first equation and replace the first equation with this new one (or add the third row to the first row and replace the first row with this new one):

System:  \begin{align*} x_1 + 0 + 0 & = -1\\ 0 + x_2 + 3x_3 & = 0\\ 0 + 0 + x_3 & = -2 \end{align*}    Augmented Matrix:  $$\left(\begin{array}{ccc|c} 1 & 0 & 0 & -1\\ 0 & 1 & 3 & 0\\ 0 & 0 & 1 & -2 \end{array}\right)$$

Now, subtract three times the third equation from the second equation and replace the second equation with this new one (or subtract three times the third row from the second row and replace the second equation with this new one):

System:  \begin{align*} x_1 + 0 + 0 & = -1\\ 0 + x_2 + 0 & = 6\\ 0 + 0 + x_3 & = -2 \end{align*}   Augmented Matrix:  $$\left(\begin{array}{ccc|c} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & 6\\ 0 & 0 & 1 & -2 \end{array}\right)$$

And now we see very clearly that the solution is $$x_1=-1$$, $$x_2 = 6$$, and $$x_3 = -2$$.

Observe that when we got the system to this form:

System:  \begin{align*} x_1 + x_2 + 2x_3 & = 1\\ 0 + x_2 + 3x_3 & = 0\\ 0 + 0 + x_3 & = -2 \end{align*}    Augmented Matrix:  $$\left(\begin{array}{ccc|c} 1 & 1 & 2 & 1\\ 0 & 1 & 3 & 0\\ 0 & 0 & 1 & -2 \end{array}\right)$$

We could then solve the system using "back–substitution''. The first non–zero entry of each row are called the "pivot positions''. A matrix is said to be in row echelon form if non–zero rows (that is, a row that contains at least one non–zero element) are above the zero rows (rows of all zeros) and if the pivot of each row is to the right of the pivot of the row above it, and each pivot is $$1$$. We can do back substitution (though at times – as we will see – this can be a little complicated) whenever a matrix is in row echelon form.

Similarly, when we got the matrix to this reduced row echelon form we can directly solve for each variable:

System: \begin{align*} x_1 + 0 + 0 & = -1\\ 0 + x_2 + 0 & = 6\\ 0 + 0 + x_3 & = -2 \end{align*}    Augmented Matrix:  $$\left(\begin{array}{ccc|c} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & 6\\ 0 & 0 & 1 & -2 \end{array}\right)$$

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Linear Algebra for Math 308: L3V7

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Video Notes

Definition 5. If $$M$$ is an $$n\times n$$ matrix, then the null space  (or kernel) of $$M$$ is the set of solutions to:
\begin{align*}
M\boldsymbol{u} = \boldsymbol{0}.
\end{align*}

This is a vector space it its own right, which means it has a basis. That is, there are linearly independent vectors $$\boldsymbol{u}_1,\ldots,\boldsymbol{u}_m$$ such that $$M\boldsymbol{u}_k = \boldsymbol{0}$$ for $$k=1,\ldots, m$$ and if any vector $$v$$ satisfies $$M\boldsymbol{u}=\boldsymbol{0}$$ then $$\boldsymbol{u}$$ can be written as a linear combination of these vectors. Finding null spaces is an important thing to know how to do, and we show how to do that here.

Understanding properties of the null space of the matrix will also help us answer existence and uniqueness
questions for equations $$M\boldsymbol{u} = \boldsymbol{b}$$.

Example. Find the null space of
\begin{align*}

=\begin{pmatrix}
2 & -3 & 1\\
1 & -2 & 1\\
1 & -3 & 2
\end{pmatrix}.
\end{align*}
Solution. We begin by row reducing (the steps are shown in the video lecture):
\begin{align*}
\left(\begin{array}{ccc|c}
2 & -3 & 1 & 0\\
1 & -2 & 1 & 0\\
1 & -3 & 2 & 0
\end{array}\right)
\sim
\left(\begin{array}{ccc|c}
1 & 0 & -1 & 0\\
0 & 1 & -1 & 0\\
0 & 0 & 0 & 0
\end{array}\right)
\end{align*}

This is different than what we saw above and this system is now two equations in three unknowns. So we will have to solve for two of the variables in terms of the other one. To decide which ones to solve for we use the following definition:

Definition 6. A pivot column is a column that contains a pivot. A basic variable is a variable corresponding to a pivot column and a free variable is a variable corresponding to a column that doesn't contain a pivot.

For example, in the reduced matrix, the first and second columns are pivot columns so $$x_1$$ and $$x_2$$ are basic variables. And the third column is not a pivot column, so $$x_3$$ is a free variable. This means we solve for $$x_1$$ and $$x_2$$ in terms of the free variable $$x_3$$. And we use a different parameter (usually letters like $$s$$ and $$t$$) to denote an aribtraty value of $$x_3$$. That is:
\begin{align*}
x_1 &= x_3 = t\\
x_2 & = x_3 = t\\
x_3 & = \textnormal{free} = t.
\end{align*}
But we really need to write this as a set of vectors. So in vector form it is:
\begin{align*}
\textnormal{null} (M) = \left\{\begin{pmatrix}t\\t\\t\end{pmatrix}:t\in\mathbb{R}\right\},
\end{align*}
which we can also write as:
\begin{align*}
\textnormal{null} (M) = \mathcal{L}\left\{\begin{pmatrix}1\\1\\1\end{pmatrix}\right\}.
\end{align*}
This notation means "all linear combinations of $$\begin{pmatrix}1\\1\\1\end{pmatrix}$$ which, in this case, means all multiples of $$\begin{pmatrix}1\\1\\1\end{pmatrix}$$.

Note. the dimension of the null space is the number of free variables, and a basis for the null space is the set of vectors that are multiplied by free variables. This part will make more sense after the next example.

Examples

1. Solve:
\begin{align*}
\begin{pmatrix}
2 & -3 & 1\\
1 & -2 & 1\\
1 & -3 & 2
\end{pmatrix}
\begin{pmatrix}x_1\\x_2\\x_2\end{pmatrix}
=\begin{pmatrix}1\\0\\0\end{pmatrix}.
\end{align*}

We reduce:
\begin{align*}
\left(\begin{array}{ccc|c}
1 & -3 & 2 & 0\\
1 & -2 & 1 & 0\\
2 & -3 & 1 & 1
\end{array}\right)
\xrightarrow[R2:= R2-R1]{R3:=R3 - 2R1}
\left(\begin{array}{ccc|c}
1 & -3 & 2 & 0\\
0 & 1 & -1 & 0\\
0 & 3 & -3 & 1
\end{array}\right)\\[4mm]
\xrightarrow{R3:=R3 - 3R2}
\left(\begin{array}{ccc|c}
1 & -3 & 2 & 0\\
0 & 1 & -1 & 0\\
0 & 0 & 0 & 1
\end{array}\right)
\end{align*}
The bottom equation says that $$0x_1 + 0x_2 + 0x_3 = 1$$; that is $$0=1$$. There are no values of $$x_1,x_2,x_3$$ that make this equation true, so there are no solutions to this system of equations.

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Linear Algebra for Math 308: L3V9

2. Find the null space of
\begin{align*}
\begin{pmatrix}
1&-3&1\\
-1 & 3 & -1\\
2 & -6 & 2
\end{pmatrix}.
\end{align*}

We begin by row--reducing:
\begin{align*}
\left(\begin{array}{ccc|c}
1&-3&1 & 0\\
-1 & 3 & -1 & 0\\
2 & -6 & 2 & 0
\end{array}\right)
\xrightarrow[{R2:= R2 + R1}]{R3:= R3 - 2R1}
\left(\begin{array}{ccc|c}
1&-3&1 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right)
\end{align*}

So $$x_1$$ is the only basic variable and $$x_2$$ and $$x_3$$ are free variables. So the solution is:
\begin{align*}
x_1 = 3x_2 - x_3\\
x_2 = s\\
x_3 = t.
\end{align*}
So in vector form, we can write this as:
\begin{align*}
\textnormal{null}(M)
&=\left\{\begin{pmatrix}3s - t\\s\\t\end{pmatrix}:s,t\in\mathbb{R}\right\}\\[4mm]
&=\left\{s\begin{pmatrix}3\\1\\0\end{pmatrix} + t\begin{pmatrix}-1\\0\\1\end{pmatrix}:s,t\in\mathbb{R}\right\}
\end{align*}
or
\begin{align*}
\textnormal{null}(M)
=\mathcal{L}\left\{\begin{pmatrix}3\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\end{pmatrix}\right\}.
\end{align*}
There are two free variables so the dimenison of the null space is $$2$$.

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Linear Algebra for Math 308: L3V10

3. Given that the eigenvalues are $$0$$ and $$1$$, find the associated eigenvectors.
\begin{align*}

=\begin{pmatrix}
2 & -3 & 1\\
1 & -2 & 1\\
1 & -3 & 2
\end{pmatrix}
\end{align*}

First, recall that $$\boldsymbol{v}$$ is an eigenvector associated to the eigenvalue $$\lambda$$ if $$M\boldsymbol{v} =\lambda\boldsymbol{v}$$ which can be written as $$(M-\lambda I\boldsymbol{v}) = \boldsymbol{0}$$. That is, we want to find the null space of $$M-\lambda I$$.

For the eigenvector associated to the eigenvalue $$0$$, we want to find the null space of
\begin{align*}
\begin{pmatrix}
2 & -3 & 1\\
1 & -2 & 1\\
1 & -3 & 2
\end{pmatrix}
-0\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}
=\begin{pmatrix}
2 & -3 & 1\\
1 & -2 & 1\\
1 & -3 & 2
\end{pmatrix}
\end{align*}
We've already done this, and we know that the null space is
\begin{align*}
\mathcal{L}\left\{\begin{pmatrix}1\\1\\1\end{pmatrix}\right\}.
\end{align*}
So any non–zero multiple of $$\begin{pmatrix}1\\1\\1\end{pmatrix}$$ is an eigenvector of this matrix associated to the eigenvalue $$0$$.

Now to find the eigenvectors  associated to $$1$$, we solve want to find the null space of:
\begin{align*}
\begin{pmatrix}
2 & -3 & 1\\
1 & -2 & 1\\
1 & -3 & 2
\end{pmatrix}
-1\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}
=\begin{pmatrix}
1 & -3 & 1\\
1 & -3 & 1\\
1 & -3 & 1
\end{pmatrix}
\end{align*}
We do this by row reducing:
\begin{align*}
\begin{pmatrix}
1 & -3 & 1\\
1 & -3 & 1\\
1 & -3 & 1
\end{pmatrix}
\xrightarrow[R2:= R2 - R1]{R3:= R3 - R1}
\begin{pmatrix}
1 & -3 & 1\\
0 & 0 & 0\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
Note this is similar to a previous example. And we see that the null space of this matrix is:
\begin{align*}
\mathcal{L}\left\{\begin{pmatrix}3\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\end{pmatrix}\right\}.
\end{align*}
That is, any linear combination of these vectors is an eigenvector of the matrix associated to the eigenvalue $$1$$. Typically, when reporting an answer (especially if we need to use them to solve an ODE or something) we pick a particular basis and write something like:
$$\hspace{.5in}\lambda_1 = 0, \hspace{.25in} \boldsymbol{v}_1 = \begin{pmatrix}1\\1\\1\end{pmatrix} \hspace{.5in}$$$$\lambda_2 = 1, \hspace{.25in} \boldsymbol{v}_2 = \begin{pmatrix}3\\1\\0\end{pmatrix} \hspace{.5in}$$$$\lambda_3 = 1, \hspace{.25in} \boldsymbol{v}_3 = \begin{pmatrix}-1\\0\\1\end{pmatrix}.$$

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Linear Algebra for Math 308: L3V11

4. Determine if the vectors
\begin{align*}
\begin{pmatrix}1\\2\\-1\end{pmatrix},
\begin{pmatrix}2\\1\\3\end{pmatrix},
\begin{pmatrix}-4\\1\\-11\end{pmatrix}
\end{align*}
are linearly independent.

We want to find solutions to:
\begin{align*}
\begin{pmatrix}
1 & 2 & -4\\
2 & 1 & 1\\
-1 & 3 & -11
\end{pmatrix}
\begin{pmatrix}c_1\\c_2\\c_2\end{pmatrix}
= \begin{pmatrix}0\\0\\0\end{pmatrix}
\end{align*}
Row reduce:
\begin{align*}
\begin{pmatrix}
1 & 2 & -4\\
2 & 1 & 1\\
-1 & 3 & -11
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 2 & -4\\
0 & -3 & 9\\
0 & 0 & 0
\end{pmatrix}
\end{align*}
This is:
\begin{align*}
c_1 + 2c_2 -4c_3 &= 0\\
-3c_2 + 9 c_3 &= 0
\end{align*}
So:
\begin{align*}
c_2 &= 3c_3\\
c_1 &= -2c_3
\end{align*}
That is, for any $$t$$, the values $$c_1,c_2,c_3$$ below:
\begin{align*}
c_1 = -2t \hspace{.5in} c_2 = 3t \hspace{.5in} c_3 = t,
\end{align*}
are a solution to:
\begin{align*}
c_1\begin{pmatrix}1\\2\\-1\end{pmatrix}+
c_2\begin{pmatrix}2\\1\\3\end{pmatrix}+
c_3\begin{pmatrix}-4\\1\\-11\end{pmatrix}
= \begin{pmatrix}0\\0\\0\end{pmatrix}.
\end{align*}
Since there are non-trivial solutions, the vectors are linearly dependent.

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Linear Algebra for Math 308: L3V12

Exercises

1. Solve the system:
\begin{align*}
x_1 + x_2 + 2x_3 &= 1\\
2x_1 + x_2 + x_3 &= 2\\
x_1 + x_2 + x_3  &= 3
\end{align*}

First, we write this as an augmented matrix:
\begin{align*}
\begin{pmatrix}
1 & 1 & 2 & 1\\
2 & 1 & 1 & 2\\
1 & 1 & 1 & 3
\end{pmatrix}
\end{align*}
Then perform the Gauss-Jordan elimination algorithm:
\begin{align*}
\begin{pmatrix}
1 & 1 & 2 & 1\\
2 & 1 & 1 & 2\\
1 & 1 & 1 & 3
\end{pmatrix}
\xrightarrow{R2 := R2 - 2R1}
\begin{pmatrix}
1 & 1 & 2 & 1\\
0 & -1 & -3 & 0\\
1 & 1 & 1 & 3
\end{pmatrix}\\[4mm]
\xrightarrow{R3 := R3 - R1}
\begin{pmatrix}
1 & 1 & 2 & 1\\
0 & -1 & -3 & 0\\
0 & 0 & -1 & 2
\end{pmatrix}\\
\end{align*}
Now the matrix is in row echelon form. We can solve at this point using back–substitution. But, it is usually just as easy (or easier) to put the matrix in reduced row echelon form and solve "by inspection''. We now work from the right–most pivot to the left–most.
\begin{align*}
\begin{pmatrix}
1 & 1 & 2 & 1\\
0 & 1 & 3 & 0\\
0 & 0 & 1 & -2
\end{pmatrix}
\xrightarrow{R2 := R2 - 3R3}
\begin{pmatrix}
1 & 1 & 2 & 1\\
0 & 1 & 0 & 6\\
0 & 0 & 1 & -2
\end{pmatrix}\\[4mm]
\xrightarrow{R1 := R1 - 2R2}
\begin{pmatrix}
1 & 1 & 0 & 5\\
0 & 1 & 0 & 6\\
0 & 0 & 1 & -2
\end{pmatrix}
\end{align*}
Now we work with the $$1$$ in the second row and second column:
\begin{align*}
\begin{pmatrix}
1 & 1 & 0 & 5\\
0 & 1 & 0 & 6\\
0 & 0 & 1 & -2
\end{pmatrix}
\xrightarrow{R1 := R1 - R2}
\begin{pmatrix}
1 & 0 & 0 & -1\\
0 & 1 & 0 & 6\\
0 & 0 & 1 & -2
\end{pmatrix}
\end{align*}
Now, this matrix is the same as this system:
\begin{align*}
x_1 + 0 + 0 &= -1\\
0 + x_2 + 0 &= 6\\
0 + 0 + x_3  &= -2
\end{align*}
And we can see "by inspection'' that the solution is $$x_1 = 1$$, $$x_2 = 6$$ and $$x_3 = -2$$. In vector form, this is:
\begin{align*}
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}
=\begin{pmatrix}1\\6\\-2\end{pmatrix}
\end{align*}

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Linear Algebra for Math 308: L3E1rr

2. Solve the following system using Gauss–Jordan Elimination.
\begin{align*}
x_1 + x_2 + 2x_3 + 2x_4 &= 3\\
2x_2 + 2x_3 + 2x_4 &= 2\\
x_1 + x_3 &= 2
\end{align*}

As an augmented matrix, this is:
\begin{align*}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & 2 & 2 & 2 & 2\\
1 & 0 & 1 & 0 & 2
\end{array}\right)
\end{align*}
We start by putting the matrix into echelon form. There are no rows of all zeros, so we don't need to do step 1. However, the pivot (the $$1$$) in the third row and first column is to the left of the pivot in the row above (the $$2$$ in the second row and second column) so we do need to do step 2:
\begin{align*}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & 2 & 2 & 2 & 2\\
1 & 0 & 1 & 0 & 2
\end{array}\right)
\xrightarrow{R2 \leftrightarrow R3}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
1 & 0 & 1 & 0 & 2\\
0 & 2 & 2 & 2 & 2
\end{array}\right)
\end{align*}
We'll start with the pivot in the first row and first column and work to the right:
\begin{align*}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
1 & 0 & 1 & 0 & 2\\
0 & 2 & 2 & 2 & 2
\end{array}\right)
\xrightarrow{R2 = R2 - R1}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & -1 & -1 & -2 & -1\\
0 & 2 & 2 & 2 & 2
\end{array}\right)
\end{align*}
Now we work with the next pivot to the right (the $$-1$$ in the second row and second column):
\begin{align*}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & -1 & -1 & -2 & -1\\
0 & 2 & 2 & 2 & 2
\end{array}\right)
&\xrightarrow{R2:=-R2}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & 1 & 1 & 2 & 1\\
0 & 2 & 2 & 2 & 2
\end{array}\right)\\[4mm]
&\xrightarrow{R3 := R3 - 2R2}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & 1 & 1 & 2 & 1\\
0 & 0 & 0 & -2 & 0
\end{array}\right)
\end{align*}
Now the next pivot to the right is the $$-2$$:
\begin{align*}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & 1 & 1 & 2 & 1\\
0 & 0 & 0 & -2 & 0
\end{array}\right)
\xrightarrow{R3 := -\frac{1}{2}R3}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & 1 & 1 & 2 & 1\\
0 & 0 & 0 & 1 & 0
\end{array}\right)
\end{align*}
The matrix is in echelon form; we put it in reduced echelon form. We start with the rightmost pivot which is the $$1$$ in the third row and fourth column and work to the left:
\begin{align*}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 2 & 3\\
0 & 1 & 1 & 2 & 1\\
0 & 0 & 0 & 1 & 0
\end{array}\right)
\xrightarrow[{R2 := R2 - 2R3}]{R1 := R1 - 2R3}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 0 & 3\\
0 & 1 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0
\end{array}\right)
\end{align*}
(observe how we combined two steps here). The next pivot to the left is the $$1$$ in the second row and second column:
\begin{align*}
\left(\begin{array}{llll|l}
1 & 1 & 2 & 0 & 3\\
0 & 1 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0
\end{array}\right)
\xrightarrow{R1 : R1 - R2}
\left(\begin{array}{llll|l}
1 & 0 & 1 & 0 & 2\\
0 & 1 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 0
\end{array}\right)
\end{align*}
This is the same as this system:
$$\hspace{.5in}\begin{array}{ll} x_1 + x_3 &= 2\\ x_2 + x_3 &= 1\\ x_4 &= 0 \end{array}$$    Which is the same as     $$\begin{array}{ll} x_1 &= 2 - x_3\\ x_2 &= 1 - x_3\\ x_3 &= \\ x_4 &= 0\end{array}$$

As we have seen above, this means that the equations place no restrictions on $$x_3$$. That is, for every value of $$x_3$$, there is a solution and so we call it a "free variable''.  (This is discussed more below). The solution can be written in vector form:
\begin{align}
\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}
=\begin{pmatrix}2\\1\\0\\0\end{pmatrix} + t\begin{pmatrix}-1\\-1\\1\\0\end{pmatrix}
\end{align}
Note that we have use "$$t$$'' instead of $$x_3$$''. This is because "$$x_3$$'' is really only a place holder and also appears on the left hand side of the previous equation. Also, it is usually customary to write the set of solutions in "set builder notation'':
\begin{align*}
\left\{\begin{pmatrix}2\\1\\0\\0\end{pmatrix} + t\begin{pmatrix}-1\\-1\\1\\0\end{pmatrix} : t\in\mathbb{R}\right\}
\end{align*}

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Linear Algebra for Math 308: L3E2rr

3. Solve the following system.
\begin{align*}
x_1 + x_2 &= 3\\
x_1 + 2x_2 &= 5\\
x_1 + 3x_2 &= 7
\end{align*}

The augmented matrix is:
\begin{align*}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
1 & 2 & 5\\
1 & 3 & 7
\end{array}\right)
\end{align*}
We put the matrix in echelon form. The $$1$$ in the first row and first column is the left most pivot:
\begin{align*}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
1 & 2 & 5\\
1 & 3 & 7
\end{array}\right)
\xrightarrow[{R2 := R2 - R1}]{R3 := R3 - R1}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
0 & 1 & 2\\
0 & 2 & 4
\end{array}\right)
\end{align*}
The next pivot to the right is the $$1$$ in the second row and second column:
\begin{align*}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
0 & 1 & 2\\
0 & 2 & 4
\end{array}\right)
\xrightarrow{R3 := R3 - 2R2}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
0 & 1 & 2\\
0 & 0 & 0
\end{array}\right)
\end{align*}
The right most pivot is the $$1$$ in the second row and second column:
\begin{align*}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
0 & 1 & 2\\
0 & 0 & 0
\end{array}\right)
\xrightarrow{R1 := R1 - R2}
\left(\begin{array}{ll|l}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & 0
\end{array}\right)
\end{align*}
And now we can solve "by inspection'' and get the solution is:
\begin{align*}
\begin{pmatrix}x_1\\x_2\end{pmatrix} = \begin{pmatrix}1\\2\end{pmatrix}.
\end{align*}

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Linear Algebra for Math 308: L3E3rr

4. Solve the following system.
\begin{align*}
x_1 + x_2 &= 3\\
x_1 + 2x_2 &= 5\\
x_1 + 3x_2 &= 6
\end{align*}

The augmented matrix is:
\begin{align*}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
1 & 2 & 5\\
1 & 3 & 6
\end{array}\right)
\end{align*}
We are going to put it in RREF:
\begin{align*}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
1 & 2 & 5\\
1 & 3 & 6
\end{array}\right)
\xrightarrow[{R2 := R2 - R1}]{R3 := R3 - R1}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
0 & 1 & 2\\
0 & 2 & 3
\end{array}\right)\\[4mm]
\xrightarrow{R3 : \frac{1}{2} R3}
\left(\begin{array}{ll|l}
1 & 1 & 3\\
0 & 1 & 2\\
0 & 1 & \frac{3}{2}
\end{array}\right)\\[4mm]
\xrightarrow[{R2 := R2 - R3}]{R1 := R1 - R3}
\left(\begin{array}{ll|l}
1 & 0 & \frac{3}{2}\\
0 & 0 & \frac{1}{2}\\
0 & 1 & \frac{3}{2}
\end{array}\right)
\end{align*}
At this point, the second equation says:
\begin{align*}
0x_1 + 0x_2 = \frac{1}{2},
\end{align*}
and there is no solution to this equation. Thus, there is no solution to the system.

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Linear Algebra for Math 308: L3E4rr

5. Solve the following system.\begin{align*}
x_1 - x_2 + x_3 &=1\\
x_1 + 3x_2 + 5x_3 &=0\\
3x_1 + 2x_2 + x_3 &=1
\end{align*}

\begin{align*}
\left(\begin{array}{ccc|c}
1 & -1 & 1 & 1\\
1 & 3 & 5 & 0\\
3 & 2 & 1 & 1
\end{array}\right)
&\xrightarrow[{R2 := \frac{1}{4}(R2 - R1)}]{R3 := R3 - 3R1}
\left(\begin{array}{ccc|c}
1 & -1 & 1 & 1\\
0 & 1 & 1 & -1/4\\
0 & 5 & -2 & -2
\end{array}\right)\\[4mm]
&\xrightarrow{R3 := -\frac{1}{7}(R3 - 5R2)}
\left(\begin{array}{ccc|c}
1 & -1 & 1 & 28/28\\
0 & 1 & 1 & -7/28\\
0 & 0 & 1 & 3/28
\end{array}\right)
\end{align*}
This is in REF; we now put it in RREF.
\begin{align*}
\left(\begin{array}{ccc|c}
1 & -1 & 1 & 28/28\\
0 & 1 & 1 & -7/28\\
0 & 0 & 1 & 3/28
\end{array}\right)
&\xrightarrow[{R1 := R1 - R3}]{R2 := R2 - R3}
\left(\begin{array}{ccc|c}
1 & -1 & 0 & 25/28\\
0 & 1 & 0 & -10/28\\
0 & 0 & 1 & 3/28
\end{array}\right)\\[4mm]
&\xrightarrow{R1 := R1 + R2}
\left(\begin{array}{ccc|c}
1 & 0 & 0 & 15/28\\
0 & 1 & 0 & -10/28\\
0 & 0 & 1 & 3/28
\end{array}\right)
\end{align*}
So then:
\begin{align*}
\left(\begin{array}{ccc|c}
x_1\\x_2\\x_3
\end{array}\right)
=\frac{1}{28}\begin{pmatrix}15\\-10\\3\end{pmatrix}
\end{align*}

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Linear Algebra for Math 308: L3E5rr

6. Solve the following system.\begin{align*}
x_1 - x_2 + x_3 &=1\\
x_1 + 3x_2 + 5x_3 &=0\\
3x_1 + 2x_2 + 8x_3 &=1
\end{align*}

\begin{align*}
\left(\begin{array}{ccc|c}
1 & -1 & 1 & 1\\
1 & 3 & 5 & 0\\
3 & 2 & 8 & 1
\end{array}\right)
&\xrightarrow[R2:=(\frac{1}{4}(R2-R1)]{R3:=\frac{1}{5}(R3-3R1)}
\left(\begin{array}{ccc|c}
1 & -1 & 1 & 1\\
0 & 1 & 1 & -5/20\\
0 & 1 & 1 & -8/20
\end{array}\right)\\[4mm]
&\xrightarrow{R3:=R3-R2}
\left(\begin{array}{ccc|c}
1 & -1 & 1 & 1\\
0 & 1 & 1 & -5/10\\
0 & 0 & 0 & -3/20
\end{array}\right)\\[4mm]
&\xrightarrow{R1:=R1+R2}
\left(\begin{array}{ccc|c}
1 & 0 & 2 & 5/10\\
0 & 1 & 1 & -5/10\\
0 & 0 & 0 & -3/20
\end{array}\right)
\end{align*}
The last line indicates $$0 = -\frac{3}{20}$$ which is never true. So there are no solutions.

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Linear Algebra for Math 308: L3E6rr

7. Solve the following system.\begin{align*}
x_1 + 2x_2 + 3x_3 + 5x_4 &= 5\\
2x_1 + x_2   + 3x_3 + 4x_4 &= 5\\
3x_1 - x_2   + 2x_3  +x_4 &= 2
\end{align*}

\begin{align*}
\left(\begin{array}{cccc|c}
1&2&3&5&5\\
2&1&3&4&5\\
3&-1&2&1&2
\end{array}\right)
&\xrightarrow[R2:=(-1/3)(R2 - 2R1)]{R3:=(-1/7)(R3-3R1)}
\left(\begin{array}{cccc|c}
1&2&3&5&5\\
0&1&1&2&5/3\\
0&1&1&2&13/7
\end{array}\right)\\[4mm]
&\xrightarrow{R3:=R3-R2}
\left(\begin{array}{cccc|c}
1&2&3&5&5\\
0&1&1&2&5/3\\
0&0&0&0&4/21
\end{array}\right)
\end{align*}
The last row says $$0=4/21$$ so there are no solutions.

Video Errata: At the end of the video, the presenter should have $$-\frac{4}{3}$$, not $$-\frac{34}{3}$$.

​To see the full video page and find related videos, click the following link.
Linear Algebra for Math 308: L3E7rr