1. What is a system of linear equations?

A linear system is a collection of linear equations. Before giving the formal definition of a linear system, let us start with linear equations.

Definition 1: A linear equation in $n$ unknown variables $x_1, x_2,\ldots, x_n$
is an equation of the form
\begin{eqnarray}\label{linear-eq}
\begin{array}{ccccccccccc}
a_{1}x_1 & +& a_{2}x_2& +& a_{3}x_3& +& \cdots & +&a_{n}x_n&=&b
\end{array}
\end{eqnarray}

Note from the definition of a linear equation that the variables appear only with a degree $1$. So in a linear equation

• we cannot have a term of the form $x_i^a$ with $a\neq 1$,
• we cannot have a term of the form $x_ix_j$,
• we cannot have a term of the form $\cos(x_i)$, $\sin(x_i)$, $\tan(x_i)$, $e^{x_i}$, $\ln(x_i)$ etc.

For instance, the equation $2x_1-3x_2^2=7$ is not linear because the term $x_2^2$ appears with a degree $2$. Similarly, the equation $\sqrt{x_1}+2x_2+5x_3=3$ is not a linear equation because of the term $\sqrt{x_1}$ which appears with a degree $\dfrac{1}{2}$.

Example 1: In each of the following, determine whether the equation is linear in $x_1$, $x_2$, and $x_3$.

1. $3x_1+x_2-\sqrt{2}x_3=4$
2. $x_1=-x_2+x_3$
3. $x_1^{\frac{3}{2}}-x_2+3x_3=21$
4. $x_1+x_1x_2+x_3=4$
5. $x_1^{-5}+3x_2+2x_3=1$
6. $x_1+\cos(x_2)+x_3=0$
7. $\cos(6)x_1+x_2+7x_3=14$

Solution of Example 1:

1. The equation $3x_1+x_2-\sqrt{2}x_3=4$ can be written in the form $a_1x_1+a_2x_2+a_3x_3=b$ with $a_1=3$, $a_2=1$, $a_3=-\sqrt{2}$, and $b=4$. Therefore the equation $3x_1+x_2-\sqrt{2}x_3=4$ is linear in the variables $x_1$, $x_2$, and $x_3$.
2. The equation $x_1=-x_2+x_3$ is equivalent to $x_1+x_2-x_3=0$. This can be written in the form $a_1x_1+a_2x_2+a_3x_3=b$ with $a_1=1$, $a_2=1$, $a_3=-1$, and $b=0$. Therefore the equation $x_1=-x_2+x_3$ is linear in the variables $x_1$, $x_2$, and $x_3$.
3. The equation $x_1^{\frac{3}{2}}-x_2+3x_3=21$ is not linear because of the term $x_1^{\frac{3}{2}}$, where the variable $x_1$ appears with an exponent $\frac{3}{2}\neq 1$.
4. The equation $x_1+x_1x_2+x_3=4$ is not linear because of the term $x_1x_2$, which is the product of two variables.
5. The equation $x_1^{-5}+3x_2+2x_3=1$ is not linear because of the term $x_1^{-5}$ where the variable $x_1$ appears with an exponent $-5\neq 1$.
6. The equation $x_1+\cos(x_2)+x_3=0$ is not linear because of the term $\cos(x_2)$.
7. The equation $\cos(6)x_1+x_2+7x_3=14$ can be written in the form $a_1x_1+a_2x_2+a_3x_3=b$ with $a_1=\cos(6)$, $a_2=1$, $a_3=7$, and $b=14$. Therefore the equation $\cos(6)x_1+x_2+7x_3=14$ is linear in the variables $x_1$, $x_2$, and $x_3$.

2. Exercise 1

In each of the following, determine whether the equation is linear in $x_1$, $x_2$, and $x_3$.

1. Select the correct answers.

 The equation $\sin(2)x_1+x_2-\sqrt{2}x_3=4$ is linear.

 The equation $\sin(2)x_1+x_2-\sqrt{2}x_3=4$ is linear.

  The equation $x_1=2-x_2+x_3$ is not linear.

  The equation $x_1=2-x_2+x_3$ is linear.

2. Select the correct answers.

 The equation $x_1-x_2^{\frac{3}{2}}+3x_3=1$ is linear.

 The equation $x_1-x_2^{\frac{3}{2}}+3x_3=1$ is not linear.

 The equation $x_1+\ln(x_2)+x_3=4$ is linear

 The equation $x_1+\ln(x_2)+x_3=4$ is not linear

3. Select the correct answers.

 The equation $x_1^{2}+2x_3=10$ is linear.

 The equation $x_1^{2}+2x_3=10$ is not linear.

 The equation $x_1+\sin(x_2)+x_3=0$ is linear

 The equation $x_1+\sin(x_2)+x_3=0$ is not linear

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Definition 2: A system of $m$ linear equations in $n$ unknown variables $x_1, x_2,\ldots, x_n$
is a collection of m equations of the form
\begin{eqnarray}\label{linear-system}
\begin{array}{ccccccccccc}
a_{11}x_1 & +& a_{12}x_2& +& a_{13}x_3& +& \cdots & +&a_{1n}x_n&=&b_1\\
a_{21}x_1 & +& a_{22}x_2& +& a_{23}x_3& +& \cdots & +&a_{2n}x_n&=&b_2\\
a_{31}x_1 & +& a_{32}x_2& +& a_{33}x_3& +& \cdots & +&a_{3n}x_n&=&b_3\\
& \vdots & & \vdots & & \vdots & & \vdots & &\vdots &\\
a_{m1}x_1 & +& a_{m2}x_2& +& a_{m3}x_3& +& \cdots & +&a_{mn}x_n&=&b_m
\end{array}
\end{eqnarray}

The numbers $a_{ij}$ are called the coefficients of the linear system.

Example 2:  Consider the following system of equations:
\begin{eqnarray*}
\begin{array}{ccccccc}
2x_1&+&x_2&-&x_3&=&1\\
3x_1&-&2x_2&+&5x_3&=&4\\
\sqrt{3}x_1&+&\pi x_2&+&2 x_3&=&\ln(4)
\end{array}
\end{eqnarray*}

This is a system of $3$ linear equations in $3$ unknown variables $x_1$, $x_2$, and $x_3$. The coefficients are: $a_{11}=2$, $a_{12}=1$, $a_{13}=-1$, $a_{21}=3$, $a_{22}=-2$, $a_{23}=5$, $a_{31}=\sqrt{3}$, $a_{32}=\pi$, $a_{33}=2$, $b_1=1$, $b_2=4$, and $b_3=\ln(4)$.

Example 3:  Consider the following system of equations:
\begin{eqnarray*}
\begin{array}{ccccccc}
21x_1&+&x_2& & &=&23\\
x_1&+&2x_2&+&5x_3&=&10
\end{array}
\end{eqnarray*}

This is a system of $2$ linear equations in $3$ unknown variables $x_1$, $x_2$, and $x_3$. The coefficients are: $a_{11}=21$, $a_{12}=1$, $a_{13}=0$, $a_{21}=1$, $a_{22}=2$, and $a_{23}=5$.

3. Solution of a linear system

The main problem with a linear system is to find a list of $n$ numbers $(s_1,s_2,\ldots,s_n)$ such that if we substitute the list of numbers $(s_1,s_2,\ldots,s_n)$ for the unknown
variables $(x_1, x_2, \ldots , x_n )$ in equation \eqref{linear-system} then the left-hand side of the ith equation will
equal $b_i$.

Definition 3:  A solution to the linear system
\begin{eqnarray*}
\begin{array}{ccccccccccc}
a_{11}x_1 & +& a_{12}x_2& +& a_{13}x_3& +& \cdots & +&a_{1n}x_n&=&b_1\\
a_{21}x_1 & +& a_{22}x_2& +& a_{23}x_3& +& \cdots & +&a_{2n}x_n&=&b_2\\
a_{31}x_1 & +& a_{32}x_2& +& a_{33}x_3& +& \cdots & +&a_{3n}x_n&=&b_3\\
& \vdots & & \vdots & & \vdots & & \vdots & &\vdots &\\
a_{m1}x_1 & +& a_{m2}x_2& +& a_{m3}x_3& +& \cdots & +&a_{mn}x_n&=&b_m
\end{array}
\end{eqnarray*}
is a list of $n$ numbers $(s_1,s_2,\ldots,s_n)$ such that
\begin{eqnarray}\label{solution}
\begin{array}{ccccccccccc}
a_{11}s_1 & +& a_{12}s_2& +& a_{13}s_3& +& \cdots & +&a_{1n}s_n&=&b_1\\
a_{21}s_1 & +& a_{22}s_2& +& a_{23}s_3& +& \cdots & +&a_{2n}s_n&=&b_2\\
a_{31}s_1 & +& a_{32}s_2& +& a_{33}s_3& +& \cdots & +&a_{3n}s_n&=&b_3\\
& \vdots & & \vdots & & \vdots & & \vdots & &\vdots &\\
a_{m1}s_1 & +& a_{m2}s_2& +& a_{m3}s_3& +& \cdots & +&a_{mn}s_n&=&b_m
\end{array}
\end{eqnarray}
When the linear system \eqref{linear-system} has \textbf{at least one} solution, we say that the linear system is consistent, and when the linear system has no solutions, we say that it is inconsistent.

The set of all solutions is called the solution set.

Example 4: Verify that $(1, 1, 2)$ is a solution to the system of equations
\begin{eqnarray*}
\begin{array}{ccccccc}
2x_1&+&x_2&- &x_3 &=&1\\
3x_1&-&2x_2&+&5x_3&=&11
\end{array}
\end{eqnarray*}

A solution to Example 4: Substituting $x_1=1$, $x_2=1$ and $x_3=2$ in the equations above gives
\begin{eqnarray*}
\mbox{Left side of Equation (1)}&=&
2\cdot (1)+(1)-(2)=2+1-2\\
&=&1=\mbox{Right side of Equation (1)}\\
\mbox{Left side of Equation (2)}&=&3\cdot (1)-2\cdot (1)+5\cdot (2)=3-2+10\\
&=&11=\mbox{Right side of Equation (2)}
\end{eqnarray*}
So both equations are satisfied. Therefore $(1, 1, 2)$ is a solution.
\end{whitebox}

Example 5: Verify that $(1, -1, 1)$ is not a solution to the system of equations
\begin{eqnarray*}
\begin{array}{ccccccc}
x_1&+&x_2&+ &x_3 &=&1\\
-x_1&+&2x_2&+&x_3&=&3
\end{array}
\end{eqnarray*}

A solution to Example 5: Substituting $x_1=1$, $x_2=-1$ and $x_3=1$ in the equations above gives
\begin{eqnarray*}
\mbox{Left side of Equation (1)}&=&1\cdot (1)+(-1)+(1)=1-1+1\\
&=&1=\mbox{Right side of Equation (1)}\\
\mbox{Left side of Equation (2)}&=&-(1)+2\cdot (-1)+ (1)=-1-2+1\\
&=&-2\neq \mbox{Right side of Equation (2)}
\end{eqnarray*}
Because the second equation is not satisfied, $(1, 1, 2)$ is not a solution.

Example 6:  Show that the system of linear equations
\begin{eqnarray*}
\begin{array}{ccccccc}
x_1&-&x_2&=&7\\
-x_1&+&x_2&=&3
\end{array}
\end{eqnarray*}
is inconsistent (that is, it does not have a solution).

A solution to Example 6: If we add the two equations, we get
\begin{eqnarray*}
x_1-x_2&=& 7\\
-x_1+x_2& =& 3 \\
———-&-&——\\
0 &=& 10
\end{eqnarray*}
The equality $0=10$ is a contradiction. Therefore, there does not exist a list $(s_1 , s_2 )$ that satisfies the
system because this would lead to the contradiction $0 = 10$. This proves that the linear system above is inconsistent as desire.

Example 7: Show that the system of linear equations
\begin{eqnarray*}
\begin{array}{ccccccc}
x_1&+&x_2&=&1\\
2x_1&+&2x_2&=&8
\end{array}
\end{eqnarray*}
is inconsistent (that is, it does not have a solution).

A solution to Example 7: Multiplying the second equation by $-\dfrac{1}{2}$, the linear system becomes
\begin{eqnarray*}
\begin{array}{ccccccc}
x_1&+&x_2&=&1\\
-x_1&-&x_2&=&-4\end{array}
\end{eqnarray*}
Now add the two equations yield
\begin{eqnarray*}
x_1+x_2&=&1\\
-x_1-x_2&=&-4\\
———–&-&———\\
0&=&-3
\end{eqnarray*}
The equality $0=-3$ is a contradiction. Therefore, there does not exist a list $(s_1 , s_2 )$ that satisfies the system because this would lead to the contradiction $0 = -3$. This proves that the linear system above is inconsistent as desire.

Example 8: Solve the system of linear equations
\begin{eqnarray*}
\begin{array}{ccccccc}
x_1&-&x_2&=&9\\
2x_1&+&x_2&=&0
\end{array}
\end{eqnarray*}

A solution to Example 8: Adding the first equation to the second, we obtain
\begin{eqnarray*}
\begin{array}{ccccccc}
x_1&-&x_2&=&9\\
3x_1& & &=&9
\end{array}
\end{eqnarray*}
The equation $3x_1=9$ gives $x_1=\dfrac{9}{3}=3$. Now letting $x_1=3$ in the first equation yields
$x_2=3-9=-6$. Hence the linear system has the unique solution
$$x_1=3,\qquad x_2=-6.$$

Example 9: Solve the linear system
\begin{eqnarray*}
\begin{array}{ccccccc}
x&-&y&=&2\\
2x&-&2y&=&4
\end{array}
\end{eqnarray*}

A solution to Example 9: We can eliminate $x$ from the second equation by adding $-2$ times the first equation to the second. This yields the simplified system
\begin{eqnarray*}
\begin{array}{ccccccc}
x&-&y&=&2\\
& &0&=&0
\end{array}
\end{eqnarray*}
The second equation does not impose any restrictions on $x$ and $y$ and hence can be omitted. Thus, the solutions of the system are those values of $x$ and $y$ that satisfy the single equation
\begin{eqnarray*}
\begin{array}{ccccccc}
x&-&y&=&2
\end{array}
\end{eqnarray*}
So we can solve for $x$ in terms of $y$ to obtain $x=2+y$, and assign an arbitrary value to $y$. Letting $y=t$ ($t$ is called a parameter), then $x=2+t$. This allows us to express the solution by the pair of equations (called \textbf{parametric equations})
$$x=2+t,\qquad y=t,\qquad\qquad \mbox{$t$ is arbitrary}.$$

Example 10: Use parametric equations to describe the solution set of the linear system in the unknown variables $x$, $y$, $z$
\begin{eqnarray*}
\begin{array}{cccccccc}
x&+&2y&-&3z=&4\\
3x&+&6y&-&9z=&12\\
-x&-&2y&+&3z=&-4\\
\end{array}
\end{eqnarray*}

A solution to Example 10: Adding $-3$ times the first equation to the second and adding the first equation to the third yields the simplified system
\begin{eqnarray*}
\begin{array}{cccccccc}
x&+&2y&-&3z=&4\\
& & & &0=&0\\
& & & &0=&0\\
\end{array}
\end{eqnarray*}
The second and third equations do not impose restrictions on $x$, $y$, and $z$, which can be omitted. Thus, the solutions of the system are those values of $x$, $y$, and $z$ that satisfy the
single equation
\begin{eqnarray*}
\begin{array}{cccccccc}
x&+&2y&-&3z=&4\\
\end{array}
\end{eqnarray*}
So we can solve for $x$ in terms of $y$ and $z$ to obtain $x=4-2y+3z$, and assign an arbitrary value to $y$ and $z$. Now letting $y=t$ and $z=s$, then $x=4-2t+3s$. This allows us to express the solution by parametric equations
$$x=4-2t+3s,\qquad y=t,\qquad z=s \qquad (\mbox{$t$ and $s$ are arbitrary real numbers}).$$

Example 11: Under what conditions on $a$ and $b$ will the following linear system have no solutions, one solution, or infinitely many solutions?
\begin{eqnarray*}
\begin{array}{ccccccc}
x&+&2y&=&a\\
3x&+&6y&=&b
\end{array}
\end{eqnarray*}

A solution to Example 11: Adding $-3$ times the first equation in the second yields the simplified system
\begin{eqnarray*}
\begin{array}{ccccccc}
x&+&2y&=&a\\
&&0&=&b-3a
\end{array}
\end{eqnarray*}
From the second equation, we see that if $b-3a\neq 0$ (that is, $b\neq 3a$), the equation $0=b-3a$ gives a contradiction. In this case, the system is inconsistent.

Now if $b-3a=0$ (that is $b=3a$), then
\begin{eqnarray*}
\begin{array}{ccccccc}
x&+&2y&=&a\\
&&0&=&0
\end{array}
\end{eqnarray*}

The equation $0=0$ does not impose any restrictions on $x$ and $y$ so it can be omitted. Hence we obtain the single equation
\begin{eqnarray*}
\begin{array}{ccccccc}
x&+&2y&=&a
\end{array}
\end{eqnarray*}
So we can solve for $x$ in terms of $y$ to obtain $x=a-2y$. Letting $y=t$ be a parameter, we obtain the parametric equations
$$x=a-2t,\qquad y=t,\qquad \mbox{$t$ is arbitrary}.$$

Conclusion: The above linear system has no solution when $b\neq 3a$, has infinitely many solutions when $b=3a$, and there are no values of $a$ and $b$ for which the above linear system has a unique solution.