Q. 93 - Precalculus Enhanced with Graphing Utilities

Question

What is the solution of the system of equations $AX = 0$, if $A^{-1}$ exists? Discuss the solution of $AX = 0$ if $A^{-1}$ does not exist. 

Step-by-Step Solution

Answer

We get trivial solution to the system of equations $AX=\Phi$ if $A^{-1}$ exists. But if $A^{-1}$ does not exist then we get infinitely many solutions.

Step 1. Systems of equations A - 1 exists

Let us consider three systems of equations,

$$\begin{gathered} 2x+y-3z=0 .....\left(1\right) \\ \begin{array}{r}x-3y+2z=0 ..\dots \left(2\right)\\ 4x+2y-z=0 ..\dots \left(3\right)\end{array} \end{gathered}$$

We can write the above system of equations in matrix form as

$AX=\Phi$ where,

$A=\left[\begin{array}{ccc}2& 1& -3\\ 1& -3& 2\\ 4& 2& -1\end{array}\right],X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]$ and $\mathrm{\Phi }=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Therefore,

$$\begin{gathered} \left|A\right|=\left|\begin{array}{ccc}2& 1& -3\\ 1& -3& 2\\ 4& 2& -1\end{array}\right|=2\left|\begin{array}{cc}-3& 2\\ 2& -1\end{array}\right|-1\left|\begin{array}{cc}1& 2\\ 4& -1\end{array}\right|-3\left|\begin{array}{cc}1& -3\\ 4& 2\end{array}\right| \\ =2(3-4)-1(-1-8)-3(2+12) \\ =-35 \end{gathered}$$

Here$\left|A\right|\ne 0$, That is$A^{-1}$exists

Step 2. The solution to the system of equation

Let us consider,

$A=\left[\begin{array}{ccc}2& 1& -3\\ 1& -3& 2\\ 4& 2& -1\end{array}\right]$

Therefore,

$\begin{array}{r}{R}_1\leftrightarrow R_2=\left[\begin{array}{ccc}1& -3& 2\\ 2& 1& -3\\ 4& 2& -1\end{array}\right]\\ R_2\to R_2-2R_1=\left[\begin{array}{ccc}1& -3& 2\\ 0& 7& -7\\ 4& 2& -1\end{array}\right]\\ R_3\to R_3-4R_1=\left[\begin{array}{ccc}1& -3& 2\\ 0& 7& -7\\ 0& 14& -9\end{array}\right]\\ R_2\to \frac{1}{7}{R}_2=\left[\begin{array}{ccc}1& -3& 2\\ 0& 1& -1\\ 0& 14& -9\end{array}\right]\\ R_3\to R_3-14R_2=\left[\begin{array}{ccc}1& -3& 2\\ 0& 1& -1\\ 0& 0& 5\end{array}\right]\end{array}$

Therefore,

$\left[\begin{array}{ccc}1& -3& 2\\ 0& 1& -1\\ 0& 0& 5\end{array}\right]\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

$\begin{array}{r}x-3y+2z=0\\ y-z=0\dots \dots \left(4\right)\\ 5z=0\dots \dots \left(5\right)\end{array}$

Solving equation (5) we obtain $z=0$,

 substituting the value of $z=0$ in equation (4) we get $y=0$ and hence $x=0$.

Thus we have obtained a trivial solution to the system when $A^{-1}$ exists.

Step 3. Systems of equations A - 1 doesn't exist

Let us consider three systems of equations,

$$\begin{gathered} x+y-6z=0 \\ -3x+y+2z=0 \\ x-y+2z=0.\dots .\left(7\right) \end{gathered}$$

We can write the above system of equations in matrix form as$AX=\mathrm{\Phi }$

Where,

$A=\left[\begin{array}{ccc}1& 1& -6\\ -3& 1& 2\\ 1& -1& 2\end{array}\right],X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]$ and$\mathrm{\Phi }=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Here,

$$\begin{gathered} \left|A\right|=\left|\begin{array}{ccc}1& 1& -6\\ -3& 1& 2\\ 1& -1& 2\end{array}\right|=1\left|\begin{array}{cc}1& 2\\ -1& 2\end{array}\right|-1\left|\begin{array}{cc}-3& 2\\ 1& 2\end{array}\right|-6\left|\begin{array}{cc}-3& 1\\ 1& -1\end{array}\right| \\ =1(2+2)-1(-6-2)-6(3-1)=0 \end{gathered}$$

Since$\left|A\right|=0$, $A^{-1}$ doesn't exist.

Step 4, The solution to the system of equation

Let us consider,

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 1& -6\\ -3& 1& 2\\ 1& -1& 2\end{array}\right] \\ \begin{array}{r}{R}_2\to R_2+3R_1=\left[\begin{array}{ccc}1& 1& -6\\ 0& 4& -16\\ 1& -1& 2\end{array}\right]\\ R_3\to R_3-R_1=\left[\begin{array}{ccc}1& 1& -6\\ 0& 4& -16\\ 0& -2& 8\end{array}\right]\end{array} \\ \begin{array}{r}{R}_2\to \frac{1}{2}{R}_2=\left[\begin{array}{ccc}1& 1& -6\\ 0& 2& -8\\ 0& -2& 8\end{array}\right]\\ R_3\to R_3+R_2=\left[\begin{array}{ccc}1& 1& -6\\ 0& 2& -8\\ 0& 0& 0\end{array}\right]\end{array} \end{gathered}$$

The above matrix is in row echelon form.

$\left[\begin{array}{ccc}1& 1& -6\\ 0& 2& -8\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

$\begin{array}{r}x+y-6z=0\\ 2y-8z=0\end{array}$

From the above equations, we get $y=4z$ and $x=2z$

Hence $x=2c,y=4c,z=c$ constitute the general solution of the above system of equations. Therefore the system has infinitely many solutions.