91Ó°ÊÓ

Let us consider three systems of equations,

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

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

$AX=\mathrm{Φ}$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 $\mathrm{Φ}=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Therefore,

localid="1647522676711" $$\begin{gathered} \left|A\right|=\left|\begin{array}{ccc}2& 1& −3\\ 1& −3& 2\\ 4& 2& −1\end{array}\right|=2\left|\begin{array}{cc}−3& 2\\ 2& −1\end{array}\right|−1\left|\begin{array}{cc}1& 2\\ 4& −1\end{array}\right|−3\left|\begin{array}{cc}1& −3\\ 4& 2\end{array}\right| \\ =2(3−4)−1(−1−8)−3(2+12) \\ =−35 \end{gathered}$$

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

Let us consider,

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

Therefore,

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

Therefore,

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

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

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.

Let us consider three systems of equations,

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

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

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$\mathrm{Φ}=\left[\begin{array}{l}0\\ 0\\ 0\end{array}\right]$

Here,

role="math" localid="1647523627648" $$\begin{gathered} \left|A\right|=\left|\begin{array}{ccc}1& 1& −6\\ −3& 1& 2\\ 1& −1& 2\end{array}\right|=1\left|\begin{array}{cc}1& 2\\ −1& 2\end{array}\right|−1\left|\begin{array}{cc}−3& 2\\ 1& 2\end{array}\right|−6\left|\begin{array}{cc}−3& 1\\ 1& −1\end{array}\right| \\ =1(2+2)−1(−6−2)−6(3−1)=0 \end{gathered}$$

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

Let us consider,

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

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.