Question

Solve the following system of linear equations in three variables
$x+y+z=6,y+3z=11,x+z=2y$

Answer 1

Hint:We will use the elimination method. In the elimination method, we decrease the number of variables between two equations by 1 by taking a certain linear combination of the two equations. Subtract the second equation from the first equation and hence form a linear equation in x and z. Multiply the first equation by 2 and add the resulting equation to the third equation and hence form a linear equation in x and z. Repeat a similar process on the equations in x and z and hence find the value of x and z. Substitute the value of x and z in the third equation and hence find the value of y.

Complete step-by-step answer:
We have
$\begin{align}
  & x+y+z=6\text{ }\left( i \right) \\
 & y+3z=11\text{ }\left( ii \right) \\
 & x+z=2y\text{ }\left( iii \right) \\
\end{align}$
Subtracting equation (ii) from equation (i), we get
$\begin{align}
  & x+y+z-y-3z=6-11 \\
 & \Rightarrow x-2z=-5\text{ }\left( iv \right) \\
\end{align}$
Multiplying equation (i) by 2 and adding equation (iii) to the resulting equation, we get
$\begin{align}
  & 2\left( x+y+z \right)+x+z=2\times 6+2y \\
 & \Rightarrow 2x+2y+2z+x+z=12+2y \\
 & \Rightarrow 3x+3z=12 \\
\end{align}$
Dividing both sides by 3, we get
$x+z=4\text{ }\left( v \right)$
Subtracting equation (iv) from equation (v), we get
$\begin{align}
  & x+z-x+2z=4+5 \\
 & \Rightarrow 3z=9 \\
\end{align}$
Dividing both sides by 3, we get
$z=3$
Substituting the value of z in equation (v), we get
$x+3=4$
Subtracting 3 from both sides, we get
$x=1$
Substituting the value of x and z in equation (iii), we get
$\begin{align}
  & 1+3=2y \\
 & \Rightarrow 2y=4 \\
\end{align}$
Dividing both sides by 2, we get
$y=2$
Hence x = 1, y = 2, z = 3 is the solution of the given system of the equations.

Note: Alternative Solution:
We can write the given system of equations as $AX=B$, where
$A=\left[ \begin{matrix}
   1 & 1 & 1 \\
   0 & 1 & 3 \\
   1 & -2 & 1 \\
\end{matrix} \right],X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right],B=\left[ \begin{matrix}
   6 \\
   11 \\
   0 \\
\end{matrix} \right]$
Premultiplying both sides by ${{A}^{-1}}$, we get
$X={{A}^{-1}}B$
Now, we know that ${{A}^{-1}}=\dfrac{1}{\det \left( A \right)}adj\left( A \right)$
Finding det A:
$\det \left( A \right)=1\left( 1+3\times 2 \right)-1\left( 0-3 \right)+1\left( 0-1 \right)=7+3-1=9$
Hence, we have det(A) = 9
Finding adj(A):
We have
$\begin{align}
  & {{C}_{11}}=1+6=7,{{C}_{12}}=-\left( 0-3 \right)=3,{{C}_{13}}=\left( 0-1 \right)=-1 \\
 & {{C}_{21}}=-\left( 1+2 \right)=-3,{{C}_{22}}=\left( 1-1 \right)=0,{{C}_{23}}=-\left( -2-1 \right)=3 \\
 & {{C}_{31}}=\left( 3-1 \right)=2,{{C}_{32}}=-\left( 3-0 \right)=-3,{{C}_{33}}=\left( 1-0 \right)=1 \\
\end{align}$
Hence, we have
$adj\left( A \right)={{\left[ \begin{matrix}
   7 & 3 & -1 \\
   -3 & 0 & 3 \\
   2 & -3 & 1 \\
\end{matrix} \right]}^{T}}=\left[ \begin{matrix}
   7 & -3 & 2 \\
   3 & 0 & -3 \\
   -1 & 3 & 1 \\
\end{matrix} \right]$
Hence, we have
${{A}^{-1}}=\dfrac{1}{9}\left[ \begin{matrix}
   7 & -3 & 2 \\
   3 & 0 & -3 \\
   -1 & 3 & 1 \\
\end{matrix} \right]$
Hence, we have
$X={{A}^{-1}}B=\dfrac{1}{9}\left[ \begin{matrix}
   7 & -3 & 2 \\
   3 & 0 & -3 \\
   -1 & 3 & 1 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   6 \\
   11 \\
   0 \\
\end{matrix} \right]=\dfrac{1}{9}\left[ \begin{matrix}
   42-33 \\
   18-0+0 \\
   -6+33 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]$
Hence, we have
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]\]
Hence, we have x = 1, y = 2 and z =3.