Question

Solve the following systems of homogeneous linear equations by matrix method:
\[
  x + y + z = 0 \\
  x - y - 5z = 0 \\
  x + 2y + 4z = 0 \\
\]

Answer 1

Hint: Check for number of solution after finding the determinant and then solve the variable matrix by finding the inverse of coefficient matrix

Complete step-by-step answer:
Above mentioned equation can be written as follows in matrix form:
$\left[ {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&{ - 1}&{ - 5} \\
  1&2&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0 \\
  0 \\
  0
\end{array}} \right]$
Consider the equivalent terms as $AX = O$, where $A$ is the coefficient matrix, $X$ is the variable matrix and $O$ is the right hand side.
To check the consistency, we will calculate the determinant of the matrix $A$.
$
  \left| A \right| = \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&{ - 1}&{ - 5} \\
  1&2&4
\end{array}} \right| \\
   \Rightarrow 1 \times [( - 1) \times 4 - 2 \times ( - 5)] - 1 \times [1 \times 4 - 1 \times ( - 5)] + 1 \times [1 \times 2 - 1 \times ( - 1)] \\
   \Rightarrow [ - 4 + 10] - [4 + 5] + [2 + 1] \\
   \Rightarrow 6 - 9 + 3 = 0 \\
   \Rightarrow \left| A \right| = 0 \\
$
This means there are infinitely many possible solutions to the above system of equations. To find a general solution of the infinite possibilities, we will consider a representative value $r$ for $z$ variable. So our equations will be as follows:
$
  x + y + r = 0 \\
  x - y - 5r = 0 \\
  x + 2y + 4r = 0 \\
$
Bringing the assumed variable terms on right hand side, we get,
$
  x + y = - r \\
  x - y = 5r \\
$
We can write this again in matrix form, as follows:
$\left[ {\begin{array}{*{20}{c}}
  1&1 \\
  1&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - r} \\
  {5r}
\end{array}} \right]$
Where above matrix form can be represented as $PX = B$.
To find whether solution is possible for the above reformed matrix, we will find the determinant of the new coefficient matrix.
$
  \left| P \right| = \left| {\begin{array}{*{20}{c}}
  1&1 \\
  1&{ - 1}
\end{array}} \right| = - 1 - 1 = - 2 \\
   \Rightarrow \left| P \right| \ne 0 \\
$ (1)
Thus the inverse of the matrix exists. Thus we can rearrange above matrix form as follows:
$X = {P^{ - 1}}B$ (2)
To proceed to the solution of the above reformed equations, we will find the inverse of the matrix.
${P^{ - 1}} = \dfrac{{adj(P)}}{{\left| P \right|}}$ (3)
Now, $adj(P) = \left[ {\begin{array}{*{20}{c}}
  {{C_{11}}}&{{C_{12}}} \\
  {{C_{21}}}&{{C_{22}}}
\end{array}} \right]$
Where ${C_{11}} = - 1;{C_{12}} = - 1;{C_{22}} = 1;{C_{21}} = - 1$
Substituting these values along with the value of determinant in (3), we get,
${P^{ - 1}} = \dfrac{1}{{( - 2)}}\left[ {\begin{array}{*{20}{c}}
  { - 1}&{ - 1} \\
  { - 1}&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{\dfrac{1}{2}} \\
  {\dfrac{1}{2}}&{\dfrac{{ - 1}}{2}}
\end{array}} \right]$ (4)
Substituting (4) in (2), we get,
$X = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{1}{2}}&{\dfrac{1}{2}} \\
  {\dfrac{1}{2}}&{\dfrac{{ - 1}}{2}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  { - r} \\
  {5r}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {\dfrac{{ - r}}{2} + \dfrac{{5r}}{2}} \\
  {\dfrac{{ - r}}{2} - \dfrac{{5r}}{2}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {2r} \\
  { - 3r}
\end{array}} \right]$
So, our solution is $(x,y,z) = (2r, - 3r,r)$.
Above solution is a general solution and represents infinite solutions based on the range of values of $r$.
If we assume a value of $r = 1$, we will get a particular solution, as follows:
$(x,y,z) = (2, - 3,1)$
The solution represents the intersection of the planes given by the above system of equations passing through the origin.

Note: This problem can also be solved by writing the equations in augmented matrix form reduced to upper matrix form to get the above general solution.