Question

Solve the following system of homogeneous linear equations by matrix method:
\[\begin{align}
  & 2x-y+z=0 \\
 & 3x+2y-z=0 \\
 & x+4y+3z=0 \\
\end{align}\]

Answer 1

Hint: In this question, from the conditions of a homogeneous system of equations we first need to write the given equations as product two matrices. Then if the determinant of the coefficient matrix is not equal to zero then we have a trivial solution but if it is equal to zero then we have a non-trivial solution.

Complete step-by-step answer:
Determinant: Every square matrix A is associated with a number, called its determinant and it is denoted by \[\left| A \right|\].
If \[A=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
   {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\
   {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\
\end{matrix} \right)\], then
\[\left| A \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{33}}-{{a}_{31}}{{a}_{23}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)\]
SOLUTION OF HOMOGENEOUS SYSTEM OF EQUATIONS BY MATRIX METHOD
Let \[AX=0\] is a system of n linear equations in n variables.
(a) If \[\left| A \right|\ne 0\], then it has only one solution \[X=0\], is called the trivial solution.
(b) If \[\left| A \right|=0\], then the system has infinitely many solutions, called non-trivial solutions.
Now, from the given linear equations in the question
\[\begin{align}
  & 2x-y+z=0 \\
 & 3x+2y-z=0 \\
 & x+4y+3z=0 \\
\end{align}\]
Let us now write them as a product of 2 matrices as mentioned above in the formula
\[\left( \begin{matrix}
   2 & -1 & 1 \\
   3 & 2 & -1 \\
   1 & 4 & 3 \\
\end{matrix} \right)\left( \begin{matrix}
   x & 0 & 0 \\
   y & 0 & 0 \\
   z & 0 & 0 \\
\end{matrix} \right)=0\]
Now, on comparing with the formula we get,
\[A=\left( \begin{matrix}
   2 & -1 & 1 \\
   3 & 2 & -1 \\
   1 & 4 & 3 \\
\end{matrix} \right),X=\left( \begin{matrix}
   x & 0 & 0 \\
   y & 0 & 0 \\
   z & 0 & 0 \\
\end{matrix} \right)\]
Now, to get the number of solutions we first need to find the determinant of the matrix A.
Let us now find the determinant of A using the formula
\[\left| A \right|={{a}_{11}}\left( {{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}} \right)-{{a}_{12}}\left( {{a}_{21}}{{a}_{33}}-{{a}_{31}}{{a}_{23}} \right)+{{a}_{13}}\left( {{a}_{21}}{{a}_{32}}-{{a}_{22}}{{a}_{31}} \right)\]
Now , we have the matrix as
\[A=\left( \begin{matrix}
   2 & -1 & 1 \\
   3 & 2 & -1 \\
   1 & 4 & 3 \\
\end{matrix} \right)\]
Now, on substituting the respective values in the formula of determinant we get,
\[\Rightarrow \left| A \right|=2\left( 2\times 3-\left( -1\times 4 \right) \right)-\left( -1 \right)\left( 3\times 3-\left( -1\times 1 \right) \right)+1\left( 3\times 4-2\times 1 \right)\]
Now, on multiplying the corresponding terms we get,
\[\Rightarrow \left| A \right|=2\left( 6-\left( -4 \right) \right)-\left( -1 \right)\left( 9-\left( -1 \right) \right)+1\left( 12-2 \right)\]
Now, on simplifying the terms in the bracket we get,
\[\Rightarrow \left| A \right|=2\left( 6+4 \right)-\left( -1 \right)\left( 9+1 \right)+1\left( 12-2 \right)\]
Now, on multiplying the corresponding terms and simplifying it further we get,
\[\Rightarrow \left| A \right|=20+10+10\]
Now, on further simplification we get,
\[\Rightarrow \left| A \right|=40\]
Now, from the above we get,
\[\therefore \left| A \right|\ne 0\]
Thus, the solution is \[X=0\]
Hence, the system of homogeneous equations given have a trivial solution of \[X=0\]

Note:
Instead of using the matrix method we can also solve it using rank method and many other methods and then verify the answer. Both the methods should give the same result.
It is important to note that the product of the matrices we wrote should give the linear equations given when simplified. It is also important to note that while considering the matrix we should change the positions of the values and neglect any of the term because it changes the result.