Question

Solve by matrix method:
2x + 3y + 3z = 5
x – 2y + z = -4
3x – y – 2z = 3

Answer 1

Hint: In order to solve this question, we will use the formula, $X={{A}^{-1}}B$ where, $X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right],{{A}^{-1}}=\dfrac{AdjA}{\left| A \right|}$ and $B=\left[ \begin{matrix}
   5 \\
   -4 \\
   3 \\
\end{matrix} \right]$. So we will have to find ${{A}^{-1}}$, for that, we will first find the determinant of matrix A, that is $\left| A \right|$, then we will find the cofactors of matrix A, take its transpose, and that will be $AdjA$. Therefore, we will be able to get the inverse of matrix A using ${{A}^{-1}}=\dfrac{AdjA}{\left| A \right|}$.and hence We then substitute all the obtained values of ${{A}^{-1}}$ and $B$in the main formula and do necessary calculations to get the value of x, y and z accordingly.

Complete step-by-step answer:
It is given in the question that, we have to solve the system of equations,
$2x+3y-5=0$
\[x-\text{2}y+z=-4\]
\[3x-y-2z=3\] using the matrix method.
So, first we have to covert the given equations into the matrix form. So, we can write it as follows.
\[\left[ \begin{matrix}
   2 & 3 & 3 \\
   1 & -2 & 1 \\
   3 & -1 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 \\
   -4 \\
   3 \\
\end{matrix} \right]\].
Here, \[A=\left[ \begin{matrix}
   2 & 3 & 3 \\
   1 & -2 & 1 \\
   3 & -1 & -2 \\
\end{matrix} \right],X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]\] and \[B=\left[ \begin{matrix}
   5 \\
   -4 \\
   3 \\
\end{matrix} \right]\].
Now, we will find the determinant of matrix A, that is, $\left| A \right|$. So, we get,
$\Rightarrow \left| A \right|=\left| \begin{matrix}
   2 & 3 & 3 \\
   1 & -2 & 1 \\
   3 & -1 & -2 \\
\end{matrix} \right|$.
\[\Rightarrow \left| A \right|=2\left( 4+1 \right)-3\left( -2-3 \right)+3\left( -1+6 \right)\].
\[\Rightarrow \left| A \right|=2\left( 5 \right)-3\left( -5 \right)+3\left( 5 \right)\].
\[\Rightarrow \left| A \right|=10+15+15\].
\[\Rightarrow \left| A \right|=40\].
As we have $\left| A \right|\ne 0$. It means that the system of equations is consistent and has a unique solution.
Now, we have AX = B, from this, we get, $X=\dfrac{B}{A}$, which can be written as, $X={{A}^{-1}}B$.
Now, we will find ${{A}^{-1}}$ so, we will find the adjoint of matrix A, that is Adj A. So, let us assume ${{c}_{ij}}$ as the cofactors of the elements ${{a}_{ij}}$ in $A\left[ {{a}_{ij}} \right]$. So, we get,
\[\Rightarrow {{c}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix}
   -2 & 1 \\
   -1 & -2 \\
\end{matrix} \right|=1\left( 4+1 \right)=5\].
\[\Rightarrow {{c}_{12}}={{\left( -1 \right)}^{1+2}}\left| \begin{matrix}
   1 & 1 \\
   3 & -2 \\
\end{matrix} \right|=\left( -1 \right)\left( -2-3 \right)=5\].
\[\Rightarrow {{c}_{13}}={{\left( -1 \right)}^{1+3}}\left| \begin{matrix}
   1 & -2 \\
   3 & -1 \\
\end{matrix} \right|=1\left( -1+6 \right)=5\].
\[\Rightarrow {{c}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix}
   3 & 3 \\
   -1 & -2 \\
\end{matrix} \right|=\left( -1 \right)\left( -6+3 \right)\].
\[\Rightarrow {{c}_{22}}={{\left( -1 \right)}^{2+2}}\left| \begin{matrix}
   2 & 3 \\
   3 & -2 \\
\end{matrix} \right|=1\left( -4-9 \right)=-13\].
\[\Rightarrow {{c}_{23}}={{\left( -1 \right)}^{2+3}}\left| \begin{matrix}
   2 & 3 \\
   3 & -1 \\
\end{matrix} \right|=\left( -1 \right)\left( -2-9 \right)=11\].
\[\Rightarrow {{c}_{31}}={{\left( -1 \right)}^{3+1}}\left| \begin{matrix}
   3 & 3 \\
   -2 & 1 \\
\end{matrix} \right|=1\left( 3+6 \right)=9\].
\[\Rightarrow {{c}_{32}}={{\left( -1 \right)}^{3+2}}\left| \begin{matrix}
   2 & 3 \\
   1 & 1 \\
\end{matrix} \right|=\left( -1 \right)\left( 2-3 \right)=1\].
\[\Rightarrow {{c}_{33}}={{\left( -1 \right)}^{3+3}}\left| \begin{matrix}
   2 & 3 \\
   1 & -2 \\
\end{matrix} \right|=1\left( -4-3 \right)=-7\].
Therefore, we get the matrix form as, $\left[ \begin{matrix}
   5 & 5 & 5 \\
   3 & -13 & 11 \\
   9 & 1 & -7 \\
\end{matrix} \right]$.
Now, on taking the transpose of the above matrix, we will get $AdjA$ as,
$\Rightarrow AdjA={{\left[ \begin{matrix}
   5 & 5 & 5 \\
   3 & -13 & 11 \\
   9 & 1 & -7 \\
\end{matrix} \right]}^{T}}$.
$\Rightarrow AdjA=\left[ \begin{matrix}
   5 & 3 & 9 \\
   5 & -13 & 1 \\
   5 & 11 & -7 \\
\end{matrix} \right]$.

Now, we know that ${{A}^{-1}}=\dfrac{1}{\left| A \right|}AdjA$. So, we get,
$\Rightarrow {{A}^{-1}}=\dfrac{1}{40}\left[ \begin{matrix}
   5 & 3 & 9 \\
   5 & -13 & 1 \\
   5 & 11 & -7 \\
\end{matrix} \right]$.
We know that $X={{A}^{-1}}B$. So, here we have $X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right],{{A}^{-1}}=\dfrac{1}{40}\left[ \begin{matrix}
   5 & 3 & 9 \\
   5 & -13 & 1 \\
   5 & 11 & -7 \\
\end{matrix} \right]$ and $B=\left[ \begin{matrix}
   5 \\
   -4 \\
   3 \\
\end{matrix} \right]$.
Therefore, we can write,
$\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{40}\left[ \begin{matrix}
   5 & 3 & 9 \\
   5 & -13 & 1 \\
   5 & 11 & -7 \\
\end{matrix} \right]\left[ \begin{matrix}
   5 \\
   -4 \\
   3 \\
\end{matrix} \right]$.
Now, we will perform the multiplication of the two matrices on the RHS. So, we can write,
\[\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{40}\left[ \begin{matrix}
   5\left( 5 \right)+3\left( -4 \right)+9\left( 3 \right) \\
   5\left( 5 \right)+\left( -13 \right)\left( -4 \right)+1\left( 3 \right) \\
   5\left( 5 \right)+11\left( -4 \right)+\left( -7 \right)\left( 3 \right) \\
\end{matrix} \right]\].
\[\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{40}\left[ \begin{matrix}
   25-12+27 \\
   25+52+3 \\
   25-44-21 \\
\end{matrix} \right]\].
\[\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{40}\left[ \begin{matrix}
   40 \\
   80 \\
   -40 \\
\end{matrix} \right]\].
Now, we will take $\dfrac{1}{40}$ and multiply it with the terms inside the matrix, as it is a constant and we know that constants can be multiplied with the terms inside a matrix. So, we will get,
$\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   40\times \dfrac{1}{40} \\
   80\times \dfrac{1}{40} \\
   -40\times \dfrac{1}{40} \\
\end{matrix} \right]$.
$\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   -1 \\
\end{matrix} \right]$.
Therefore, we get the values of x = 1, y = 2 and z = -1.

Note: We should know that transpose is formed by just interchanging the rows and columns of the given matrix. We should solve every step of this problem carefully as it involves lengthy calculations which may lead to committing the mistake. Similarly, we can expect problems to solve the given linear set of equations by using Cramer’s method, Gauss Jordan method or by elementary row transformations.