Question

How do you solve the system of equations $x + 2y + z = 2$, $2x + 3y + 3z = - 3$ and $2x + 3y + 2z = 2$?

Answer 1

Hint: In this question we have to solve the given system of equations, to solve this we have to use the formula $x = {A^{ - 1}}B$where $x = \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right]$, ${A^{ - 1}} = \dfrac{{AdjA}}{{\left| A \right|}}$ and $B$ is the matrix with numbers on the right hand side of the equation, to find $AdjA$ we will find the cofactors of matrix A and then take its transpose and then divide this by the determinant of matrix A, then we will find the inverse of the matrix, now substituting the values in the formula we will get the required values for $x$,$y$ and $z$.

Complete step by step solution:
Given system of equations are,
$x + 2y + z = 2$,
$2x + 3y + 3z = - 3$ and ,
$2x + 3y + 2z = 2$,
Now we will write the equations in the matrix form i.e.,
$\left[ {\begin{array}{*{20}{c}}
  1&2&1 \\
  2&3&3 \\
  2&3&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  2 \\
  { - 3} \\
  2
\end{array}} \right]$,
Here $A = \left[ {\begin{array}{*{20}{c}}
  1&2&1 \\
  2&3&3 \\
  2&3&2
\end{array}} \right]$, $x = \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right]$and $B = \left[ {\begin{array}{*{20}{c}}
  2 \\
  { - 3} \\
  2
\end{array}} \right]$,
 $ \Rightarrow Ax = B$,
Now taking A to right hand side we get,
$ \Rightarrow x = {A^{ - 1}}B$,
Now we know that inverse of a matrix is given by the formula, ${A^{ - 1}} = \dfrac{{AdjA}}{{\left| A \right|}}$,
Now we will find the determinant of matrix A i.e.,
$ \Rightarrow $$\left| A \right| = \left| {\left[ {\begin{array}{*{20}{c}}
  1&2&1 \\
  2&3&3 \\
  2&3&2
\end{array}} \right]} \right|$,
Now applying determinant we get,
$ \Rightarrow \left| A \right| = 1\left( {6 - 9} \right) - 2\left( {4 - 6} \right) + 1\left( {6 - 6} \right)$,
Now simplifying we get,
$ \Rightarrow \left| A \right| = - 3 - 2\left( { - 2} \right) + 1\left( 0 \right)$,
Now again simplifying we get,
$ \Rightarrow \left| A \right| = - 3 + 4 = 1$,
So, the determinant is not equal to 0, it means the system of equations is consistent and has a unique solution.
Now we have to inverse of matrix A, for this we have to find the adjoint of matrix A, that is Adj A, let us take ${c_{ij}}$as the cofactors of the elements ${a_{ij}}$in $A\left[ {{a_{ij}}} \right]$, we get,
${c_{11}} = {\left( { - 1} \right)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  3&3 \\
  3&2
\end{array}} \right| = {\left( { - 1} \right)^2}\left( {6 - 9} \right) = 1\left( { - 3} \right) = - 3$,
\[{c_{12}} = {\left( { - 1} \right)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  2&3 \\
  2&2
\end{array}} \right| = {\left( { - 1} \right)^3}\left( {4 - 6} \right) = - 1\left( { - 2} \right) = 2\],
\[{c_{13}} = {\left( { - 1} \right)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  2&3 \\
  2&3
\end{array}} \right| = {\left( { - 1} \right)^4}\left( {6 - 6} \right) = 1\left( 0 \right) = 0\]
\[{c_{21}} = {\left( { - 1} \right)^{2 + 1}}\left| {\begin{array}{*{20}{c}}
  2&1 \\
  3&2
\end{array}} \right| = {\left( { - 1} \right)^3}\left( {4 - 3} \right) = - 1\left( 1 \right) = - 1\],
\[{c_{22}} = {\left( { - 1} \right)^{2 + 2}}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  2&2
\end{array}} \right| = {\left( { - 1} \right)^4}\left( {2 - 2} \right) = 1\left( 0 \right) = 0\],
\[{c_{23}} = {\left( { - 1} \right)^{2 + 3}}\left| {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right| = {\left( { - 1} \right)^5}\left( {3 - 4} \right) = - 1\left( { - 1} \right) = 1\],
\[{c_{31}} = {\left( { - 1} \right)^{3 + 1}}\left| {\begin{array}{*{20}{c}}
  2&1 \\
  3&3
\end{array}} \right| = {\left( { - 1} \right)^4}\left( {6 - 3} \right) = 1\left( 3 \right) = 3\],
\[{c_{32}} = {\left( { - 1} \right)^{3 + 2}}\left| {\begin{array}{*{20}{c}}
  1&1 \\
  2&3
\end{array}} \right| = {\left( { - 1} \right)^5}\left( {3 - 2} \right) = - 1\left( 1 \right) = - 1\],
\[{c_{33}} = {\left( { - 1} \right)^{3 + 3}}\left| {\begin{array}{*{20}{c}}
  1&2 \\
  2&3
\end{array}} \right| = {\left( { - 1} \right)^6}\left( {3 - 4} \right) = 1\left( { - 1} \right) = - 1\],
Now we get the matrix of the form, \[\left[ {\begin{array}{*{20}{c}}
  { - 3}&2&0 \\
  { - 1}&0&1 \\
  3&{ - 1}&{ - 1}
\end{array}} \right]\],
Now taking the transpose of the matrix we get,
\[ \Rightarrow AdjA = {\left[ {\begin{array}{*{20}{c}}
  { - 3}&2&0 \\
  { - 1}&0&1 \\
  3&{ - 1}&{ - 1}
\end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 1}&3 \\
  2&0&{ - 1} \\
  0&1&{ - 1}
\end{array}} \right]\],
Now using the formula of inverse of matrix i.e., ${A^{ - 1}} = \dfrac{{AdjA}}{{\left| A \right|}}$, we get,
\[{A^{ - 1}} = \dfrac{1}{1}\left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 1}&3 \\
  2&0&{ - 1} \\
  0&1&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 1}&3 \\
  2&0&{ - 1} \\
  0&1&{ - 1}
\end{array}} \right]\],
We know that $x = {A^{ - 1}}B$, now multiplying the matrices we get,
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 3}&{ - 1}&3 \\
  2&0&{ - 1} \\
  0&1&{ - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2 \\
  { - 3} \\
  2
\end{array}} \right]\],
Now multiplying we get,
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 3\left( 2 \right) + \left( { - 1} \right)\left( { - 3} \right) + 3\left( 2 \right)} \\
  {2\left( 2 \right) + 0\left( { - 3} \right) + \left( { - 1} \right)\left( 2 \right)} \\
  {0\left( 2 \right) + 1\left( { - 3} \right) + \left( { - 1} \right)\left( 2 \right)}
\end{array}} \right]\],
Now simplifying we get,
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 6 + 3 + 6} \\
  {4 + 0 - 2} \\
  {0 - 3 - 2}
\end{array}} \right]\],
Now simplifying we get,
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  3 \\
  2 \\
  { - 5}
\end{array}} \right]\],
So, the values are \[x = 3\], \[y = 2\] and \[z = - 5\].


\[\therefore \]When the system of equations $x + 2y + z = 2$, $2x + 3y + 3z = - 3$ and $2x + 3y + 2z = 2$are solved then the values of \[x\], \[y\]and \[z\]are 3, 2 and -5.

Note:
It is possible to solve this system using the elimination or substitution method, but it is also possible to do it with a matrix operation. Before we start setting up the matrices, it is important to do the following:
Make sure that all of the equations are written in a similar manner, meaning the variables need to all be in the same order.
Make sure that one side of the equation is only variables and their coefficients, and the other side is just constants.