Solve the following system of equation by matrix method:
4x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

Hint:We will use the formula, $X={{A}^{-1}}B$ to solve this question using the matrix method, where, $X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right],{{A}^{-1}}=\dfrac{AdjA}{\left| A \right|}$ and $B=\left[ \begin{matrix}
   16 \\
   19 \\
   25 \\
\end{matrix} \right]$. So, to find ${{A}^{-1}}$, 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 ${{A}^{-1}}=\dfrac{AdjA}{\left| A \right|}$ and then we will solve the question.

Complete step-by-step answer:
It is given in the question that, we have to solve the system of equations, 4x + 3y + z = 16, 2x + y + 3z = 19, x + 2y + 4z = 25 using the matrix method.
So, first we have to convert the given equations into the matrix form. So, we can write it as follows.
\[\left[ \begin{matrix}
   5 & 3 & 1 \\
   2 & 1 & 3 \\
   1 & 2 & 4 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   16 \\
   19 \\
   25 \\
\end{matrix} \right]\]
Here, \[A=\left[ \begin{matrix}
   5 & 3 & 1 \\
   2 & 1 & 3 \\
   1 & 2 & 4 \\
\end{matrix} \right],X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]\] and $B=\left[ \begin{matrix}
   16 \\
   19 \\
   25 \\
\end{matrix} \right]$.
Now, we will find the determinant of matrix A, that is, $\left| A \right|$. So, we get,
$A=\left| \begin{matrix}
   5 & 3 & 1 \\
   2 & 1 & 3 \\
   1 & 2 & 4 \\
\end{matrix} \right|$
= 5 (4 - 6) – 3 (8 - 3) + 1 (4 - 1)
= 5 (-2) – 3 (5) + (3)
= -10 – 15 + 3
= -22
Now, 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,
\[\begin{align}
  & {{c}_{11}}={{\left( -1 \right)}^{1+1}}\left| \begin{matrix}
   1 & 3 \\
   2 & 4 \\
\end{matrix} \right|=1\left( 4-6 \right)=-2 \\
 & {{c}_{12}}={{\left( -1 \right)}^{1+2}}\left| \begin{matrix}
   2 & 3 \\
   1 & 4 \\
\end{matrix} \right|=\left( -1 \right)\left( 8-3 \right)=-5 \\
 & {{c}_{13}}={{\left( -1 \right)}^{1+3}}\left| \begin{matrix}
   2 & 1 \\
   1 & 2 \\
\end{matrix} \right|=1\left( 4-1 \right)=3 \\
 & {{c}_{21}}={{\left( -1 \right)}^{2+1}}\left| \begin{matrix}
   3 & 1 \\
   2 & 4 \\
\end{matrix} \right|=\left( -1 \right)\left( 12-2 \right)=-10 \\
 & {{c}_{22}}={{\left( -1 \right)}^{2+2}}\left| \begin{matrix}
   5 & 1 \\
   1 & 4 \\
\end{matrix} \right|=1\left( 20-1 \right)=19 \\
 & {{c}_{23}}={{\left( -1 \right)}^{2+3}}\left| \begin{matrix}
   5 & 3 \\
   1 & 2 \\
\end{matrix} \right|=\left( -1 \right)\left( 10-3 \right)=-7 \\
 & {{c}_{31}}={{\left( -1 \right)}^{3+1}}\left| \begin{matrix}
   3 & 1 \\
   1 & 3 \\
\end{matrix} \right|=1\left( 9-1 \right)=8 \\
 & {{c}_{32}}={{\left( -1 \right)}^{3+2}}\left| \begin{matrix}
   5 & 1 \\
   2 & 3 \\
\end{matrix} \right|=\left( -1 \right)\left( 15-2 \right)=-13 \\
 & {{c}_{33}}={{\left( -1 \right)}^{3+3}}\left| \begin{matrix}
   5 & 3 \\
   2 & 1 \\
\end{matrix} \right|=1\left( 5-6 \right)=-1 \\
\end{align}\]
Therefore, we get the cofactor matrix as, $\left[ \begin{matrix}
   -2 & -5 & 3 \\
   -10 & 19 & -7 \\
   8 & -13 & -1 \\
\end{matrix} \right]$
Now, on taking the transpose of the above matrix, we will get $AdjA$ as,
$\begin{align}
  & AdjA={{\left[ \begin{matrix}
   -2 & -5 & 3 \\
   -10 & 19 & -7 \\
   8 & -13 & -1 \\
\end{matrix} \right]}^{T}} \\
 & AdjA=\left[ \begin{matrix}
   -2 & -10 & 8 \\
   -5 & 19 & -3 \\
   3 & -7 & -1 \\
\end{matrix} \right] \\
\end{align}$

Now, we know that ${{A}^{-1}}=\dfrac{1}{\left| A \right|}AdjA$. So, we get,
${{A}^{-1}}=\dfrac{1}{-22}\left[ \begin{matrix}
   -2 & -10 & 8 \\
   -5 & 19 & -3 \\
   3 & -7 & -1 \\
\end{matrix} \right]$
We know that $X={{A}^{-1}}B$. So, here we have,
$X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right],{{A}^{-1}}=\dfrac{1}{-22}\left[ \begin{matrix}
   -2 & -10 & 8 \\
   -5 & 19 & -3 \\
   3 & -7 & -1 \\
\end{matrix} \right]$ and $B=\left[ \begin{matrix}
   16 \\
   19 \\
   25 \\
\end{matrix} \right]$.
Therefore, we can write,
$\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{-22}\left[ \begin{matrix}
   -2 & -10 & 8 \\
   -5 & 19 & -3 \\
   3 & -7 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
   16 \\
   19 \\
   25 \\
\end{matrix} \right]$
Now, we will perform the multiplication of the two matrices on the RHS. So, we can write,
\[\begin{align}
  & \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{-22}\left[ \begin{matrix}
   -32-190+200 \\
   -80+361-325 \\
   48-133-25 \\
\end{matrix} \right] \\
 & \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{-22}\left[ \begin{matrix}
   -22 \\
   -44 \\
   -110 \\
\end{matrix} \right] \\
\end{align}\]
Now, we will take $\dfrac{1}{-22}$ 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,
$\begin{align}
  & \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   -22\times \dfrac{1}{-22} \\
   -44\times \dfrac{1}{-22} \\
   -110\times \dfrac{1}{-22} \\
\end{matrix} \right] \\
 & \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   5 \\
\end{matrix} \right] \\
\end{align}$
Therefore, we get the values of x = 1, y = 2 and z = 5.

Note: The most common mistake that the students make in this question is, by not taking the transpose of the matrix formed after finding the cofactors of matrix A. This will lead to getting the wrong answers. Also, while finding the determinant of matrix A, some students may make mistakes by changing the signs. And the students should also know how to carry out multiplication of matrices as there are chances of calculation mistakes also. In the last step we can also take -22 common from the matrix as shown below.
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{-22}\left[ \begin{matrix}
   -22 \\
   -44 \\
   -110 \\
\end{matrix} \right]\]
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{-22}\times -22\left[ \begin{matrix}
   1 \\
   2 \\
   5 \\
\end{matrix} \right]\]
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 \\
   2 \\
   5 \\
\end{matrix} \right]\]
Therefore, we get the values of x = 1, y = 2 and z = 5.