Question

Using matrices solve the following system of equations:
\[x + y - z = 3\];\[2x + 3y + z = 10\];\[3x - y - 7z = 1\].

Answer 1

Hint: To solve the question, at first we have to find out the coefficients of x, y and z from the system of equation and represent it in matrix P. x, y, and z in the system of linear equations are represented in the column matrix X and the constants in the system of equations are represented in the column matrix Q. By solving the matrix equation that is \[PX = Q\]we can get the respective values of x, y and z.

Complete step by step answer:
The system of equations are given by,
\[x+y-z=3\]…………….. (1)
\[2x+3y+z=10\]………………… (2)
And \[3x-y-7z=1\] ………………….. (3)
Now we must represent the above system of equations in matrices. The three sets of coefficients \[1,1,-1\]; \[2,3,1\] and\[3,-1,-7\]must occupy first, second and third rows respectively in the matrix P which is given by
\[P=\left[ \begin{matrix}
   1 & 1 & -1 \\
   2 & 3 & 1 \\
   3 & -1 & -7 \\
\end{matrix} \right]\]……………… (4)
And let the matrix
\[X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]\]………………… (5)
And\[Q=\left[ \begin{matrix}
   3 \\
   10 \\
   1 \\
\end{matrix} \right]\]…….. (6)

We know the system of equation can be written in matrix form
\[\Rightarrow PX=Q\]
Or, \[\Rightarrow X={{P}^{-1}}Q\]…………………… (7)
Where \[{{P}^{-1}}\] is the inverse matrix of P.? Now we will determine\[{{P}^{-1}}\]. To find out the inverse at first let’s find out the determinant of P and cofactors of each element in the matrix P.
The determinant of P is given by
\[\left| P \right|=\left| \begin{matrix}
   1 & 1 & -1 \\
   2 & 3 & 1 \\
   3 & -1 & -7 \\
\end{matrix} \right|\]
\[=1\times \left| \begin{matrix}
   3 & 1 \\
   -1 & -7 \\
\end{matrix} \right|-1\times \left| \begin{matrix}
   2 & 1 \\
   3 & -7 \\
\end{matrix} \right|+(-1)\left| \begin{matrix}
   2 & 3 \\
   3 & -1 \\
\end{matrix} \right|\]
On solving, we get
\[\begin{align}
  & =-21+1-(-14-3)-(-2-9) \\
 & =-20+17+11
\end{align}\]
\[=8\]……..(8)
Let the cofactor matrix of P be\[\left[ \begin{matrix}
   {{C}_{11}} & {{C}_{12}} & {{C}_{13}} \\
   {{C}_{21}} & {{C}_{22}} & {{C}_{23}} \\
   {{C}_{31}} & {{C}_{32}} & {{C}_{33}} \\
\end{matrix} \right]\].
Now
\[{{C}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix}
   3 & 1 \\
   -1 & -7 \\
\end{matrix} \right|=1\left( -21+1 \right)=-20\]
\[{{C}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix}
   2 & 1 \\
   3 & -7 \\
\end{matrix} \right|=\left( -1 \right)\left( -14-3 \right)=17\]
\[{{C}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix}
   2 & 3 \\
   3 & -1 \\
\end{matrix} \right|=1\left( -2-9 \right)=-11\]
\[{{C}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix}
   1 & -1 \\
   -1 & -7 \\
\end{matrix} \right|=\left( -1 \right)\left( -7-1 \right)=8\]
\[{{C}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix}
   1 & -1 \\
   3 & -7 \\
\end{matrix} \right|=1\left( -7+3 \right)=-4\]
\[{{C}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix}
   1 & 1 \\
   3 & -1 \\
\end{matrix} \right|=\left( -1 \right)\left( -1-3 \right)=4\]
\[{{C}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix}
   1 & -1 \\
   3 & 1 \\
\end{matrix} \right|=1\left( 1+3 \right)=4\]
\[{{C}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix}
   1 & -1 \\
   2 & 1 \\
\end{matrix} \right|=\left( -1 \right)\left( 1+2 \right)=-3\]
\[{{C}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix}
   1 & 1 \\
   2 & 3 \\
\end{matrix} \right|=1\left( 3-2 \right)=1\]
The adjoint of matrix P is the transpose of cofactor matrix of P which is defined by
\[adjP={{\left[ \begin{matrix}
   {{C}_{11}} & {{C}_{12}} & {{C}_{13}} \\
   {{C}_{21}} & {{C}_{22}} & {{C}_{23}} \\
   {{C}_{31}} & {{C}_{32}} & {{C}_{33}} \\
\end{matrix} \right]}^{T}}\]
\[=\left[ \begin{matrix}
   {{C}_{11}} & {{C}_{21}} & {{C}_{31}} \\
   {{C}_{12}} & {{C}_{22}} & {{C}_{32}} \\
   {{C}_{13}} & {{C}_{23}} & {{C}_{33}} \\
\end{matrix} \right]\]………………………… (9)
Substituting the cofactor values in eq. (9) we will get
\[adjP=\left[ \begin{matrix}
   -20 & 8 & 4 \\
   17 & -4 & -3 \\
   -11 & 4 & 1 \\
\end{matrix} \right]\] ……………………… (10)
Now we now the inverse matrix of P is defined by
\[{{P}^{-1}}=\dfrac{adjP}{\left| P \right|}\] ………………………………… (11)
Substituting the values of eq. (8) and (10) in eq. (11) we will get
\[{{P}^{-1}}=\dfrac{adjP}{\left| P \right|}\]
\[=\dfrac{1}{8}\left[ \begin{matrix}
   -20 & 8 & 4 \\
   17 & -4 & -3 \\
   -11 & 4 & 1 \\
\end{matrix} \right]\]…………………………. (12)
Substituting the values of eq. (5), (6) and (12) in eq. (7) we will get
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{8}\left[ \begin{matrix}
   -20 & 8 & 4 \\
   17 & -4 & -3 \\
   -11 & 4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   3 \\
   10 \\
   1 \\
\end{matrix} \right]\]
\[\begin{align}
  & =\dfrac{1}{8}\left[ \begin{matrix}
   -20\times 3+8\times 10+4\times 1 \\
   17\times 3+(-4)\times 10+(-3)\times 1 \\
   (-11)\times 3+4\times 10+1\times 1 \\
\end{matrix} \right] \\
 & =\dfrac{1}{8}\left[ \begin{matrix}
   24 \\
   8 \\
   8 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \dfrac{24}{8} \\
   \dfrac{8}{8} \\
   \dfrac{8}{8} \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   3 \\
   1 \\
   1 \\
\end{matrix} \right]
\end{align}\]
So, \[\Rightarrow \left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   3 \\
   1 \\
   1 \\
\end{matrix} \right]\] ………………………………. (12)
Equating the rows of the matrices of both the sides we will get
\[x=3,y=1,z=1\]

So, the correct answer is “Option A”.

Note: If the determinant of a matrix is zero then its inverse does not exist. While representing a system of equations in matrix form, it should be observed that the coefficients are placed in an order. The product of matrices is not commutative therefore \[X = {P^{ - 1}}Q \ne Q{P^{ - 1}}\].