Question

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

Answer 1

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

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{aligned} & =-21+1-(-14-3)-(-2-9) \\\ & =-20+17+11 \end{aligned}$$ $$=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{aligned} & =\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{aligned}$$ So, $$\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}}$$.