Browse Questions

# Examine the consistency of the system of equations: $\quad 3x -y -2z = 2$ $\quad 2y -z = -1$ $\quad 3x -5y = 3$

Toolbox:
• (i) A matrix is said to be singular if |A|=0.
• (ii)If (adj A)B$\neq 0$,then there is no solution and the system is said the system is said to be in consistent.
•  If A is a non-singular matrix,AX=B,then X=A−1B.
Using this we can solve the system of equation which has unique solution.
This can be written in the form AX=B

$\begin{bmatrix}3 & -1& -2\\0 & 2& -1\\3 & -5 &0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\-1\\3\end{bmatrix}$

Here $A=\begin{bmatrix}3 & -1& -2\\0 & 2& -1\\3 & -5&0\end{bmatrix}\;X=\begin{bmatrix}x\\y\\z\end{bmatrix}\;B=\begin{bmatrix}2\\-1\\3\end{bmatrix}$

Let us evaluate the value of determinant |A| by expanding along $R_1$

|A|=3(0-5)-(-1)(0+3)-2(0-6)=-15+3+12=0

Hence it is a singular matrix.

Let us calculate (adj A).This can be done by finding the minors and cofactors.

$M_{11}=\begin{vmatrix}2 & -1\\-5 & 0\end{vmatrix}=0-5=-5$

$M_{12}=\begin{vmatrix}0 & -1\\3 & 0\end{vmatrix}=0+3=3$

$M_{13}=\begin{vmatrix}0 & 2\\3 & -5\end{vmatrix}=0-6=-6$

$M_{21}=\begin{vmatrix}-1 & -2\\-5 & 0\end{vmatrix}=0-10=-10$

$M_{22}=\begin{vmatrix}3 & -2\\3 & 0\end{vmatrix}=0+6=6$

$M_{23}=\begin{vmatrix}3 & -1\\3 & -5\end{vmatrix}=-15+3=-12$

$M_{31}=\begin{vmatrix}-1 & -2\\2 & -1\end{vmatrix}=1+4=5$

$M_{32}=\begin{vmatrix}3 & -2\\0 & -1\end{vmatrix}=-3+0=-3$

$M_{33}=\begin{vmatrix}3 & -1\\0 & 2\end{vmatrix}=6+0=6$

$A_{11}=(-1)^{1+1}.-5=-5.$

$A_{12}=(-1)^{1+2}.3=-3.$

$A_{13}=(-1)^{1+3}.(-6)=-6.$

$A_{21}=(-1)^{2+1}.(-10)=10.$

$A_{22}=(-1)^{2+2}.(6)=6.$

$A_{23}=(-1)^{2+3}.(-12)=12.$

$A_{31}=(-1)^{3+1}.(5)=5.$

$A_{32}=(-1)^{3+2}.(-3)=3.$

$A_{33}=(-1)^{3+3}.(6)=6.$

$(adj A)=\begin{bmatrix}A_{11} &A_{21} &A_{31}\\A_{12} &A_{22} &A_{32}\\A_{13} &A_{23} &A_{33}\end{bmatrix}$

$\;\;\;\quad=\begin{bmatrix}-5 &10 &5\\-3 &6 &3\\-6 &12 &6\end{bmatrix}$

(adj A)B$=\begin{bmatrix}-5 &10 &5\\-3 &6 &3\\-6 &12 &6\end{bmatrix}\begin{bmatrix}2 \\-1\\3\end{bmatrix}$

Matrix multiplication can be done by multiplying the rows of A by the column B.

Hence (adj A)B$=\begin{bmatrix}-10-10+15\\-6-6+9\\-12-12+18\end{bmatrix}=\begin{bmatrix}-5 \\-3\\-6\end{bmatrix}\neq 0$

Thus (adj A)B$\neq 0$

Hence the given system of equation has no solution and it is inconsistent.

edited Feb 27, 2013