Question

Let system of linear equations $a_{1}x+b_{1}y+c_{1}z=d_1{a}_{2}x+b_{2}y+c_{2}z=d_2$ &$a_{3}x+b_{3}y+c_{3}z=d_3$ can be expressed in the form $AX=B....(\ast )$ where $A=\left(\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right)$ ,$B=\left(\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right)$ $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$ The above system of equations (*) is said to be consistent with unique solution if A is non singular & the values of the variables x, y, z can be determined by using the equation $X=A^{-1}B$ and if A is singular then system of equations are either consistent with infinitely many solutions or inconsistent with no solution accordingly $\left(adjA\right)\left(B\right)=0$ and $\left(adjA\right)\left(B\right)\ne 0$ where (adj A) is the transpose of cofactor matrix of A Now Assume $A=\left(\begin{array}{ccc}a& 1& 0\\ 1& b& d\\ 1& b& c\end{array}\right)$, $B=\left(\begin{array}{ccc}a& 1& 1\\ 0& d& c\\ f& g& h\end{array}\right)$ $P=\left(\begin{array}{c}f\\ g\\ h\end{array}\right)$,$Q=\left(\begin{array}{c}{a}^2\\ 0\\ 0\end{array}\right)$,$X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$The system $AX=P$ possesses consistency with infinitely many solutions if?

• A. $ab=1,c=d$
• B. $c=d,h=g$
• C. $ab=1,h=g$
• D. $c=d,h=g,ab=1$

Answer 1

Answer: (B), (D)

The correct options are
B $c=d,h=g$
D $c=d,h=g,ab=1$

$AX=P$ has either no solution or infinite solutions if $\left|A\right|=0$

$\left|\begin{array}{ccc}a& 1& 0\\ 1& b& d\\ 1& b& c\end{array}\right|=0$

$\Rightarrow a(bc-bd)-1(c-d)=0$

$\Rightarrow (ab-1)(c-d)=0$

$\therefore ab=1,c=d$

Now cofactor matrix of $A=\left[\begin{array}{ccc}bc-bd& d-c& 0\\ -c& ac& 1-ab\\ d& -da& ab-1\end{array}\right]$

$\therefore adj.A=\left[\begin{array}{ccc}bc-bd& -c& d\\ d-c& ac& -da\\ 0& 1-ab& ab-1\end{array}\right]$

Now for infinite solution $\left(adiA\right)\left(P\right)=O$
$\therefore (adj.A)\left(P\right)=\left[\begin{array}{ccc}bc-bd& -c& d\\ d-c& ac& -da\\ 0& 1-ab& ab-1\end{array}\right]$ $\left[\begin{array}{c}f\\ g\\ h\end{array}\right]$$=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$
$\Rightarrow bf(c-d)-cg+dh=0,f(d-c)+acg-dha=0$ and $(ab-1)(h-g)=0$

So all the above holds if $d=c,g=h,$ whether $ab=1,ab\ne 1$

$\therefore$ Choice (b) (d) are correct.