Question

Given $a,b\in \{0,1,2,3,\mathrm{4.....20}\}$. Consider the system of equations
x + y + 2z =5
2x + y + 3z = 6
x + 2y + az =b
Let
'A' denotes number of ordered pairs (a,b) so that the system of equations has unique solution.
'B' denotes number of ordered pairs (a,b) so that the system of equations has no solution.
'C' denotes number of ordered pairs (a,b) so that the sytem of equations has infinite solutions.
then which of the following is/are correct?

• A. A = 400
• B. B = 20
• C. C = 21
• D. A+ B = 440

Answer 1

Answer: (B), (D)

The correct options are
B B = 20
D A+ B = 440

$\begin{array}{cc}\Delta =\left|\begin{array}{ccc}1& 1& 2\\ 2& 1& 3\\ 1& 2& a\end{array}\right|=(3-a)& {\Delta }_2=\left|\begin{array}{ccc}1& 5& 2\\ 2& 6& 3\\ 1& b& a\end{array}\right|=-4a+b+3\\ {\Delta }_1=\left|\begin{array}{ccc}5& 1& 2\\ 6& 1& 3\\ b& 2& a\end{array}\right|=-a+b-6& {\Delta }_3=\left|\begin{array}{ccc}1& 1& 5\\ 2& 1& 6\\ 1& 2& b\end{array}\right|=(9-b)\end{array}$
$A:\Delta \ne 0\Rightarrow a\ne 3$,
So, A has $(20\times 21)=420$ ways.
$B:\Delta =0\Rightarrow a=3,{\Delta }_1=b-9,{\Delta }_2=b-9,{\Delta }_3=9-b\ne 0$
So, B has $(1\times 20)=20$ ways.
$C:\Delta =0,{\Delta }_1=0\Rightarrow b=9$
a = 3 b = 9
So, C has only 1 way.