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3 and 4 .Determinants and Matrices
normal
Number of triplets of $a, b \, \& \,c$ for which the system of equations,$ax - by = 2a - b$ and $(c + 1) x + cy = 10 - a + 3 b$ has infinitely many solutions and $x = 1, y = 3$ is one of the solutions, is :
A
exactly one
B
exactly two
C
exactly three
D
infinitely many
Solution
put $x = 1\, $ and $\, y = 3$ in $1^{st}$ equation ==> $a$ = $- 2b$ and from $2^{nd}$ equation
$c =$ $ \frac{{9\,\, + \,\,5\,b}}{4}$ ;
Now use $\frac{a}{{c\,\, + \,\,1}}$ $=$ $-\frac{b}{c}$ $=$ $\frac{{2\,a\,\, – \,\,b}}{{10\,\, – \,\,a\,\, + \,\,3\,b}}$ ;
from first two $b = 0$ or $c = 1$ ;
if $b = 0 ==> a = 0\, $ and $ \, c = 9/4$ ;
if $c = 1 ; b = – 1 ; a = 2$
Standard 12
Mathematics
