- Home
- Standard 12
- Mathematics
3 and 4 .Determinants and Matrices
hard
If the following system of linear equations
$2 x+y+z=5$
$x-y+z=3$
$x+y+a z=b$
has no solution, then :
A
$\mathrm{a}=-\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}$
B
$a \neq \frac{1}{3}, b=\frac{7}{3}$
C
$\mathrm{a} \neq-\frac{1}{3}, \mathrm{~b}=\frac{7}{3}$
D
$\mathrm{a}=\frac{1}{3}, \mathrm{~b} \neq \frac{7}{3}$
(JEE MAIN-2021)
Solution
Here $\mathrm{D}=\left|\begin{array}{ccc}2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & \mathrm{a}\end{array}\right|$
$=2(-\mathrm{a}-1)-1(\mathrm{a}-1)+1+1$
$=1-3 \mathrm{a}$
$\mathrm{D}_{3}=\left|\begin{array}{ccc}2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & \mathrm{~b}\end{array}\right|$
$=2(-b-3)-1(b-3)+5(1+1)$
$=7-3 b$
for $a=\frac{1}{3}, b \neq \frac{7}{3}$, system has no solutions
Standard 12
Mathematics
