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Matrices and Determinants question

2020 · 4 Sep · Shift 2 · Q24

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Matrices and Determinants
2020 · 4 Sep · Shift 2 · Q24

Matrices and Determinants question

2020 · 4 Sep · Shift 2 · Q24

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1

If the system of equations x+y+z=2 2x+4y–z=6 3x+2y+

\lambda

z=

\mu

has infinitely many solutions, then

A
2 $\lambda$ - $\mu$ = 5
B
$\lambda$ - 2 $\mu$ = -5
C
2 $\lambda$ + $\mu$ = 14
D
$\lambda$ + 2 $\mu$ = 14

View written solutionFree

Correct answer: C

For the system to have infinitely many solutions, the third equation must be a linear combination of the first two, and the coefficient matrix must be singular.

The system is:

\begin{aligned} x+y+z &= 2 \quad ...(1)\\ 2x+4y-z &= 6 \quad ...(2)\\ 3x+2y+\lambda z &= \mu \quad ...(3) \end{aligned}

First, write the coefficient rows:

R_1=(1,1,1), \quad R_2=(2,4,-1), \quad R_3=(3,2,\lambda)

For infinitely many solutions, $R_3$ must be expressible as

R_3=aR_1+bR_2

for some constants $a,b$ .

So,

a(1,1,1)+b(2,4,-1)=(3,2,\lambda)

Equating components:

a+2b=3 \quad ...(i)

a+4b=2 \quad ...(ii)

a-b=\lambda \quad ...(iii)

Solve for $a,b$ using (i) and (ii): Subtract (i) from (ii):

2b=-1 \Rightarrow b=-\frac12

Then from (i):

a+2\left(-\frac12\right)=3

a-1=3 \Rightarrow a=4

Now find $\lambda$ from (iii):

\lambda=a-b=4-\left(-\frac12\right)=\frac92

For consistency with infinitely many solutions, the constant term must also satisfy the same linear combination:

\mu=a\cdot 2+b\cdot 6

\mu=4\cdot 2+\left(-\frac12\right)\cdot 6=8-3=5

Thus,

\lambda=\frac92, \qquad \mu=5

Check the options:

A: $2\lambda-\mu=2\cdot \frac92-5=9-5=4 \ne 5$
B: $\lambda-2\mu=\frac92-10=-\frac{11}{2} \ne -5$
C: $2\lambda+\mu=9+5=14$ ✅
D: $\lambda+2\mu=\frac92+10=\frac{29}{2} \ne 14$

Therefore, the correct option is C.

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