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Matrices and Determinants question

2024 · 1 Feb · Shift 2 · Q35

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2024 · 1 Feb · Shift 2 · Q35

Matrices and Determinants question

2024 · 1 Feb · Shift 2 · Q35

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1

Let the system of equations

x+2 y+3 z=5,2 x+3 y+z=9,4 x+3 y+\lambda z=\mu

have infinite number of solutions. Then

\lambda+2 \mu

is equal to :

A
22
B
17
C
15
D
28

View written solutionFree

Correct answer: B

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 equations are:

\begin{aligned} x+2y+3z &= 5 \ 2x+3y+z &= 9 \ 4x+3y+\lambda z &= \mu \end{aligned}

Let the third equation be obtained as:

a(x+2y+3z) + b(2x+3y+z) = 4x+3y+\lambda z

Comparing coefficients:

a+2b = 4 \quad \text{(coefficient of }x)

2a+3b = 3 \quad \text{(coefficient of }y)

Solve these two equations: From

a+2b=4

we get

a=4-2b

Substitute into

2a+3b=3:

2(4-2b)+3b=3

8-4b+3b=3

8-b=3

b=5

Then

a=4-2(5)=-6

Now compare the coefficient of $z$ :

\lambda = 3a+b = 3(-6)+5=-18+5=-13

Compare the constants on the right-hand side:

\mu = 5a+9b = 5(-6)+9(5)=-30+45=15

Therefore,

\lambda + 2\mu = -13 + 2(15) = -13+30 = 17

Hence the correct option is:

\boxed{17}

which is Option B.

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