If the system of equation 2x + y + 3z = 5 3x + 2y - z = 7 4x + 5y + z = 9 has infinitely many solutions, then ^2 + ^2 is equal to:

Given: aligned 2x + y + 3z &= 5 3x + 2y - z &= 7 4x + 5y + z &= 9 aligned Find: ^2 + ^2 when the system has infinitely many solutions. For infinitely many solutions, the determinants must satisfy = 0 and z = 0. The coefficient determinant is = vmatrix 2 & & 3 3 & 2 & -1 4 & 5 & vmatrix = 0 Expanding along the first row, 2(2 - (-5)) - (3 - (-4)) + 3(15 - 8) = 0 2(2 + 5) - (3 + 4) + 21 = 0 4 + 10 - 3 - 4 + 21 = 0 4 - 4 - 3 + 31 = 0 Now compute z = vmatrix 2 & & 5 3 & 2 & 7 4 & 5 & 9 vmatrix = 0 Expanding along the first row, 2(2 9 - 7 5) - (3 9 - 7 4) + 5(3 5 - 2 4) = 0 2(18 - 35) - (27 - 28) + 5(15 - 8) = 0 2(-17) - (-1) + 5(7) = 0 -34 + + 35 = 0 = -1 Substitute = -1 into 4 - 4 - 3 + 31 = 0 Then 4 - 4(-1) - 3(-1) + 31 = 0 4 + 4 + 3 + 31 = 0 7 + 35 = 0 = -5 Therefore, ^2 + ^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26 The correct option is C.

If the system of equation 2x + y + 3z = 5 3x + 2y - z = 7 4x + 5y… — Solution | JEE Mathematics

MCQMediumJEE 2025Solving Linear Equations (Matrix Method)

If the system of equation $2x + \lambda y + 3z = 5$ $3x + 2y - z = 7$ $4x + 5y + \mu z = 9$ has infinitely many solutions, then $\lambda^2 + \mu^2$ is equal to:

A
$22$
B
$18$
C
$26$
D
$30$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

\begin{aligned} 2x + \lambda y + 3z &= 5 \\ 3x + 2y - z &= 7 \\ 4x + 5y + \mu z &= 9 \end{aligned}

Find: $\lambda^2 + \mu^2$ when the system has infinitely many solutions.

For infinitely many solutions, the determinants must satisfy $\Delta = 0$ and $\Delta_z = 0$ .

The coefficient determinant is

\Delta = \begin{vmatrix} 2 & \lambda & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \mu \end{vmatrix} = 0

Expanding along the first row,

2(2\mu - (-5)) - \lambda(3\mu - (-4)) + 3(15 - 8) = 0

2(2\mu + 5) - \lambda(3\mu + 4) + 21 = 0

4\mu + 10 - 3\lambda\mu - 4\lambda + 21 = 0

4\mu - 4\lambda - 3\lambda\mu + 31 = 0

Using $$\Delta_z = 0$$ to find the parameters

Now compute

\Delta_z = \begin{vmatrix} 2 & \lambda & 5 \\ 3 & 2 & 7 \\ 4 & 5 & 9 \end{vmatrix} = 0

Expanding along the first row,

2(2 \cdot 9 - 7 \cdot 5) - \lambda(3 \cdot 9 - 7 \cdot 4) + 5(3 \cdot 5 - 2 \cdot 4) = 0

2(18 - 35) - \lambda(27 - 28) + 5(15 - 8) = 0

2(-17) - \lambda(-1) + 5(7) = 0

-34 + \lambda + 35 = 0

\lambda = -1

Substitute and conclude

Substitute $\lambda = -1$ into

4\mu - 4\lambda - 3\lambda\mu + 31 = 0

Then

4\mu - 4(-1) - 3(-1)\mu + 31 = 0

4\mu + 4 + 3\mu + 31 = 0

7\mu + 35 = 0

\mu = -5

Therefore,

\lambda^2 + \mu^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26

The correct option is C.

Common mistakes

Setting only $\Delta = 0$ is not sufficient for infinitely many solutions. That condition gives a singular system, but consistency must also be checked using determinants like $\Delta_z$ . Use both conditions, not only the coefficient determinant.
Making an error while expanding the determinant is common, especially with the minus sign in the middle term. In the cofactor expansion, the term involving $\lambda$ carries a negative sign. Keep the cofactor signs as $+,-,+$ .
After finding $\lambda = -1$ , students may substitute it incorrectly into $4\mu - 4\lambda - 3\lambda\mu + 31 = 0$ . Carefully replace both occurrences of $\lambda$ by $-1$ before simplifying.

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