When does a system of 3 equations have infinitely many solutions?

When the determinant of the coefficient matrix is 0 (singular) and the augmented matrix has the same rank as the coefficient matrix.

What is det(A) for this system?

det(A) = 1(2λ−0)−1(λ−0)+1(3−4) = 2λ−λ−1 = λ−6. Wait — expanding: det=1(2λ−0)−1(λ−0)+1(3−4)=2λ−λ−1=λ−6. Setting to 0 gives λ=6.

Setting det(A)=0: λ−6=0, so λ=6.

With λ=6, R3 = R1+R2 gives 2x+3y+6z=16, so μ=16.

How to verify the answer?

With λ=6 and μ=16, the third equation 2x+3y+6z=16 equals (first equation)+(second equation): (x+y+z)+(x+2y+5z)=2x+3y+6z=6+10=16. Consistent — infinitely many solutions confirmed.

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MathMatrices

Q MCQ Matrices & Determinants

If the system of linear equations:

$x + y + z = 6$
$x + 2y + 5z = 10$
$2x + 3y + \lambda z = \mu$

has infinitely many solutions, then $\lambda + \mu$ equals:

A) $12$ B) $16$ C) $22$ D) $28$

✅ Correct Answer

C) 22

✓

Solution

Write the coefficient matrix and set det = 0

$$A=\begin{pmatrix}1&1&1\\1&2&5\\2&3&\lambda\end{pmatrix}$$

For infinitely many solutions, $\det(A)=0$:

$\det(A)=1(2\lambda-15)-1(\lambda-10)+1(3-4)$

$=2\lambda-15-\lambda+10-1$

$=\lambda-6=0$

$\therefore\quad\lambda=6$

Find μ using consistency condition

With $\lambda=6$, the third equation becomes $2x+3y+6z=\mu$.

Check if $R_3 = R_1 + R_2$:

$R_1+R_2$: $(1+1)x+(1+2)y+(1+5)z=6+10$

$\Rightarrow\quad 2x+3y+6z=16$

This matches $R_3$ when $\mu=16$.

Compute λ + μ

$$\lambda+\mu=6+16=\boxed{22}$$

Verify: Solve the consistent system

From $R_1$: $x+y+z=6$ and $R_2$: $x+2y+5z=10$.

$R_2-R_1$: $y+4z=4$. Let $z=t$, then $y=4-4t$, $x=6-y-z=6-(4-4t)-t=2+3t$.

$x=2+3t,\quad y=4-4t,\quad z=t\quad(t\in\mathbb{R})$

Infinitely many solutions ✅

📘 Key Concept

For $AX=B$ to have infinitely many solutions: (1) $\det(A)=0$ — singular matrix, and (2) the augmented matrix $[A|B]$ must have the same rank as $A$ — consistent. Find $\lambda$ from condition 1, then $\mu$ from condition 2 by expressing $R_3$ as a linear combination of $R_1$ and $R_2$.

💡

Important Concepts

Condition for Infinite Solutions $\det(A)=0$ is necessary but not sufficient. We also need the augmented matrix $[A|B]$ to have rank equal to rank of $A$. This ensures the system is consistent.

Row Combination Method Once $\det(A)=0$, check which linear combination of existing rows gives $R_3$. The same combination of the RHS gives $\mu$. Here $R_3=R_1+R_2$ gives $\mu=6+10=16$.

No Solution vs Infinite Solutions If $\det(A)=0$ but the augmented matrix has higher rank, the system has no solution. E.g., if $\mu\neq16$ here, no solution would exist.

Determinant Expansion Expand along Row 1: $1\cdot M_{11}-1\cdot M_{12}+1\cdot M_{13}$ where $M_{ij}$ are $2\times2$ minors. Always expand along the row with most zeros for speed.

FAQs

Why set det(A)=0 for infinite solutions?

⌄

If $\det(A)\neq0$, $A$ is invertible and the system has exactly one unique solution. For infinite or no solutions, the matrix must be singular — hence $\det(A)=0$.

How to expand det(A) correctly here?

⌄

Along Row 1: $1\cdot(2\lambda-15)-1\cdot(\lambda-10)+1\cdot(3-4)=2\lambda-15-\lambda+10-1=\lambda-6$. Setting $\lambda-6=0$ gives $\lambda=6$.

What if μ were 15 instead of 16?

⌄

Then the system would be inconsistent — no solution. The third equation $2x+3y+6z=15$ cannot be satisfied simultaneously with the first two since their sum gives 16.

Can we solve for actual values of x, y, z?

⌄

Not uniquely. The solution is a family: $x=2+3t$, $y=4-4t$, $z=t$ for any $t\in\mathbb{R}$. Infinite solutions form a line in 3D space.

Is this from JEE Main 2026?

⌄

Yes, this question appeared in JEE Main 2026.

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