Why can't we just use the determinant method ($\Delta = 0$)?

You absolutely can! The determinant method is a powerful tool to check for consistency. However, the determinant method only tells you *if* the system is consistent or inconsistent. It doesn't always easily give you the specific values of $\lambda$ and $\mu$ without extra work. The linear combination method is often faster and gives you deeper insight into the geometric structure of the equations, which is exactly what JEE examiners love to test.

JEE Main 2019 (10 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $x + y + z = 5$ , $x + 2 y + 2 z = 6$ , $x + 3 y + z = μ$ , $(λ, μ \in R)$ , has infinitely many solutions, then the value of $λ + μ$ is :

Q: Is it possible for $p$ or $q$ to be zero?

Yes, absolutely! If $p$ or $q$ were zero, it would mean the third plane is parallel to or identical to one of the other planes. The method of linear combination is robust enough to handle those cases naturally. If the math leads you to a zero, trust the algebra—it is telling you something specific about the orientation of the planes.

Select Answer:

Progressive Hints

Hint 1

Visualize the three equations not as abstract lines, but as three distinct planes in 3D space. When a system has infinitely many solutions, these planes do not just meet at a single point; they share a common line of intersection, much like the pages of an open book meeting at the spine. Your first task is to accept that these planes are not independent.

Hint 2

To force these planes to share a common line, the third equation must be 'redundant'—it must be a linear combination of the first two. Think of it as saying the third plane is just a specific blend of the first two planes. If you can express the third equation as a weighted sum of the first two, you have unlocked the secret to the system's dependency.

Hint 3

Set up the identity:

(x + 3 y + λ z - μ) = p (x + y + z - 5) + q (x + 2 y + 2 z - 6)

. Now, equate the coefficients of

x

y

z

, and the constant terms on both sides to solve for the constants

p

and

q

. Once you have

p

and

q

, finding

λ

and

μ

becomes a simple matter of arithmetic.

Visualized Solution

$System of Equations$

Given System:
$L_{1} : x + y + z = 5$
$L_{2} : x + 2 y + 2 z = 6$
$L_{3} : x + 3 y + λ z = μ$
Condition: Infinitely many solutions.

$Condition for Infinite Solutions$

For infinite solutions, the planes intersect at a common line.
The third equation must be a linear combination of the first two.
$L_{3} = p L_{1} + q L_{2}$
where $p$ and $q$ are real constants.

$Linear Combination Setup$

Substitute the equations into the condition:
$x + 3 y + λ z - μ = p (x + y + z - 5) + q (x + 2 y + 2 z - 6)$
This identity holds true for all $x, y, z$ .

$Comparing Coefficients of x$

Compare the coefficient of $x$ on both sides:
LHS coefficient = $1$
RHS coefficient = $p (1) + q (1) = p + q$
$\Rightarrow p + q = 1 \dots (1)$

$Comparing Coefficients of y$

Compare the coefficient of $y$ on both sides:
LHS coefficient = $3$
RHS coefficient = $p (1) + q (2) = p + 2 q$
$\Rightarrow p + 2 q = 3 \dots (2)$

$Solving for p and q$

Solve the system of two equations:
$(p + 2 q) - (p + q) = 3 - 1$
$\Rightarrow q = 2$
Substitute $q = 2$ in $(1)$ :
$p + 2 = 1 \Rightarrow p = - 1$

$Finding λ$

Compare the coefficient of $z$ on both sides:
$λ = p (1) + q (2)$
Substitute $p = - 1$ and $q = 2$ :
$λ = - 1 + 2 (2) = 3$

$Finding μ$

Compare the constant terms on both sides:
$- μ = p (- 5) + q (- 6)$
$\Rightarrow μ = 5 p + 6 q$
Substitute $p = - 1$ and $q = 2$ :
$μ = 5 (- 1) + 6 (2) = 7$

$Final Calculation: λ + μ$

Calculate the required sum:
$λ + μ = 3 + 7$
$\Rightarrow λ + μ = 10$
Final Answer: $10$

00:00 / 00:00

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Solution Diagram

Analyzing the Setup

We are given a system of three linear equations representing planes in 3D space: 1.

x + y + z = 5

x + 2 y + 2 z = 6

x + 3 y + λ z = μ

For the system to have infinitely many solutions, the three planes must intersect along a common line. This implies that the third plane is linearly dependent on the first two.

The Master Equation

We can express the third plane,

L_{3}

, as a linear combination of the first two planes,

L_{1}

and

L_{2}

. We define this relationship as:

L_{3} = p L_{1} + q L_{2}

Substituting the given equations, we form the following identity:

(x + 3 y + λ z - μ) = p (x + y + z - 5) + q (x + 2 y + 2 z - 6)

This identity must hold true for all values of

x

y

, and

z

on the line of intersection.

Equating Coefficients

By comparing the coefficients of

x

y

, and

z

, and the constant terms on both sides, we generate a system of linear equations:

For

x

p + q = 1

For

y

p + 2 q = 3

Subtracting the first equation from the second yields:

(p + 2 q) - (p + q) = 3 - 1 \Rightarrow q = 2

Substituting

q = 2

into

p + q = 1

, we find:

p + 2 = 1 \Rightarrow p = - 1

Final Calculation

Now, we determine the values of

λ

and

μ

using the constants

p = - 1

and

q = 2

For

z

λ = p (1) + q (2)

λ = - 1 (1) + 2 (2) = - 1 + 4 = 3

For the constant term:

- μ = p (- 5) + q (- 6)

- μ = - 1 (- 5) + 2 (- 6) = 5 - 12 = - 7

μ = 7

The problem asks for the sum

λ + μ

. Substituting our results:

λ + μ = 3 + 7 = 10

The final answer is 10.

Common Traps & FAQs

Why can't we just use the determinant method (

Δ = 0

You absolutely can! The determinant method is a powerful tool to check for consistency. However, the determinant method only tells you *if* the system is consistent or inconsistent. It doesn't always easily give you the specific values of

λ

and

μ

without extra work. The linear combination method is often faster and gives you deeper insight into the geometric structure of the equations, which is exactly what JEE examiners love to test.

Wait, why do we assume the third equation is a linear combination of the first two?

That is the definition of linear dependence in a system of equations. If the system has infinitely many solutions, the rank of the augmented matrix must be less than the number of variables. This implies one equation is essentially 'built' from the others. By writing

L_{3} = p L_{1} + q L_{2}

, we are mathematically asserting that the third plane is not adding any new information to the system, which is the only way to have a line of intersection rather than a single point.

Is it possible for

p

q

to be zero?

Yes, absolutely! If

p

q

were zero, it would mean the third plane is parallel to or identical to one of the other planes. The method of linear combination is robust enough to handle those cases naturally. If the math leads you to a zero, trust the algebra—it is telling you something specific about the orientation of the planes.

Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $x + y + z = 5$ , $x + 2 y + 2 z = 6$ , $x + 3 y + z = μ$ , $(λ, μ \in R)$ , has infinitely many solutions, then the value of $λ + μ$ is :

Select Answer:

Visualized Solution

$System of Equations$

$Condition for Infinite Solutions$

$Linear Combination Setup$

$Comparing Coefficients of x$

$Comparing Coefficients of y$

$Solving for p and q$

$Finding λ$

$Finding μ$

$Final Calculation: λ + μ$

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Master Equation

Equating Coefficients

Final Calculation

Similar Questions

If the system of equations $x + 2 y - 3 z = 2$ $2 x + λ y + 5 z = 5$ $14 x + 3 y + μ z = 33$ has infinitely many solutions, then $λ + μ$ is equal to:

If the system of equations $(λ - 1) x + (λ - 4) y + λ z = 5$ $λ x + (λ - 1) y + (λ - 4) z = 7$ $(λ + 1) x + (λ + 2) y - (λ + 2) z = 9$ has infinitely many solutions, then $λ^{2} + λ$ is equal to

Let the system of equations $x + 2 y + 3 z = 5, 2 x + 3 y + z = 9, 4 x + 3 y + λ z = μ$ have infinite number of solutions. Then $λ + 2 μ$ is equal to :

If the system of equations $x + y + z = 6$ , $2 x + 5 y + α z = β$ , $x + 2 y + 3 z = 14$ has infinitely many solutions, then $α + β$ is equal to :

Let $λ, μ \in R$ . If the system of equations $3 x + 5 y + λ z = 3$ , $7 x + 11 y - 9 z = 2$ , $97 x + 155 y - 189 z = μ$ has infinitely many solutions, then $μ + 2 λ$ is equal to :

If the system of equations $x + y + z = 2 2 x + 4 y - z = 6 3 x + 2 y + λ z = μ$ has infinitely many solutions, then:

If the system of equations $x + 4 y - z = λ, 7 x + 9 y + μ z = - 3, 5 x + y + 2 z = - 1$ has infinitely many solutions, then $(2 μ + 3 λ)$ is equal to :

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

If the system of linear equations $7 x + 11 y + a z = 13$ , $5 x + 4 y + 7 z = β$ , $175 x + 194 y + 57 z = 361$ has infinitely many solutions, then $a + β + 2$ is equal to

If the system of linear equations $2 x + y - z = 3$ , $x - y - z = α$ , $3 x + 3 y + β z = 3$ has infinitely many solution, then $α + β - α β$ is equal to .

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System of Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Infinite Solutions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Linear Combination Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing Coefficients of x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing Coefficients of y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for p and q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding λ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding μ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation: λ+μ

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Master Equation

Equating Coefficients

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations x+2y−3z=2 2x+λy+5z=5 14x+3y+μz=33 has infinitely many solutions, then λ+μ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations (λ−1)x+(λ−4)y+λz=5 λx+(λ−1)y+(λ−4)z=7 (λ+1)x+(λ+2)y−(λ+2)z=9 has infinitely many solutions, then λ2+λ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the system of equations x+2y+3z=5,2x+3y+z=9,4x+3y+λz=μ have infinite number of solutions. Then λ+2μ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations x+y+z=6, 2x+5y+αz=β, x+2y+3z=14 has infinitely many solutions, then α+β is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let λ,μ∈R. If the system of equations 3x+5y+λz=3, 7x+11y−9z=2, 97x+155y−189z=μ has infinitely many solutions, then μ+2λ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations x+y+z=22x+4y−z=63x+2y+λz=μ has infinitely many solutions, then:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations x+4y−z=λ,7x+9y+μz=−3,5x+y+2z=−1 has infinitely many solutions, then (2μ+3λ) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equation 2x+λy+3z=5, 3x+2y−z=7, 4x+5y+μz=9 has infinitely many solutions, then (λ2+μ2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of linear equations 7x+11y+az=13, 5x+4y+7z=β, 175x+194y+57z=361 has infinitely many solutions, then a+β+2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of linear equations 2x+y−z=3, x−y−z=α, 3x+3y+βz=3 has infinitely many solution, then α+β−αβ is equal to .

$System of Equations$

$Condition for Infinite Solutions$

$Linear Combination Setup$

$Comparing Coefficients of x$

$Comparing Coefficients of y$

$Solving for p and q$

$Finding λ$

$Finding μ$

$Final Calculation: λ + μ$

If the system of equations $x + 2 y - 3 z = 2$ $2 x + λ y + 5 z = 5$ $14 x + 3 y + μ z = 33$ has infinitely many solutions, then $λ + μ$ is equal to:

If the system of equations $(λ - 1) x + (λ - 4) y + λ z = 5$ $λ x + (λ - 1) y + (λ - 4) z = 7$ $(λ + 1) x + (λ + 2) y - (λ + 2) z = 9$ has infinitely many solutions, then $λ^{2} + λ$ is equal to

Let the system of equations $x + 2 y + 3 z = 5, 2 x + 3 y + z = 9, 4 x + 3 y + λ z = μ$ have infinite number of solutions. Then $λ + 2 μ$ is equal to :

If the system of equations $x + y + z = 6$ , $2 x + 5 y + α z = β$ , $x + 2 y + 3 z = 14$ has infinitely many solutions, then $α + β$ is equal to :

Let $λ, μ \in R$ . If the system of equations $3 x + 5 y + λ z = 3$ , $7 x + 11 y - 9 z = 2$ , $97 x + 155 y - 189 z = μ$ has infinitely many solutions, then $μ + 2 λ$ is equal to :

If the system of equations $x + y + z = 2 2 x + 4 y - z = 6 3 x + 2 y + λ z = μ$ has infinitely many solutions, then:

If the system of equations $x + 4 y - z = λ, 7 x + 9 y + μ z = - 3, 5 x + y + 2 z = - 1$ has infinitely many solutions, then $(2 μ + 3 λ)$ is equal to :

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

If the system of linear equations $7 x + 11 y + a z = 13$ , $5 x + 4 y + 7 z = β$ , $175 x + 194 y + 57 z = 361$ has infinitely many solutions, then $a + β + 2$ is equal to

If the system of linear equations $2 x + y - z = 3$ , $x - y - z = α$ , $3 x + 3 y + β z = 3$ has infinitely many solution, then $α + β - α β$ is equal to .