JEE Main 2020 - 7 Jan (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $2 x + 2 a y + a z = 0 2 x + 3 b y + b z = 0 2 x + 4 cy + cz = 0$ , where $a, b, c \in R$ are non-zero and distinct ; has a non-zero solution, then

Q: Wait, why does a non-zero solution force the determinant to be zero?

In a homogeneous system $AX = 0$, if the determinant $|A|$ is non-zero, the matrix $A$ is invertible. Multiplying by $A^{-1}$ gives $X = A^{-1}0 = 0$, which is the trivial solution. To get a non-zero solution, we must force the system to be dependent, which happens only when $|A| = 0$.

Q: Why did we divide by $abc$ at the end? It feels like a random step.

It is a classic technique! We arrived at $2ac = ab + bc$. Dividing by $abc$ is the standard trick to transform a product-based equation into a reciprocal-based one. This reveals the structure $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$, which is the definition of an Arithmetic Progression.

Q: Can I just use Cramer's Rule instead of row operations?

Cramer's Rule is great for finding specific values of $x, y,$ and $z$, but here we are looking for a condition on the variables $a, b,$ and $c$. Using row operations to simplify the determinant is much faster and less prone to algebraic errors than expanding a massive determinant directly.

Select Answer:

Visualized Solution

Homogeneous System Condition

The given system is of the form $A X = 0$ .
This is a homogeneous system of linear equations.
For a non-trivial (non-zero) solution, the determinant of the coefficient matrix must be zero: $∣ A ∣ = 0$ .

Setting up $Δ = 0$

Extract the coefficients of $x, y, z$ to form the determinant.
$222 2 a 3 b 4 c a b c = 0$

Factoring out from $C_{1}$

Notice that all elements in the first column ( $C_{1}$ ) are $2$ .
Factor out $2$ from $C_{1}$ :
$2 111 2 a 3 b 4 c a b c = 0$
Divide by $2$ to simplify.

Row Transformations

To create zeros in $C_{1}$ , apply row operations:
$R_{2} \to R_{2} - R_{1}$
$R_{3} \to R_{3} - R_{1}$
$100 2 a 3 b - 2 a 4 c - 2 a a b - a c - a = 0$

Expanding along $C_{1}$

Expand the determinant along the first column ( $C_{1}$ ):
$1 \cdot [(3 b - 2 a) (c - a) - (4 c - 2 a) (b - a)] = 0$
$(3 b - 2 a) (c - a) - (4 c - 2 a) (b - a) = 0$

Algebraic Expansion

Multiply the brackets carefully:
$(3 b c - 3 ab - 2 a c + 2 a^{2}) - (4 b c - 4 a c - 2 ab + 2 a^{2}) = 0$
Distribute the negative sign:
$3 b c - 3 ab - 2 a c + 2 a^{2} - 4 b c + 4 a c + 2 ab - 2 a^{2} = 0$

Simplifying the Equation

Cancel out $2 a^{2}$ and combine like terms:
$(3 b c - 4 b c) + (- 3 ab + 2 ab) + (- 2 a c + 4 a c) = 0$
$- b c - ab + 2 a c = 0$
Rearrange to get positive terms:
$2 a c = ab + b c$

Dividing by $ab c$

We have the relation: $2 a c = ab + b c$
Divide the entire equation by $ab c$ (since $a, b, c$ are non-zero):
$\frac{2 a c}{ab c} = \frac{ab}{ab c} + \frac{b c}{ab c}$
$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$

Condition for Arithmetic Progression

The equation $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ is the standard condition for an Arithmetic Progression.
Therefore, the terms $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P.
*(Note: This also means $a, b, c$ are in Harmonic Progression, H.P.)*

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

Solution Diagram

Analyzing the Setup

A homogeneous system of linear equations possesses a non-trivial solution if and only if the determinant of its coefficient matrix is equal to zero. Given the system:

222 2 a 3 b 4 c a b c = 0

This condition is the geometric key to unlocking the relationship between the variables

a

b

, and

c

The Art of Simplification

First, we factor out the constant

2

from the first column to simplify the determinant:

2 \cdot 111 2 a 3 b 4 c a b c = 0

To further simplify, we apply row operations

R_{2} \to R_{2} - R_{1}

and

R_{3} \to R_{3} - R_{1}

100 2 a 3 b - 2 a 4 c - 2 a a b - a c - a = 0

Expanding along the first column, we obtain the following equation:

(3 b - 2 a) (c - a) - (4 c - 2 a) (b - a) = 0

The Algebraic Dance

Expanding the terms within the brackets, we get:

(3 b c - 3 ab - 2 a c + 2 a^{2}) - (4 b c - 4 a c - 2 ab + 2 a^{2}) = 0

Notice that the

2 a^{2}

terms cancel out perfectly. Combining the remaining terms yields:

(3 b c - 4 b c) + (- 3 ab + 2 ab) + (- 2 a c + 4 a c) = 0

This simplifies elegantly to:

- b c - ab + 2 a c = 0 \Rightarrow 2 a c = ab + b c

The Final Revelation

To reveal the underlying structure, we divide the entire equation by

ab c

(given $a, b, c eq 0$):

\frac{2 a c}{ab c} = \frac{ab}{ab c} + \frac{b c}{ab c}

This results in the final relationship:

\frac{2}{b} = \frac{1}{c} + \frac{1}{a}

This equation confirms that the reciprocals of $a, b,$ and $c$ are in Arithmetic Progression (A.P.).

Common Traps & FAQs

Wait, why does a non-zero solution force the determinant to be zero?

In a homogeneous system

A X = 0

, if the determinant

∣ A ∣

is non-zero, the matrix

A

is invertible. Multiplying by

A^{- 1}

gives

X = A^{- 1} 0 = 0

, which is the trivial solution. To get a non-zero solution, we must force the system to be dependent, which happens only when

∣ A ∣ = 0

Why did we divide by

ab c

at the end? It feels like a random step.

It is a classic technique! We arrived at

2 a c = ab + b c

. Dividing by

ab c

is the standard trick to transform a product-based equation into a reciprocal-based one. This reveals the structure

\frac{2}{b} = \frac{1}{c} + \frac{1}{a}

, which is the definition of an Arithmetic Progression.

Can I just use Cramer's Rule instead of row operations?

Cramer's Rule is great for finding specific values of

x, y,

and

z

, but here we are looking for a condition on the variables

a, b,

and

c

. Using row operations to simplify the determinant is much faster and less prone to algebraic errors than expanding a massive determinant directly.

Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $2 x + 2 a y + a z = 0 2 x + 3 b y + b z = 0 2 x + 4 cy + cz = 0$ , where $a, b, c \in R$ are non-zero and distinct ; has a non-zero solution, then

Select Answer:

Visualized Solution

Homogeneous System Condition

Setting up $Δ = 0$

Factoring out from $C_{1}$

Row Transformations

Expanding along $C_{1}$

Algebraic Expansion

Simplifying the Equation

Dividing by $ab c$

Condition for Arithmetic Progression

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

Analyzing the Setup

The Art of Simplification

The Algebraic Dance

The Final Revelation

Similar Questions

If the system of linear equations $2 x + 2 a y + a z = 0, 2 x + 3 b y + b z = 0, 2 x + 4 cy + cz = 0$ , where $a, b, c \in R$ are non-zero and distinct; has non-zero solution, then

If the system of linear equations $x + 2 a y + a z = 0$ ; $x + 3 b y + b z = 0$ ; $x + 4 cy + cz = 0$ ; has a non-zero solution, then $a, b, c$

If the system of linear equations $2 x + 2 y + 3 z = a$ , $3 x - y + 5 z = b$ , $x - 3 y + 2 z = c$ where $a, b, c$ are non-zero real numbers, has more than one solution, then :

If the system of equations $2 x + 3 y - z = 0, x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non-trivial solution $(x, y, z)$ , then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to :-

The following system of linear equations $7 x + 6 y - 2 z = 0$ , $3 x + 4 y + 2 z = 0$ , $x - 2 y - 6 z = 0$ , has

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The system of equations $λ x + y + z = 0$ , $- x + λ y + z = 0$ , $- x - y + λ z = 0$ will have a non-zero solution if real values of $λ$ are given by .........

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Let the system of linear equations $4 x + λ y + 2 z = 0, 2 x - y + z = 0, μx + 2 y + 3 z = 0, λ, μ \in R$ has a non-trivial solution. Then which of the following is true?

For a real number $a$ , if the system $1 a a^{2} a 1 a a^{2} a 1 x y z = 1 - 1 1$ of linear equations, has infinitely many solutions, then $1 + a + a^{2} =$

Select Answer:

Visualized Solution

Homogeneous System Condition

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ=0

Factoring out from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Row Transformations

Expanding along .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Algebraic Expansion

Simplifying the Equation

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

Condition for Arithmetic Progression

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

Analyzing the Setup

The Art of Simplification

The Algebraic Dance

The Final Revelation

Similar Questions

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Expanding along $C_{1}$

Dividing by $ab c$

If the system of linear equations $2 x + 2 a y + a z = 0, 2 x + 3 b y + b z = 0, 2 x + 4 cy + cz = 0$ , where $a, b, c \in R$ are non-zero and distinct; has non-zero solution, then

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If the system of linear equations $2 x + 2 y + 3 z = a$ , $3 x - y + 5 z = b$ , $x - 3 y + 2 z = c$ where $a, b, c$ are non-zero real numbers, has more than one solution, then :

If the system of equations $2 x + 3 y - z = 0, x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non-trivial solution $(x, y, z)$ , then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to :-

The following system of linear equations $7 x + 6 y - 2 z = 0$ , $3 x + 4 y + 2 z = 0$ , $x - 2 y - 6 z = 0$ , has

The system of linear equations $x + λ y - z = 0$ , $λ x - y - z = 0$ , $x + y - λ z = 0$ has a non-trivial solution for:

The system of equations $λ x + y + z = 0$ , $- x + λ y + z = 0$ , $- x - y + λ z = 0$ will have a non-zero solution if real values of $λ$ are given by .........

Consider the following system of equations: $x + 2 y - 3 z = a, 2 x + 6 y - 11 z = b, x - 2 y + 7 z = c$ where $a, b$ and $c$ are real constants. Then the system of equations :

Let the system of linear equations $4 x + λ y + 2 z = 0, 2 x - y + z = 0, μx + 2 y + 3 z = 0, λ, μ \in R$ has a non-trivial solution. Then which of the following is true?

For a real number $a$ , if the system $1 a a^{2} a 1 a a^{2} a 1 x y z = 1 - 1 1$ of linear equations, has infinitely many solutions, then $1 + a + a^{2} =$