What if $\Delta = 0$ and $\Delta_x = 0$? Does that mean infinite solutions?

Not necessarily! If $\Delta = 0$ and $\Delta_x = 0$, it is a strong indicator of infinite solutions, but you must also check $\Delta_y$ and $\Delta_z$. If all of them are zero, the system is likely dependent. If even one of them is non-zero, the system remains inconsistent. Always check all determinants before concluding.

JEE Main 2021 (24 February Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: For the system of linear equations: $x - 2 y = 1, x - y + k z = - 2, k y + 4 z = 6, k \in R$ consider the following statements: (A) The system has unique solution if $k \neq = 2, k \neq = - 2$ . (B) The system has unique solution if $k = - 2$ . (C) The system has unique solution if $k = 2$ . (D) The system has no-solution if $k = 2$ . (E) The system has infinite number of solutions if $k \neq = - 2$ . Which of the following statements are correct?

Q: Why do we check $\Delta_x$ when $\Delta = 0$? Can't we just say it has no solution?

That is a fantastic question! When $\Delta = 0$, the system is 'singular'. It could be inconsistent (meaning the planes do not intersect at a common point) or dependent (meaning the planes intersect along a line or are coincident, leading to infinite solutions). We check $\Delta_x$ (or $\Delta_y$ or $\Delta_z$) to distinguish between these two. If any of these are non-zero, the system is guaranteed to be inconsistent.

Select Answer:

Progressive Hints

Hint 1

The determinant of the coefficient matrix, denoted as

Δ

, is the gatekeeper of your system. If

Δ \neq = 0

, the system is well-behaved and has a unique solution. If

Δ = 0

, the system is either inconsistent (no solution) or dependent (infinite solutions). Start by finding

Δ

in terms of

k

Hint 2

Once you have

Δ = 4 - k^{2}

, you know the system is unique when

k \neq = \pm 2

. But what happens at the boundaries? You must use Cramer's Rule to check the consistency of the system when

Δ = 0

. Specifically, calculate

Δ_{x}

by replacing the first column of the coefficient matrix with the constants

[1, - 2, 6]^{T}

Hint 3

For

k = 2

, substitute the value into your matrix and calculate

Δ_{x}

. If you find that

Δ_{x} \neq = 0

while

Δ = 0

, the system is inconsistent. The starting equation for your determinant calculation is

Δ_{x} = 1 - 2 6 - 2 - 1 2 024

Visualized Solution

System of Linear Equations

Given system of equations:
$x - 2 y + 0 z = 1$
$x - y + k z = - 2$
$0 x + k y + 4 z = 6$

Cramer's Rule and $Δ$

To determine the nature of solutions, we use Cramer's Rule.
We first calculate the determinant of the coefficient matrix, denoted as $Δ$ .
If $Δ \neq = 0$ , the system has a unique solution.

Constructing $Δ$

Extract coefficients of $x, y, z$ to form $Δ$ .
$Δ = 110 - 2 - 1 k 0 k 4$

Expanding $Δ$

Expand along the first row ( $R_{1}$ ):
$Δ = 1 (- 4 - k^{2}) - (- 2) (4 - 0) + 0$

Simplifying $Δ$

$Δ = - 4 - k^{2} + 8$
$Δ = 4 - k^{2}$

Condition for Unique Solution

For a unique solution: $Δ \neq = 0$
$4 - k^{2} \neq = 0 ⟹ k^{2} \neq = 4$
$k \neq = 2$ and $k \neq = - 2$
Statement (A) is correct.

Investigating $k = 2$

What if $k = 2$ ?
Then $Δ = 4 - (2)^{2} = 0$ .
If $Δ = 0$ , we must check $Δ_{x}, Δ_{y}, Δ_{z}$ .
If any of them is non-zero, the system has no solution.

Constructing $Δ_{x}$ for $k = 2$

Replace the first column of $Δ$ with the constant terms $[1, - 2, 6]^{T}$ .
Substitute $k = 2$ in the remaining columns.
$Δ_{x} = 1 - 2 6 - 2 - 1 2 024$

Calculating $Δ_{x}$

Expand along the first row:
$Δ_{x} = 1 (- 4 - 4) - (- 2) (- 8 - 12) + 0$
$Δ_{x} = 1 (- 8) + 2 (- 20)$
$Δ_{x} = - 8 - 40 = - 48$

Conclusion for $k = 2$

For $k = 2$ , we found $Δ = 0$ and $Δ_{x} = - 48 \neq = 0$ .
Therefore, the system has no solution.
Statement (D) is correct.
Final Answer: Statements (A) and (D) are correct.

00:00 / 00:00

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

Solution Diagram

Analyzing the System Setup

We are examining a system of three linear equations in three variables:

x - 2 y + 0 z = 1

x - y + k z = - 2

0 x + k y + 4 z = 6

To understand the behavior of these planes, we first construct the coefficient matrix and calculate its determinant, denoted as

Δ

. This determinant acts as the "gatekeeper" for the existence of a unique solution.

Δ = 110 - 2 - 1 k 0 k 4

The Master Equation

Expanding the determinant along the first row, we perform the following calculation:

Δ = 1 ((- 1) (4) - (k) (k)) - (- 2) ((1) (4) - (0) (k)) + 0

Δ = 1 (- 4 - k^{2}) + 2 (4)

Δ = - 4 - k^{2} + 8 = 4 - k^{2}

If $\Delta eq 0$, the system is guaranteed to have a unique solution. This condition holds whenever $k^2 eq 4$, which implies $k eq \pm 2$. This confirms that the system possesses a unique solution for all values of

k

except

2

and

- 2

Investigating Singular Cases

When

k = 2

, the determinant

Δ

becomes zero, rendering the system singular. To determine if the system is inconsistent or has infinitely many solutions, we apply Cramer's Rule by calculating

Δ_{x}

We replace the first column of the coefficient matrix with the constants from the right-hand side of the equations:

Δ_{x} = 1 - 2 6 - 2 - 1 2 024

Expanding this determinant:

Δ_{x} = 1 ((- 1) (4) - (2) (2)) - (- 2) ((- 2) (4) - (6) (2)) + 0

Δ_{x} = 1 (- 4 - 4) + 2 (- 8 - 12)

Δ_{x} = - 8 + 2 (- 20) = - 8 - 40 = - 48

Final Conclusion

Since

Δ = 0

and $\Delta_x eq 0$ when

k = 2

, the system is inconsistent.

This means there is no point in space where these three planes intersect when

k = 2

. We have successfully determined that the system has a unique solution when $k eq \pm 2$ and no solution when $k = 2$ .

Common Traps & FAQs

Why do we check

Δ_{x}

when

Δ = 0

? Can't we just say it has no solution?

That is a fantastic question! When

Δ = 0

, the system is 'singular'. It could be inconsistent (meaning the planes do not intersect at a common point) or dependent (meaning the planes intersect along a line or are coincident, leading to infinite solutions). We check

Δ_{x}

(or

Δ_{y}

Δ_{z}

) to distinguish between these two. If any of these are non-zero, the system is guaranteed to be inconsistent.

Is Cramer's Rule the only way to solve this?

Absolutely not! You could use Gaussian elimination or row reduction to reach the same conclusion. However, for a

3 \times 3

system with a parameter, Cramer's Rule is often the most direct and elegant path to finding the conditions for unique vs. non-unique solutions. It turns the problem into a simple determinant calculation.

What if

Δ = 0

and

Δ_{x} = 0

? Does that mean infinite solutions?

Not necessarily! If

Δ = 0

and

Δ_{x} = 0

, it is a strong indicator of infinite solutions, but you must also check

Δ_{y}

and

Δ_{z}

. If all of them are zero, the system is likely dependent. If even one of them is non-zero, the system remains inconsistent. Always check all determinants before concluding.

Select Answer:

Visualized Solution

System of Linear Equations

Cramer's Rule and $Δ$

Constructing $Δ$

Expanding $Δ$

Simplifying $Δ$

Condition for Unique Solution

Investigating $k = 2$

Constructing $Δ_{x}$ for $k = 2$

Calculating $Δ_{x}$

Conclusion for $k = 2$

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

Analyzing the System Setup

The Master Equation

Investigating Singular Cases

Final Conclusion

Similar Questions

Let the system of linear equations $x + y + k z = 2$ , $2 x + 3 y - z = 1$ , $3 x + 4 y + 2 z = k$ have infinitely many solutions. Then the system $(k + 1) x + (2 k - 1) y = 7$ , $(2 k + 1) x + (k + 5) y = 10$ has :

Consider the system of equations $x - 2 y + 3 z = - 1$ , $- x + y - 2 z = k$ , $x - 3 y + 4 z = 1$ . STATEMENT - 1 : The system of equations has no solution for $k \neq = 3$ and STATEMENT - 2 : The determinant $1 - 1 1 3 - 2 4 - 1 k 1 \neq = 0$ , for $k \neq = 3$ .

Let $a, λ, μ \in R$ . Consider the system of linear equations $a x + 2 y = λ, 3 x - 2 y = μ$ . Which of the following statement(s) is (are) correct?

For the system of linear equations $2 x + 4 y + 2 a z = b$ , $x + 2 y + 3 z = 4$ , $2 x + 5 y + 2 z = 8$ which of the following is NOT correct?

Consider the following system of equations $α x + 2 y + z = 1$ , $2 α x + 3 y + z = 1$ , $3 x + α y + 2 z = β$ For some $α, β \in R$ . Then which of the following is NOT correct.

If the system of equations $x - k y - z = 0$ , $k x - y - z = 0$ , $x + y - z = 0$ has a non-zero solution, then the possible values of $k$ are

The system of linear equations $λ x + 2 y + 2 z = 5$ , $2 λ x + 3 y + 5 z = 8$ , $4 x + λ y + 6 z = 10$ has:

For the system of linear equations $a x + y + z = 1$ , $x + a y + z = 1$ , $x + y + a z = β$ , which one of the following statements is NOT correct ?

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

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 :

Select Answer:

Visualized Solution

System of Linear Equations

Cramer's Rule and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

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

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

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

Condition for Unique Solution

Investigating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=2

Constructing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δx​ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=2

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

Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=2

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

Analyzing the System Setup

The Master Equation

Investigating Singular Cases

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the system of linear equations x+y+kz=2, 2x+3y−z=1, 3x+4y+2z=k have infinitely many solutions. Then the system (k+1)x+(2k−1)y=7, (2k+1)x+(k+5)y=10 has :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,λ,μ∈R. Consider the system of linear equations ax+2y=λ,3x−2y=μ. Which of the following statement(s) is (are) correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the system of linear equations 2x+4y+2az=b, x+2y+3z=4, 2x+5y+2z=8 which of the following is NOT correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the following system of equations αx+2y+z=1, 2αx+3y+z=1, 3x+αy+2z=β For some α,β∈R. Then which of the following is NOT correct.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the system of equations x−ky−z=0, kx−y−z=0, x+y−z=0 has a non-zero solution, then the possible values of k are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The system of linear equations λx+2y+2z=5, 2λx+3y+5z=8, 4x+λy+6z=10 has:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the system of linear equations ax+y+z=1, x+ay+z=1, x+y+az=β, which one of the following statements is NOT correct ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The following system of linear equations 7x+6y−2z=03x+4y+2z=0x−2y−6z=0, has

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider the following system of equations: x+2y−3z=a,2x+6y−11z=b,x−2y+7z=c where a,b and c are real constants. Then the system of equations :

Cramer's Rule and $Δ$

Constructing $Δ$

Expanding $Δ$

Simplifying $Δ$

Investigating $k = 2$

Constructing $Δ_{x}$ for $k = 2$

Calculating $Δ_{x}$

Conclusion for $k = 2$

Let the system of linear equations $x + y + k z = 2$ , $2 x + 3 y - z = 1$ , $3 x + 4 y + 2 z = k$ have infinitely many solutions. Then the system $(k + 1) x + (2 k - 1) y = 7$ , $(2 k + 1) x + (k + 5) y = 10$ has :

Let $a, λ, μ \in R$ . Consider the system of linear equations $a x + 2 y = λ, 3 x - 2 y = μ$ . Which of the following statement(s) is (are) correct?

For the system of linear equations $2 x + 4 y + 2 a z = b$ , $x + 2 y + 3 z = 4$ , $2 x + 5 y + 2 z = 8$ which of the following is NOT correct?

Consider the following system of equations $α x + 2 y + z = 1$ , $2 α x + 3 y + z = 1$ , $3 x + α y + 2 z = β$ For some $α, β \in R$ . Then which of the following is NOT correct.

If the system of equations $x - k y - z = 0$ , $k x - y - z = 0$ , $x + y - z = 0$ has a non-zero solution, then the possible values of $k$ are

The system of linear equations $λ x + 2 y + 2 z = 5$ , $2 λ x + 3 y + 5 z = 8$ , $4 x + λ y + 6 z = 10$ has:

For the system of linear equations $a x + y + z = 1$ , $x + a y + z = 1$ , $x + y + a z = β$ , which one of the following statements is NOT correct ?

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

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 :