JEE Main 2020 - 3 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $S$ be the set of all integer solutions, $(x, y, z)$ , of the system of equations $x - 2 y + 5 z = 0 - 2 x + 4 y + z = 0 - 7 x + 14 y + 9 z = 0$ such that $15 < x^{2} + y^{2} + z^{2} < 150$ . Then, the number of elements in the set $S$ is equal to

Q: Wait, why is $z$ always zero? Can't we have non-zero $z$?

Great question! In a homogeneous system, if the determinant of the coefficient matrix is zero, we have infinitely many solutions. Here, the equations are linearly dependent. When we perform elimination, we find that $z$ must be zero to satisfy the system. It's not that $z$ *can't* be non-zero, it's that the system *forces* it to be zero.

Q: Why do we use $k$ as an integer? Does the problem say $x, y, z$ are integers?

Yes, the problem explicitly asks for integer solutions $(x, y, z)$. Since $x=2k$ and $y=k$, for $x$ and $y$ to be integers, $k$ must be an integer. This is a crucial step!

Q: What if $k^2$ wasn't a perfect square? Do we just ignore those?

Exactly! We are looking for integer solutions. If $k^2$ is not a perfect square, then $k$ would not be an integer, which would mean $x$ and $y$ would not be integers. So, we only care about the perfect squares that fall within our range.

Enter Numerical Value:

Visualized Solution

Analyze the System

Given system of homogeneous equations:
1) $x - 2 y + 5 z = 0$
2) $- 2 x + 4 y + z = 0$
3) $- 7 x + 14 y + 9 z = 0$
Constraint: $15 < x^{2} + y^{2} + z^{2} < 150$

Solve for $z$

Multiply Eq (1) by $2$ : $2 x - 4 y + 10 z = 0$
Add to Eq (2): $- 2 x + 4 y + z = 0$
$(2 x - 4 y + 10 z) + (- 2 x + 4 y + z) = 0$
$11 z = 0 ⟹ z = 0$

Find Relation between $x$ and $y$

Substitute $z = 0$ into Eq (1):
$x - 2 y + 5 (0) = 0$
$x - 2 y = 0 ⟹ x = 2 y$
General solution: $(x, y, z) = (2 k, k, 0)$ where $k \in Z$

Apply the Magnitude Constraint

Constraint: $15 < x^{2} + y^{2} + z^{2} < 150$
Substitute $x = 2 k, y = k, z = 0$ :
$15 < (2 k)^{2} + (k)^{2} + 0^{2} < 150$
$15 < 4 k^{2} + k^{2} < 150$
$15 < 5 k^{2} < 150$

Simplify the Inequality

Divide the entire inequality by $5$ :
$\frac{15}{5} < \frac{5 k ^{2}}{5} < \frac{150}{5}$
$3 < k^{2} < 30$
We need integer values for $k$ , so $k^{2}$ must be a perfect square.

Find Integer Values of $k$

Possible perfect squares between $3$ and $30$ :
$k^{2} \in {4, 9, 16, 25}$
For $k^{2} = 4 ⟹ k = \pm 2$
For $k^{2} = 9 ⟹ k = \pm 3$
For $k^{2} = 16 ⟹ k = \pm 4$
For $k^{2} = 25 ⟹ k = \pm 5$

Final Conclusion

Total valid values for $k$ :
$k \in {- 5, - 4, - 3, - 2, 2, 3, 4, 5}$
Number of elements in set $S = 8$
The system has exactly $8$ integer solutions.

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

The given system of equations is: (1)

x - 2 y + 5 z = 0

(2)

- 2 x + 4 y + z = 0

(3)

- 7 x + 14 y + 9 z = 0

Since the constant term on the right side of every equation is zero, this is a homogeneous system. Such systems always possess the trivial solution

(0, 0, 0)

, but we are tasked with finding non-trivial integer solutions satisfying the constraint

15 < x^{2} + y^{2} + z^{2} < 150

Unmasking the Line

Observe that the coefficients of

x

and

y

in equation (2) are exactly

- 2

times the coefficients in equation (1). This indicates that the equations are linearly dependent.

To simplify, multiply equation (1) by

2

2 x - 4 y + 10 z = 0

Now, add this to equation (2):

(2 x - 4 y + 10 z) + (- 2 x + 4 y + z) = 0

11 z = 0 ⟹ z = 0

With

z = 0

, the system reduces to

x - 2 y = 0

, or

x = 2 y

. Thus, any solution must take the parametric form

(2 k, k, 0)

for some integer

k

The Geometric Constraint

We apply the magnitude constraint

15 < x^{2} + y^{2} + z^{2} < 150

. Substituting our parametric form

(2 k, k, 0)

into this inequality:

15 < (2 k)^{2} + k^{2} + 0^{2} < 150

15 < 4 k^{2} + k^{2} < 150

15 < 5 k^{2} < 150

Dividing the entire inequality by

5

, we obtain:

3 < k^{2} < 30

The Integer Hunt

We must identify all integers

k

such that

k^{2}

is a perfect square between

3

and

30

. The possible values for

k^{2}

are

4, 9, 16,

and

25

This yields the following possibilities for

k

: $k^{2} = 4 ⟹ k = \pm 2$

k^{2} = 9 ⟹ k = \pm 3

$k^{2} = 16 ⟹ k = \pm 4$

k^{2} = 25 ⟹ k = \pm 5

Each value of

k

corresponds to a unique integer triplet

(x, y, z)

. Since there are

4

positive values and

4

negative values for

k

, there are a total of 8 distinct integer solutions.

Common Traps & FAQs

Wait, why is

z

always zero? Can't we have non-zero

z

Great question! In a homogeneous system, if the determinant of the coefficient matrix is zero, we have infinitely many solutions. Here, the equations are linearly dependent. When we perform elimination, we find that

z

must be zero to satisfy the system. It's not that

z

*can't* be non-zero, it's that the system *forces* it to be zero.

Why do we use

k

as an integer? Does the problem say

x, y, z

are integers?

Yes, the problem explicitly asks for integer solutions

(x, y, z)

. Since

x = 2 k

and

y = k

, for

x

and

y

to be integers,

k

must be an integer. This is a crucial step!

What if

k^{2}

wasn't a perfect square? Do we just ignore those?

Exactly! We are looking for integer solutions. If

k^{2}

is not a perfect square, then

k

would not be an integer, which would mean

x

and

y

would not be integers. So, we only care about the perfect squares that fall within our range.

Enter Numerical Value:

Visualized Solution

Analyze the System

Solve for $z$

Find Relation between $x$ and $y$

Apply the Magnitude Constraint

Simplify the Inequality

Find Integer Values of $k$

Final Conclusion

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

Analyzing the Setup

Unmasking the Line

The Geometric Constraint

The Integer Hunt

Similar Questions

Let the system of equations $x + 5 y - z = 1$ , $4 x + 3 y - 3 z = 7$ , $24 x + y + λ z = μ$ , $λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, If $x, y, z$ are integers and satisfy $7 \leq x + y + z \leq 77$ , is

Let $(x, y, z)$ be points with integer coordinates satisfying the system of homogeneous equations : $3 x - y - z = 0; - 3 x + z = 0; - 3 x + 2 y + z = 0$ . Then the number of such points for which $x^{2} + y^{2} + z^{2} \leq 100$ is .........

Let $S$ be the set of all values of $θ \in [- π, π]$ for which the system of linear equations $x + y + 3 z = 0$ , $- x + (tan θ) y + 7 z = 0$ , $x + y + (tan θ) z = 0$ has non-trivial solution. Then $120 \sum_{θ \in S} θ$ 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 :

The system of equations $x + y + z = 6$ $x + 2 y + 5 z = 9,$ $x + 5 y + λ z = μ,$ has no solution if

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 $S_{1}$ and $S_{2}$ be respectively the sets of all $a \in R - {0}$ for which the system of linear equations $a x + 2 a y - 3 a z = 1$ , $(2 a + 1) x + (2 a + 3) y + (a + 1) z = 2$ , $(3 a + 5) x + (a + 5) y + (a + 2) z = 3$ has unique solution and infinitely many solutions. Then

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

Let $[λ]$ be the greatest integer less than or equal to $λ$ . The set of all values of $λ$ for which the system of linear equations $x + y + z = 4, 3 x + 2 y + 5 z = 3, 9 x + 4 y + (28 + [λ]) z = [λ]$ has a solution is:

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:

Enter Numerical Value:

Visualized Solution

Analyze the System

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

Find Relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Apply the Magnitude Constraint

Simplify the Inequality

Find Integer Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Final Conclusion

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

Analyzing the Setup

Unmasking the Line

The Geometric Constraint

The Integer Hunt

Similar Questions

.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; } The system of equations x+y+z=6 x+2y+5z=9, x+5y+λz=μ, has no solution if

.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; } If the system of equations 2x+y−z=5, 2x−5y+λz=μ, x+2y−5z=7 has infinitely many solutions, then (λ+μ)2+(λ−μ)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 equations x+y+z=22x+4y−z=63x+2y+λz=μ has infinitely many solutions, then:

Solve for $z$

Find Relation between $x$ and $y$

Find Integer Values of $k$

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 :

The system of equations $x + y + z = 6$ $x + 2 y + 5 z = 9,$ $x + 5 y + λ z = μ,$ has no solution if

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

If the system of equations $2 x + y - z = 5$ , $2 x - 5 y + λ z = μ$ , $x + 2 y - 5 z = 7$ has infinitely many solutions, then $(λ + μ)^{2} + (λ - μ)^{2}$ 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: