Why can't I just solve for $x$ directly?

Solving for $x$ directly would involve messy algebraic manipulation that obscures the underlying symmetry. By substituting, we reveal the structure of the system, making the solution path clear and intuitive.

Wait, why do we get 8 solutions? Shouldn't it be just one?

That's a classic trap! When you have $x^2 = a^2$, you have two roots: $x = a$ and $x = -a$. Since each variable $x, y, z$ has two possible values, the total number of combinations is $2 \times 2 \times 2 = 8$.

Is there a case where there are no solutions?

In this specific system, the linear equations are consistent, leading to $X=1, Y=1, Z=1$. If the system were inconsistent, we would have no solutions. Always check the consistency of your linear system first!

JEE Advanced 1995

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $a, b, c$ be the real numbers. Then following system of equations in $x, y$ and $z$ $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} - \frac{z ^{2}}{c ^{2}} = 1$ , $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} + \frac{z ^{2}}{c ^{2}} = 1$ , $- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} + \frac{z ^{2}}{c ^{2}} = 1$ has

Select Answer:

Visualized Solution

Analyze the System Structure

Given system of equations:
$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} - \frac{z ^{2}}{c ^{2}} = 1$
$\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} + \frac{z ^{2}}{c ^{2}} = 1$
$- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} + \frac{z ^{2}}{c ^{2}} = 1$
Observe the symmetry in the terms involving $x, y,$ and $z$ .

Apply Variable Substitution

To simplify, let's substitute the repeating terms.
Let $X = \frac{x ^{2}}{a ^{2}}$
Let $Y = \frac{y ^{2}}{b ^{2}}$
Let $Z = \frac{z ^{2}}{c ^{2}}$

The Transformed Linear System

The system transforms into:
$(1) X + Y - Z = 1$
$(2) X - Y + Z = 1$
$(3) - X + Y + Z = 1$

Solve for $X$

Adding Eq $(1)$ and Eq $(2)$ :
$(X + Y - Z) + (X - Y + Z) = 1 + 1$
$2 X = 2 ⟹ X = 1$

Solve for $Z$

Adding Eq $(2)$ and Eq $(3)$ :
$(X - Y + Z) + (- X + Y + Z) = 1 + 1$
$2 Z = 2 ⟹ Z = 1$

Solve for $Y$

Adding Eq $(1)$ and Eq $(3)$ :
$(X + Y - Z) + (- X + Y + Z) = 1 + 1$
$2 Y = 2 ⟹ Y = 1$

Back-Substitution for $x, y, z$

We found $X = 1, Y = 1, Z = 1$ .
Now, substitute back the original expressions:
$X = 1 ⟹ \frac{x ^{2}}{a ^{2}} = 1$
$Y = 1 ⟹ \frac{y ^{2}}{b ^{2}} = 1$
$Z = 1 ⟹ \frac{z ^{2}}{c ^{2}} = 1$

Finding the Roots

$\frac{x ^{2}}{a ^{2}} = 1 ⟹ x^{2} = a^{2} ⟹ x = \pm a$
$\frac{y ^{2}}{b ^{2}} = 1 ⟹ y^{2} = b^{2} ⟹ y = \pm b$
$\frac{z ^{2}}{c ^{2}} = 1 ⟹ z^{2} = c^{2} ⟹ z = \pm c$

Counting the Total Solutions

Each variable has $2$ possible values.
Total solutions are formed by combinations of $(\pm a, \pm b, \pm c)$ .
Number of solutions $= 2 \times 2 \times 2 = 8$ .

Final Conclusion

The system has exactly $8$ distinct solutions.
Since $8$ is a finite number, the system has finitely many solutions.
Correct Option: (d)

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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:

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} - \frac{z ^{2}}{c ^{2}} = 1

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} + \frac{z ^{2}}{c ^{2}} = 1

- \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} + \frac{z ^{2}}{c ^{2}} = 1

At first glance, this system appears chaotic. However, the key to mastering JEE problems is identifying the hidden symmetry rather than relying on brute force.

The Power of Substitution

To simplify the algebra, we introduce a smart substitution. Let:

X = \frac{x ^{2}}{a ^{2}}, Y = \frac{y ^{2}}{b ^{2}}, Z = \frac{z ^{2}}{c ^{2}}

Substituting these into the original equations, the system transforms into a linear set:

X + Y - Z = 1

X - Y + Z = 1

- X + Y + Z = 1

Solving the Linear System

We can now solve for

X, Y,

and

Z

using simple elimination. Adding equation (1) and equation (2) yields:

2 X = 2 \Rightarrow X = 1

Adding equation (2) and equation (3) yields:

2 Z = 2 \Rightarrow Z = 1

Finally, adding equation (1) and equation (3) yields:

2 Y = 2 \Rightarrow Y = 1

Thus, we have determined that

X = 1

Y = 1

, and

Z = 1

The Trap of the Roots

We must now reverse the substitution to find the values of

x, y,

and

z

. Since

X = \frac{x ^{2}}{a ^{2}} = 1

, we have

x^{2} = a^{2}

, which implies

x = \pm a

Similarly, we find

y = \pm b

and

z = \pm c

A solution is defined as an ordered triple

(x, y, z)

. Since each variable has two possible values, we apply the fundamental principle of counting:

2 \times 2 \times 2 = 8

The system has exactly 8 distinct solutions. Because this is a finite number, the system possesses finitely many solutions.

Select Answer:

Visualized Solution

Analyze the System Structure

Apply Variable Substitution

The Transformed Linear System

Solve for $X$

Solve for $Z$

Solve for $Y$

Back-Substitution for $x, y, z$

Finding the Roots

Counting the Total Solutions

Final Conclusion

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

Analyzing the Setup

The Power of Substitution

Solving the Linear System

The Trap of the Roots

Similar Questions

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 :

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.

The equation $x + 2 y + 2 z = 1$ and $2 x + 4 y + 4 z = 9$ have

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 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 .........

If $α + β + γ = 2 π$ , then the system of equations $x + (cos γ) y + (cos β) z = 0$ , $(cos γ) x + y + (cos α) z = 0$ , $(cos β) x + (cos α) y + z = 0$ has :

The following system of linear equations $3 x + 3 y + 2 z = 9, 3 x + 2 y + 2 z = 9, x - y + 4 z = 8$ has:

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 :

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 $α, β$ and $γ$ be real numbers. consider the following system of linear equations $x + 2 y + z = 7$ $x + α z = 11$ $2 x - 3 y + β z = γ$ Match each entry in List - I to the correct entries in List-II

List-I

List-II

Select Answer:

Visualized Solution

Analyze the System Structure

Apply Variable Substitution

The Transformed Linear System

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

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

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

Back-Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x,y,z

Finding the Roots

Counting the Total Solutions

Final Conclusion

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

Analyzing the Setup

The Power of Substitution

Solving the Linear System

The Trap of the Roots

Similar Questions

.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 :

.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; } The equation x+2y+2z=1 and 2x+4y+4z=9 have

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α+β+γ=2π, then the system of equations x+(cosγ)y+(cosβ)z=0, (cosγ)x+y+(cosα)z=0, (cosβ)x+(cosα)y+z=0 has :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The following system of linear equations 3x+3y+2z=9,3x+2y+2z=9,x−y+4z=8 has:

.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+2y+3z=a, 3x−y+5z=b, x−3y+2z=c where a,b,c are non-zero real numbers, has more than one solution, then :

.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 :

List-I

List-II

Solve for $X$

Solve for $Z$

Solve for $Y$

Back-Substitution for $x, y, z$

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 :

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.

The equation $x + 2 y + 2 z = 1$ and $2 x + 4 y + 4 z = 9$ have

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 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 .........

If $α + β + γ = 2 π$ , then the system of equations $x + (cos γ) y + (cos β) z = 0$ , $(cos γ) x + y + (cos α) z = 0$ , $(cos β) x + (cos α) y + z = 0$ has :

The following system of linear equations $3 x + 3 y + 2 z = 9, 3 x + 2 y + 2 z = 9, x - y + 4 z = 8$ has:

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 :

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 :