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Animated Solution for Mathematics - Matrices and Determinants: Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x = cy + b z, y = a z + c x$ , and $z = b x + a y$ . Then $a^{2} + b^{2} + c^{2} + 2 ab c$ is equal to

Q: Why does a non-trivial solution imply that the determinant must be zero?

Great question! A homogeneous system $AX = 0$ always has the trivial solution $x=y=z=0$. If the determinant of $A$ is non-zero, the matrix $A$ is invertible, meaning we can multiply by $A^{-1}$ to get $X = A^{-1}0 = 0$. Thus, a non-zero determinant forces the only solution to be the trivial one. Since we are told a non-trivial solution exists, the matrix must be singular, which means its determinant must be zero.

Select Answer:

Visualized Solution

Visualizing the System of Equations

We are given a system of three linear equations in three variables $x$ , $y$ , and $z$ .
The problem states that there exists a non-trivial solution, meaning $x, y, z$ are not all zero.
Geometrically, each equation represents a plane passing through the origin $(0, 0, 0)$ .
Since a non-trivial solution exists, these three planes must intersect along a common line rather than just a single point.

The Given System of Equations

Let's write down the given equations:
$x = cy + b z$
$y = a z + c x$
$z = b x + a y$
Constraint: $x, y, z$ are not all zero.

Rearranging into Homogeneous Form

To analyze the system, we move all terms to the left-hand side.
Equation 1: $1 x - cy - b z = 0$
Equation 2: $- c x + 1 y - a z = 0$
Equation 3: $- b x - a y + 1 z = 0$
This is now a standard homogeneous system of linear equations.

Matrix Representation $A X = 0$

We can represent this homogeneous system in matrix form as $A X = 0$ .
The coefficient matrix $A$ is:
$A = 1 - c - b - c 1 - a - b - a 1$
The variable vector $X$ is $x y z$ .

Condition for Non-Trivial Solutions

For a homogeneous system $A X = 0$ to have a non-trivial solution ( $X \neq = 0$ ):
The determinant of the coefficient matrix must be zero: $det (A) = 0$ .
If $det (A) \neq = 0$ , the system has only the unique trivial solution $x = y = z = 0$ .

Setting up the Determinant Equation

Setting the determinant of $A$ to zero:
$1 - c - b - c 1 - a - b - a 1 = 0$
We will expand this determinant to find a relationship between $a$ , $b$ , and $c$ .

Expanding along the First Row

Expanding along the first row:
$1 \cdot 1 - a - a 1 - (- c) \cdot - c - b - a 1 + (- b) \cdot - c - b 1 - a = 0$
Be extremely careful with the signs during expansion!

Evaluating the Sub-Determinants

Evaluating each $2 \times 2$ determinant:
$1 - a - a 1 = 1 - a^{2}$
$- c - b - a 1 = - c - ab$
$- c - b 1 - a = a c + b$
Substituting these back: $1 (1 - a^{2}) + c (- c - ab) - b (a c + b) = 0$

Simplifying the Equation

Distributing the terms:
$1 - a^{2} - c^{2} - ab c - ab c - b^{2} = 0$
Combine the like terms:
$1 - a^{2} - b^{2} - c^{2} - 2 ab c = 0$

Finding the Final Value

Rearranging the equation to isolate the target expression:
$1 = a^{2} + b^{2} + c^{2} + 2 ab c$
Therefore, $a^{2} + b^{2} + c^{2} + 2 ab c = 1$ .
The correct option is 1 (Option 3).

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

Analyzing the Setup

Imagine you are standing in a three-dimensional space, looking at three planes that all pass through the origin. Usually, these planes would intersect at a single point—the origin itself.

However, this problem presents a fascinating twist: we are told that there exists a non-trivial solution, meaning

x, y,

and

z

are not all zero. This implies that these three planes do not just meet at a point; they intersect along an entire line.

This geometric reality is the key to unlocking the algebraic mystery behind the expression

a^{2} + b^{2} + c^{2} + 2 ab c

The Homogeneous Transformation

We start with the given system of equations:

x = cy + b z

y = a z + c x

z = b x + a y

To see the underlying structure, we must bring these into a standard homogeneous form. By moving all terms to the left-hand side, we obtain:

x - cy - b z = 0

- c x + y - a z = 0

- b x - a y + z = 0

This is a classic homogeneous system, which we can represent in matrix form as

A X = 0

, where

A

is the coefficient matrix and

X

is the column vector of variables. The matrix

A

is:

A = 1 - c - b - c 1 - a - b - a 1

The Power of the Determinant

For a homogeneous system to have a non-trivial solution, the matrix

A

must be singular. In the language of linear algebra, this means the determinant of

A

must be exactly zero.

If it were not zero, the system would only have the trivial solution

x = y = z = 0

, which contradicts our premise. So, we set the determinant to zero:

det (A) = 1 - c - b - c 1 - a - b - a 1 = 0

Now, let us expand this determinant along the first row. Be methodical with the signs:

det (A) = 1 \cdot 1 - a - a 1 - (- c) \cdot - c - b - a 1 + (- b) \cdot - c - b 1 - a = 0

Evaluating these

2 \times 2

determinants is straightforward: 1. The first minor:

(1) (1) - (- a) (- a) = 1 - a^{2}

2. The second minor:

(- c) (1) - (- a) (- b) = - c - ab

3. The third minor:

(- c) (- a) - (1) (- b) = a c + b

Substituting these back into our expansion, we get:

1 (1 - a^{2}) + c (- c - ab) - b (a c + b) = 0

The Elegant Conclusion

Now, let us distribute the terms and simplify the expression:

1 - a^{2} - c^{2} - ab c - ab c - b^{2} = 0

Combining the like terms, we arrive at:

1 - a^{2} - b^{2} - c^{2} - 2 ab c = 0

Finally, by moving the negative terms to the right-hand side, we find the beautiful result:

$a^{2} + b^{2} + c^{2} + 2 ab c = 1$

Common Traps & FAQs

Why does a non-trivial solution imply that the determinant must be zero?

Great question! A homogeneous system

A X = 0

always has the trivial solution

x = y = z = 0

. If the determinant of

A

is non-zero, the matrix

A

is invertible, meaning we can multiply by

A^{- 1}

to get

X = A^{- 1} 0 = 0

. Thus, a non-zero determinant forces the only solution to be the trivial one. Since we are told a non-trivial solution exists, the matrix must be singular, which means its determinant must be zero.

Can I just use substitution instead of determinants?

You could, but it would be algebraically exhausting! Substitution would lead to a messy chain of equations that is prone to sign errors. Using the determinant property is the elegant, professional way to handle homogeneous systems in JEE Advanced. It directly links the existence of solutions to the structure of the matrix.

What if I get a different sign when expanding the determinant?

Sign errors are the most common trap in determinant expansion. Remember the cofactor expansion rule: the signs follow a checkerboard pattern:

+ - +

for the first row. When you expand, ensure you multiply the element by its minor and apply the correct sign from the pattern. Practice makes perfect!

Animated Solution for Mathematics - Matrices and Determinants: Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x = cy + b z, y = a z + c x$ , and $z = b x + a y$ . Then $a^{2} + b^{2} + c^{2} + 2 ab c$ is equal to

Select Answer:

Visualized Solution

Visualizing the System of Equations

The Given System of Equations

Rearranging into Homogeneous Form

Matrix Representation $A X = 0$

Condition for Non-Trivial Solutions

Setting up the Determinant Equation

Expanding along the First Row

Evaluating the Sub-Determinants

Simplifying the Equation

Finding the Final Value

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

Analyzing the Setup

The Homogeneous Transformation

The Power of the Determinant

The Elegant Conclusion

Similar Questions

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 $1 + a + a^{2} =$

Comprehension Passage

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

Let $R^{2}$ denote $R \times R$ . Let $S = {(a, b, c) : a, b, c \in R and a x^{2} + 2 b x y + c y^{2} > 0 for all (x, y) \in R^{2} - {(0, 0)}}$ . Then which of the following statements is (are) TRUE?

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

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

Consider the following linear equations $a x + b y + cz = 0; b x + cy + a z = 0; c x + a y + b z = 0$ Match the conditions/expressions in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

List-II

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 linear equations $x + y + 3 z = 0 x + 3 y + k^{2} z = 0 3 x + y + 3 z = 0$ has a non-zero solution $(x, y, z)$ for some $k \in R$ then $x + (\frac{y}{z})^{2}$ is equal to

Select Answer:

Visualized Solution

Visualizing the System of Equations

The Given System of Equations

Rearranging into Homogeneous Form

Matrix Representation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AX=0

Condition for Non-Trivial Solutions

Setting up the Determinant Equation

Expanding along the First Row

Evaluating the Sub-Determinants

Simplifying the Equation

Finding the Final Value

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

Analyzing the Setup

The Homogeneous Transformation

The Power of the Determinant

The Elegant Conclusion

Similar Questions

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Comprehension Passage

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List-I

List-II

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Matrix Representation $A X = 0$

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 $1 + a + a^{2} =$

If the point $P (a, b, c)$ , with reference to (E), lies on the plane $2 x + y + z = 1$ , then the value of $7 a + b + c$ is

Let $ω$ be a solution of $x^{3} - 1 = 0$ with $Im (ω) > 0$ , if $a = 2$ with $b$ and $c$ satisfying (E), then the value of $\frac{3}{ω ^{a}} + \frac{1}{ω ^{b}} + \frac{3}{ω ^{c}}$ is equal to

Let $b = 6$ , with $a$ and $c$ satisfying (E). If $α$ and $β$ are the roots of the quadratic equation $a x^{2} + b x + c = 0$ , then $\sum_{n = 0}^{\infty} (\frac{1}{α} + \frac{1}{β})^{n}$ is

Let $R^{2}$ denote $R \times R$ . Let $S = {(a, b, c) : a, b, c \in R and a x^{2} + 2 b x y + c y^{2} > 0 for all (x, y) \in R^{2} - {(0, 0)}}$ . Then which of the following statements is (are) TRUE?

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

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

If the system of linear equations $x + y + 3 z = 0 x + 3 y + k^{2} z = 0 3 x + y + 3 z = 0$ has a non-zero solution $(x, y, z)$ for some $k \in R$ then $x + (\frac{y}{z})^{2}$ is equal to