JEE Main 2023 (24 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Matrices and Determinants: Let $N$ denote the number that turns up when a fair die is rolled. If the probability that the system of equations $x + y + z = 1$ , $2 x + N y + 2 z = 2$ , $3 x + 3 y + N z = 3$ has unique solution is $\frac{k}{6}$ , then the sum of value of $k$ and all possible values of $N$ is

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the system of equations as three planes in 3D space. For the system to have a unique solution, these planes must intersect at exactly one point. This geometric condition is equivalent to the algebraic requirement that the determinant of the coefficient matrix, denoted as

Δ

, must be non-zero. If

Δ = 0

, the planes are either parallel or intersect along a line, leading to no solution or infinitely many solutions.

Hint 2

Hint 2: The Arsenal. Your primary tool here is the determinant of a

3 \times 3

matrix. Recall that for a matrix

A = a d g b e h c f i

, the determinant is calculated as

a (e i - f h) - b (d i - f g) + c (d h - e g)

. Apply this to your coefficient matrix where the variables are

1, 1, 1

in the first row,

2, N, 2

in the second, and

3, 3, N

in the third.

Hint 3

Hint 3: The Execution. After calculating the determinant, you will arrive at a quadratic expression in terms of

N

. Set this expression equal to zero to find the 'forbidden' values of

N

that make the determinant vanish. Once you have these values, exclude them from the set of possible outcomes of a fair die, which is

{1, 2, 3, 4, 5, 6}

. Finally, calculate the probability using the number of valid outcomes and proceed to find

k

and the final sum.

Visualized Solution

The System of Equations

Given system of equations:
$x + y + z = 1$
$2 x + N y + 2 z = 2$
$3 x + 3 y + N z = 3$
Where $N \in {1, 2, 3, 4, 5, 6}$ (Outcome of a fair die)

Condition for Unique Solution

For a unique solution, the determinant of the coefficient matrix ( $Δ$ ) must be non-zero.
Condition: $Δ \neq = 0$

Setting up the Determinant $Δ$

The coefficient matrix determinant is:
$Δ = 123 1 N 3 12 N$

Expanding the Determinant

Expanding along the first row:
$Δ = 1 (N^{2} - 6) - 1 (2 N - 6) + 1 (6 - 3 N)$

Simplifying $Δ$

$Δ = N^{2} - 6 - 2 N + 6 + 6 - 3 N$
$Δ = N^{2} - 5 N + 6$

Factorizing the Quadratic

Factorizing the quadratic expression:
$Δ = (N - 2) (N - 3)$

Applying the Unique Solution Condition

For unique solution: $Δ \neq = 0$
$(N - 2) (N - 3) \neq = 0$
Therefore, $N \neq = 2$ and $N \neq = 3$

Identifying Possible Values of $N$

Possible outcomes of a fair die: ${1, 2, 3, 4, 5, 6}$
Values of $N$ for unique solution: ${1, 4, 5, 6}$

Calculating Probability and $k$

Number of favorable outcomes = $4$
Total outcomes = $6$
Probability = $\frac{4}{6}$
Given probability = $\frac{k}{6} ⟹ k = 4$

Final Sum Calculation

Sum = $k + \sum N_{valid}$
Sum = $4 + (1 + 4 + 5 + 6)$
Sum = $4 + 16 = 20$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

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

x + y + z = 1

2 x + N y + 2 z = 2

3 x + 3 y + N z = 3

Here,

N

represents the outcome of a fair die roll, such that

N \in {1, 2, 3, 4, 5, 6}

. For the system to possess a unique solution, the determinant of the coefficient matrix, denoted by

Δ

, must be non-zero.

The Gatekeeper

Determinant Calculation

The coefficient matrix is defined as:

Δ = 123 1 N 3 12 N

Expanding this determinant along the first row, we obtain:

Δ = 1 (N^{2} - 6) - 1 (2 N - 6) + 1 (6 - 3 N)

Simplifying the expression step-by-step:

Δ = N^{2} - 6 - 2 N + 6 + 6 - 3 N

Δ = N^{2} - 5 N + 6

Factoring the resulting quadratic equation, we find:

Δ = (N - 2) (N - 3)

Identifying Forbidden Values

The condition for a unique solution is $\Delta eq 0$. Therefore, we must satisfy:

(N - 2) (N - 3) e q 0

This implies that $N eq 2$ and $N eq 3$. These are the forbidden values for

N

Given that

N

is the result of a fair die roll, the sample space is

{1, 2, 3, 4, 5, 6}

. By excluding the forbidden values, the set of valid outcomes for

N

{1, 4, 5, 6}

Probability and Final Calculation

There are exactly

4

favorable outcomes out of

6

total possibilities. Thus, the probability is:

P = \frac{4}{6}

The problem states this probability is

\frac{k}{6}

, which identifies

k = 4

The final task is to calculate the sum of

k

and all valid values of

N

Sum = k + (1 + 4 + 5 + 6)

Sum = 4 + 16 = 20

The final result is 20.

Common Traps & FAQs

Wait, why do we reject the negative root here? Or rather, why do we reject specific values of N?

Great question! We aren't rejecting values because they are negative; we are rejecting them because they violate the condition for a unique solution. The condition for a unique solution is

Δ \neq = 0

. When we set

Δ = 0

, we find the values of

N

that make the system singular (i.e., having either no solution or infinite solutions). Since the problem demands a unique solution, any

N

that makes

Δ = 0

must be excluded from our set of possibilities.

Why can't we just use Cramer's Rule for everything?

You absolutely could! Cramer's Rule is a powerful method. However, Cramer's Rule relies on the determinant

Δ

being non-zero to find the unique values of

x, y,

and

z

. If

Δ = 0

, Cramer's Rule fails to provide a unique solution. Since the question specifically asks for the condition of a unique solution, focusing on

Δ \neq = 0

is the most direct and efficient path to the answer.

Is there a shortcut to expanding the determinant?

Yes! You can use row operations to simplify the determinant before expanding. For example, you could perform

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

and

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

. This would create zeros in the first column, making the expansion much faster. Always look for ways to introduce zeros into a row or column before expanding; it reduces the chance of arithmetic errors significantly.

Select Answer:

Visualized Solution

The System of Equations

Condition for Unique Solution

Setting up the Determinant $Δ$

Expanding the Determinant

Simplifying $Δ$

Factorizing the Quadratic

Applying the Unique Solution Condition

Identifying Possible Values of $N$

Calculating Probability and $k$

Final Sum Calculation

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

Analyzing the Setup

The Gatekeeper

Identifying Forbidden Values

Probability and Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

The System of Equations

Condition for Unique Solution

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

Expanding the Determinant

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

Factorizing the Quadratic

Applying the Unique Solution Condition

Identifying Possible Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

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

Final Sum Calculation

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

Analyzing the Setup

The Gatekeeper

Identifying Forbidden Values

Probability and Final Calculation

Similar Questions

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Simplifying $Δ$

Identifying Possible Values of $N$

Calculating Probability and $k$

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The number of values of $k$ for which the linear equations $4 x + k y + 2 z = 0$ , $k x + 4 y + z = 0$ and $2 x + 2 y + z = 0$ possess a non-zero solution is

If the system of equations $x + y + z = 6$ , $2 x + 5 y + α z = β$ , $x + 2 y + 3 z = 14$ has infinitely many solutions, then $α + β$ 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 $3 x + y + 4 z = 3$ , $2 x + α y - z = - 3$ , $x + 2 y + z = 4$ has no solution, then the value of $α$ is equal to :

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 :-

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