Two fair dice are thrown. The numbers on them are taken as λ and μ, and a system of linear equations

Q: Wait, why do we only check $D_1$ for the 'no solution' case? What about $D_2$ and $D_3$?

That is a fantastic question! In a system of linear equations, if the determinant $D = 0$, the system is inconsistent (no solution) if *at least one* of the determinants $D_1, D_2,$ or $D_3$ is non-zero. If all of them are zero, the system might have infinitely many solutions. In this specific problem, checking $D_1$ is sufficient to find the condition on $\mu$ that makes the system inconsistent. If you check $D_2$ or $D_3$, you will find they lead to the same restriction on $\mu$.

Q: Why is $\lambda = 5$ the only condition for $D=0$?

When we expand the determinant $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix}$, we get $1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2)$. Simplifying this gives $2\lambda - 9 - \lambda + 3 + 1 = \lambda - 5$. Setting this to zero gives $\lambda = 5$. It is the only value that makes the determinant vanish, effectively collapsing the 3D space into a lower dimension.

Q: Can I use matrix inversion instead of Cramer's Rule?

You certainly can, but it is like using a sledgehammer to crack a nut! Matrix inversion requires calculating the inverse matrix $A^{-1}$, which involves finding the adjoint and dividing by the determinant. Since $D = \lambda - 5$, the inverse only exists when $\lambda \neq 5$. Cramer's Rule is much more direct for analyzing the nature of solutions in a $3 \times 3$ system.

Question

Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations $x + y + z = 5$, $x + 2y + 3z = \mu$, $x + 3y + \lambda z = 1$ is constructed. If $p$ is the probability that the system has a unique solution and $q$ is the probability that the system has no solution, then :

Options:

(A) $p = \frac{1}{6}$ and $q = \frac{1}{36}$

(B) $p = \frac{5}{6}$ and $q = \frac{5}{36}$

(C) $p = \frac{5}{6}$ and $q = \frac{1}{36}$

(D) $p = \frac{1}{6}$ and $q = \frac{5}{36}$

Sigma Percentile · Accepted Answer

### The Architecture of Consistency: A Journey Through Linear Systems

**Introduction**

Welcome, future engineer! Today, we are not just solving a system of linear equations; we are exploring the very architecture of consistency.

Imagine you are standing in a 3D space, and you have three planes defined by your equations. Depending on the values of $\lambda$ and $\mu$—which are determined by the roll of two fair dice—these planes might intersect at a single point, be parallel, or coincide.

Our goal is to find the probability of these different realities.

### Phase 1: The Determinant as the Gatekeeper

The heartbeat of any $3 	imes 3$ linear system is its coefficient determinant, $D$. This value tells us whether the system is 'well-behaved' (unique solution) or 'troubled' (no solution or infinite solutions).

Let us construct our matrix:

$$A = \begin{pmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & \lambda \end{pmatrix}$$

The determinant $D$ is calculated as:

$$D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & \lambda \end{vmatrix}$$

Expanding along the first row, we get:

$$D = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 2\lambda - 9 - \lambda + 3 + 1 = \lambda - 5$$

This simple expression, $D = \lambda - 5$, is the key to everything.

### Phase 2: The Unique Solution ($p$)

For a system to have a unique solution, the planes must intersect at exactly one point. This happens if and only if $D 
eq 0$.

Therefore, we need $\lambda - 5 
eq 0$, which means $\lambda 
eq 5$. Since $\lambda$ is the result of a fair die roll, it can take any value from $\{1, 2, 3, 4, 5, 6\}$.

The condition $\lambda 
eq 5$ leaves us with 5 favorable outcomes for $\lambda$: $\{1, 2, 3, 4, 6\}$. Since $\mu$ can be any of the 6 values on the second die, the total favorable outcomes for a unique solution are $5 	imes 6 = 30$.

With a total of $6 	imes 6 = 36$ possible outcomes, the probability $p$ is:

$$p = \frac{30}{36} = \frac{5}{6}$$

### Phase 3: The Inconsistency Trap ($q$)

Now, let us look at the 'no solution' case. This occurs when $D = 0$ (so $\lambda = 5$) AND the system is inconsistent.

For inconsistency, at least one of the determinants $D_1, D_2,$ or $D_3$ must be non-zero. Let us calculate $D_1$ by replacing the first column with the constants $(5, \mu, 1)$:

$$D_1 = \begin{vmatrix} 5 & 1 & 1 \ \mu & 2 & 3 \ 1 & 3 & 5 \end{vmatrix}$$

Expanding this:

$$D_1 = 5(10 - 9) - 1(5\mu - 3) + 1(3\mu - 2) = 5(1) - 5\mu + 3 + 3\mu - 2 = 6 - 2\mu$$

For the system to have no solution, we need $D_1 
eq 0$. Thus, $6 - 2\mu 
eq 0$, which implies $\mu 
eq 3$.

So, the condition for no solution is $\lambda = 5$ and $\mu 
eq 3$. For $\lambda = 5$, there is only 1 choice. For $\mu 
eq 3$, there are 5 choices ($\{1, 2, 4, 5, 6\}$).

The number of favorable outcomes is $1 	imes 5 = 5$. The probability $q$ is:

$$q = \frac{5}{36}$$

### Conclusion

We have successfully navigated the landscape of this system. We found **$p = \frac{5}{6}$** and **$q = \frac{5}{36}$**.

By breaking down the problem into the behavior of the determinant and the conditions for inconsistency, we turned a complex algebraic problem into a clear, logical path. Keep practicing, and remember: every equation has a story to tell!

Select Answer:

Visualized Solution

The Setup: Dice and Equations

Coefficient Determinant $D$

Expanding Determinant $D$

Condition for Unique Solution

Calculating Probability $p$

Condition for No Solution

Forming Determinant $D_{1}$

Expanding Determinant $D_{1}$

Applying No Solution Condition

Calculating Probability $q$

Final Conclusion

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

The Architecture of Consistency

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

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

Let $n$ be the number obtained on rolling a fair die. If the probability that the system $x - n y + z = 6$ , $x + (n - 2) y + (n + 1) z = 8$ , $(n - 1) y + z = 1$ has a unique solution is $\frac{k}{6}$ , then the sum of $k$ and all possible values of $n$ is :

Consider the system of equations $a x + b y = 0, c x + d y = 0$ , where $a, b, c, d \in {0, 1}$ . STATEMENT - 1 : The probability that the system of equations has a unique solution is $3/8$ . and STATEMENT - 2 : The probability that the system of equations has a solution is 1.

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

The system of linear equations $x + y + z = 6$ , $2 x + 5 y + a z = 36$ , $x + 2 y + 3 z = b$ has

The values of $λ$ and $μ$ such that the system of equations $x + y + z = 6$ , $3 x + 5 y + 5 z = 26$ , $x + 2 y + λ z = μ$ has no solution, are :

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:

If the system of linear equations $x + y + z = 5$ , $x + 2 y + 2 z = 6$ , $x + 3 y + z = μ$ , $(λ, μ \in R)$ , has infinitely many solutions, then the value of $λ + μ$ is :

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:

Consider the system of linear equations $x + y + z = 5, x + 2 y + λ^{2} z = 9, x + 3 y + λ z = μ$ , where $λ, μ \in R$ . Then, which of the following statement is NOT correct?

Select Answer:

Visualized Solution

The Setup: Dice and Equations

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

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

Condition for Unique Solution

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

Condition for No Solution

Forming Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D1​

Expanding Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D1​