Let $n$ be the number obtained on rolling a fair die. If the probability that the system
$x – ny + z = 6$
$x + (n-2)y + (n+1)z = 8$
$(n-1)y + z = 1$
has a unique solution is $\dfrac{k}{6}$, then the sum of $k$ and all possible values of $n$ is :
(A) 22 (B) 20 (C) 24 (D) 21
$x – ny + z = 6$
$x + (n-2)y + (n+1)z = 8$
$(n-1)y + z = 1$
has a unique solution is $\dfrac{k}{6}$, then the sum of $k$ and all possible values of $n$ is :
(A) 22 (B) 20 (C) 24 (D) 21
✓
Solution Steps
1
Write the coefficient matrix
The three equations are:
$x – ny + z = 6$
$x + (n-2)y + (n+1)z = 8$
$(n-1)y + z = 1$
The coefficient matrix $A$ is:
$$A = \begin{pmatrix} 1 & -n & 1 \\ 1 & n-2 & n+1 \\ 0 & n-1 & 1 \end{pmatrix}$$
2
Compute the determinant $D = \det(A)$
Expanding along Row 1:
$$D = 1\cdot\bigl[(n-2)\cdot1-(n+1)(n-1)\bigr]$$
$$\quad -(-n)\cdot\bigl[1\cdot1-(n+1)\cdot0\bigr]$$
$$\quad +1\cdot\bigl[1\cdot(n-1)-(n-2)\cdot0\bigr]$$
3
Simplify each cofactor
First term: $(n-2) – (n^2-1) = n – 2 – n^2 + 1 = -n^2 + n – 1$
Second term: $+n \cdot 1 = n$
Third term: $+(n-1)$
So:
$$D = (-n^2+n-1) + n + (n-1)$$
$$D = -n^2 + 3n – 2$$
$$D = -(n^2 – 3n + 2) = -(n-1)(n-2)$$
4
Condition for unique solution: $D \ne 0$
$$D \ne 0 \;\Rightarrow\; -(n-1)(n-2) \ne 0$$
$$\Rightarrow\; n \ne 1 \text{ and } n \ne 2.$$
5
Identify valid values of $n$ from die outcomes
Rolling a fair die: $n \in \{1,2,3,4,5,6\}$.
Excluding $n=1$ and $n=2$, the valid values are:
$$n \in \{3,4,5,6\} \quad \text{(4 values)}.$$
6
Find $k$ from probability $= k/6$
Favourable outcomes = 4, Total = 6.
$$P = \frac{4}{6} = \frac{k}{6} \;\Rightarrow\; k = 4.$$
7
Compute the required sum
Sum of $k$ and all possible values of $n$:
$$= k + 3 + 4 + 5 + 6$$
$$= 4 + 18 = \boxed{22}.$$
Hence the correct option is $\boxed{\text{(A) 22}}$.
💡
Key Insight
Unique solution ⟺ D = −(n−1)(n−2) ≠ 0, so n ≠ 1,2. Valid n = {3,4,5,6}, k = 4. Sum = 4+18 = 22.
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Linear Algebra Theory
1. Unique solution of a 3×3 linear system via determinant
A system of three linear equations in three unknowns $AX = B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero, i.e., $\det(A) \ne 0$. This is because $\det(A) \ne 0$ ensures that $A$ is invertible, and the unique solution is $X = A^{-1}B$. Equivalently, Cramer’s rule gives $x_i = D_i / D$ where $D = \det(A)$. If $D = 0$, the system is either inconsistent (no solution) or consistent with infinitely many solutions, depending on the right-hand side. In JEE problems involving parameters, the key task is to find all parameter values that make $D = 0$, because those are the cases where uniqueness fails. Conversely, all values of the parameter that keep $D \ne 0$ correspond to a unique solution. In this question the parameter is $n$, which must also be a die outcome, so it is restricted to integers 1 through 6.
2. Expanding a 3×3 determinant along a row
The most systematic way to compute a $3\times3$ determinant is cofactor expansion along any row or column. Expanding along the first row of matrix $A = (a_{ij})$ gives:
$\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13},$
where $C_{ij} = (-1)^{i+j} M_{ij}$ and $M_{ij}$ is the $2\times2$ minor obtained by deleting row $i$ and column $j$. In this problem, expanding along Row 1 gives three $2\times2$ determinants. The signs alternate as $+, -, +$. A helpful trick when Row 3 starts with a zero (as here, where the first entry of Row 3 is 0) is to expand along Row 3 instead, reducing the work by one term. However, expanding along Row 1 also works cleanly as demonstrated in the solution. Careful algebra during expansion is crucial, especially when entries are expressions in $n$.
3. Classical probability: counting favourable outcomes
Classical probability is defined as the ratio of the number of favourable outcomes to the total number of equally likely outcomes. When a fair die is rolled, each face (1 through 6) has equal probability $\frac{1}{6}$. If a condition is met for a specific subset of outcomes $\{n_1, n_2, \ldots, n_k\}$, then the probability is simply $\frac{k}{6}$. In this problem, the system has a unique solution for $n \in \{3,4,5,6\}$ — four outcomes — so the probability is $\frac{4}{6}$. Comparing with $\frac{k}{6}$ directly gives $k = 4$. The question then asks for the sum of $k$ plus all valid $n$ values: $4 + 3 + 4 + 5 + 6 = 22$. This type of combined algebra-probability question is common in JEE, requiring both the determinant condition and probability calculation.
4. Factoring the determinant expression and checking integer roots
After computing the determinant, you often get a polynomial in the parameter $n$. Factoring this polynomial reveals exactly which values of $n$ make $D = 0$. In this problem, $D = -n^2 + 3n – 2 = -(n-1)(n-2)$. The roots are $n = 1$ and $n = 2$. Since $n$ must be an integer from 1 to 6, only $n = 1$ and $n = 2$ cause $D = 0$. The remaining values $n = 3, 4, 5, 6$ all give $D \ne 0$. Factoring quadratics of the form $n^2 – (r+s)n + rs = (n-r)(n-s)$ is a standard algebraic skill. Recognizing this quickly saves time in JEE examinations. Always verify the factoring by re-expanding: $(n-1)(n-2) = n^2 – 3n + 2$, which confirms $D = -(n^2 – 3n + 2) = -n^2 + 3n – 2$.
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FAQs
Q
When does a linear system have a unique solution?
⌄
When the determinant of the coefficient matrix is non-zero, $\det(A) \ne 0$. Then $A$ is invertible and the system has exactly one solution.
Q
What is the coefficient matrix for this system?
⌄
$A = \begin{pmatrix}1 & -n & 1\\1 & n-2 & n+1\\0 & n-1 & 1\end{pmatrix}$.
Q
What is $\det(A)$ after simplification?
⌄
$D = -(n-1)(n-2)$. So $D = 0$ only when $n=1$ or $n=2$.
Q
For which die outcomes does the system have a unique solution?
⌄
For $n \in \{3,4,5,6\}$. The values $n=1$ and $n=2$ make $D=0$, so they are excluded.
Q
What is the probability of unique solution?
⌄
$P = 4/6$, since 4 out of 6 die outcomes give unique solution. So $k = 4$.
Q
What is the sum of all valid values of $n$?
⌄
$3+4+5+6 = 18$.
Q
What is the final answer?
⌄
$k + \text{sum of valid } n = 4 + 18 = 22$. Option (A).
Q
Why is $n=1$ not valid?
⌄
At $n=1$, $D = -(1-1)(1-2) = 0$. The matrix becomes singular so unique solution does not exist.
Q
Why is $n=2$ not valid?
⌄
At $n=2$, $D = -(2-1)(2-2) = 0$. Again the determinant is zero, so no unique solution.
Q
What is Cramer’s rule?
⌄
For $AX=B$ with $D=\det(A)\ne0$, each variable is $x_i = D_i/D$ where $D_i$ replaces the $i$-th column with $B$.
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