Question

Given $a,b \in\{0, 1, 2, 3, 4, \dots, 9, 10\}$. Consider the system of equations

$x+y+z=4$ $2x+y+3z = 6$ $x+2y+az = b$

Match each entry in List-I to the correct entries in List-II.

	List-I		List-II
(P)	The number of ordered pairs $(a, b)$ so that the system of equations has unique solution, is equal to	1	10
(Q)	The number of ordered pairs $(a, b)$ so that the system of equations has no solution, is equal to	2	1
(R)	The number of ordered pairs $(a, b)$ so that the system of equations has infinitely many solutions, is equal to	3	110
(S)	If ordered pairs $(a, b)$ is such that so that the system of equations has infinite solutions, then the value of $(a + b)$ is equal to	4	6
		5	15

Answer 1

P → 3; Q → 1; R → 2; S → 4

Answer 2

The given system of linear equations is:

1. $x+y+z=4$
2. $2x+y+3z=6$
3. $x+2y+az=b$

The parameters $a, b \in \{0, 1, 2, \dots, 10\}$. There are 11 possible values for $a$ and 11 possible values for $b$.

We can analyze the system using the determinant of the coefficient matrix or by Gaussian elimination. Let's use Gaussian elimination. The augmented matrix for the system is: $M = \begin{pmatrix} 1 & 1 & 1 & | & 4 \\ 2 & 1 & 3 & | & 6 \\ 1 & 2 & a & | & b \end{pmatrix}$

Perform row operations to simplify the matrix: $R_2 \to R_2 - 2R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 4 \\ 0 & -1 & 1 & | & -2 \\ 1 & 2 & a & | & b \end{pmatrix}$

$R_3 \to R_3 - R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 4 \\ 0 & -1 & 1 & | & -2 \\ 0 & 1 & a-1 & | & b-4 \end{pmatrix}$

$R_2 \to -R_2$: $\begin{pmatrix} 1 & 1 & 1 & | & 4 \\ 0 & 1 & -1 & | & 2 \\ 0 & 1 & a-1 & | & b-4 \end{pmatrix}$

$R_3 \to R_3 - R_2$: $\begin{pmatrix} 1 & 1 & 1 & | & 4 \\ 0 & 1 & -1 & | & 2 \\ 0 & 0 & (a-1) - (-1) & | & (b-4) - 2 \end{pmatrix}$ $\begin{pmatrix} 1 & 1 & 1 & | & 4 \\ 0 & 1 & -1 & | & 2 \\ 0 & 0 & a & | & b-6 \end{pmatrix}$

Now we analyze the system based on the last row, which corresponds to the equation $az = b-6$.

1. Unique Solution (P): A system has a unique solution if the determinant of the coefficient matrix is non-zero. From the row-echelon form, this means the coefficient of $z$ in the last equation must be non-zero. So, $a \neq 0$. Since $a \in \{0, 1, \dots, 10\}$, the possible values for $a$ are $\{1, 2, \dots, 10\}$, which is 10 choices. The value of $b$ does not affect the existence of a unique solution when $a \neq 0$. So, $b \in \{0, 1, \dots, 10\}$, which is 11 choices. The number of ordered pairs $(a, b)$ for which the system has a unique solution is $10 \times 11 = 110$. This matches with List-II entry 3. So, P → 3.

2. No Solution (Q): A system has no solution if it leads to a contradiction. In our simplified form, this occurs when $a=0$ and $b-6 \neq 0$. So, $a=0$ (1 choice). And $b-6 \neq 0 \implies b \neq 6$. Since $b \in \{0, 1, \dots, 10\}$, the possible values for $b$ are $\{0, 1, 2, 3, 4, 5, 7, 8, 9, 10\}$, which is 10 choices. The number of ordered pairs $(a, b)$ for which the system has no solution is $1 \times 10 = 10$. This matches with List-II entry 1. So, Q → 1.

3. Infinitely Many Solutions (R): A system has infinitely many solutions if it is consistent and has at least one free variable. In our simplified form, this occurs when $a=0$ and $b-6=0$. So, $a=0$ (1 choice). And $b-6=0 \implies b=6$ (1 choice). The number of ordered pairs $(a, b)$ for which the system has infinitely many solutions is $1 \times 1 = 1$. This ordered pair is $(0, 6)$. This matches with List-II entry 2. So, R → 2.

4. Value of (a+b) for infinite solutions (S): From the analysis for infinitely many solutions, the only ordered pair $(a, b)$ that yields infinite solutions is $(0, 6)$. For this pair, $a+b = 0+6 = 6$. This matches with List-II entry 4. So, S → 4.

Summary of Matches:

• (P) → 3
• (Q) → 1
• (R) → 2
• (S) → 4

The final answer is $\boxed{P \to 3; Q \to 1; R \to 2; S \to 4}$.