Let the system of equations , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy , is : [2025]

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Solution

Q.
Let the system of equations

$x + 5 y - z = 1$

$4 x + 3 y - 3 z = 7$

$24 x + y + λ z = μ$

$λ, μ \in R$ , have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy $7 \leq x + y + z \leq 77$ , is : [2025]

1 3

2 5

3 6

4 4

Ans.
(1)

For infinitely many solutions, we have D = 0

$\Rightarrow$ $| \begin{matrix} 1 & 5 & - 1 \\ 4 & 3 & - 3 \\ 24 & 1 & λ \end{matrix} |$ $= 0$

$\Rightarrow 1 (3 λ + 3) - 5 (4 λ + 72) - 1 (4 - 72) = 0$

$\Rightarrow 17 λ = - 289 \Rightarrow λ = - 17$ ... (i)

Now, $D_{1} = 0$

$\Rightarrow$ $| \begin{matrix} 1 & 5 & - 1 \\ 7 & 3 & - 3 \\ μ & 1 & - 17 \end{matrix} |$ $= 0$

$\Rightarrow 1 (- 51 + 3) - 5 (- 119 + 3 μ) - 1 (7 - 3 μ) = 0$

$\Rightarrow - 48 + 595 - 15 μ - 7 + 3 μ = 0$

$\Rightarrow 12 μ = 540$

$\Rightarrow μ = 45$ ... (ii)

Using (i) and (ii), we get

x + 5y – z = 1, 4x + 3y – 3z =7, 24x + y –17z = 45

$\Rightarrow z = x + 5 y - 1$

$∴ 4 x + 3 y - 3 x - 15 y + 3 = 7$

$\Rightarrow x - 12 y = 4$

$\Rightarrow x = 4 + 12 y and z = 4 + 12 y + 5 y - 1 = 3 + 17 y$

$∴ (x, y, z) = (4 + 12 k, k, 3 + 17 k)$ ( $∵$ Assume y = k)

Also, $7 \leq 7 + 30 k \leq 77$

$\Rightarrow 0 \leq 30 k < 70$

$\Rightarrow 0 \leq k \leq 2.3$

$\Rightarrow$ k = 0, 1, 2

Thus, there are three possible solutions.

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