Question

If $l>0, K<0$ are roots of $x^2 + x - 20 = 0$, the system of linear equations $3x - y + 4z = 3$, $x + 2y - 3z = -2$ and $6x + 5y + Pz = -3$ has?

• A. unique solution when $P = K$
• B. infinite number of solutions when $P = l$
• C. no solution when $P = K + l$
• D. unique solution when $P = l$

Hint:

• First, solve the quadratic equation to find the values of $l$ and $K$.
• For a system of linear equations $AX=B$:
  • If $\det(A) \neq 0$, there is a unique solution.
  • If $\det(A) = 0$ and $(\text{adj } A)B = 0$, there are infinitely many solutions.
  • If $\det(A) = 0$ and $(\text{adj } A)B \neq 0$ (i.e., at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero), there is no solution.
• Calculate the determinant of the coefficient matrix and set it to non-zero for a unique solution.
• Substitute the values of $l$ and $K$ into the conditions given in the options.

Answer 1

The Correct Option is D

First, we find the roots of the quadratic equation $x^2 + x - 20 = 0$.
Factoring the quadratic equation: $(x+5)(x-4) = 0$.
The roots are $x = -5$ and $x = 4$.
Given that $l>0$, we have $l = 4$.
Given that $K<0$, we have $K = -5$.

Now, consider the system of linear equations: 
1) $3x - y + 4z = 3$ 2) $x + 2y - 3z = -2$ 3) $6x + 5y + Pz = -3$ 
For the system to have a unique solution, the determinant of the coefficient matrix (let's call it $\Delta$) must be non-zero.
$\Delta = \begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & P \end{vmatrix}$ 
Expanding the determinant: $\Delta = 3(2P - (5)(-3)) - (-1)(P - (6)(-3)) + 4(1 \cdot 5 - 6 \cdot 2)$ $\Delta = 3(2P + 15) + 1(P + 18) + 4(5 - 12)$ $\Delta = 6P + 45 + P + 18 + 4(-7)$ $\Delta = 7P + 63 - 28$ $\Delta = 7P + 35$ 
For a unique solution, $\Delta \neq 0$, which means $7P + 35 \neq 0 \Rightarrow 7P \neq -35 \Rightarrow P \neq -5$.

Let's evaluate the given options: 

 

Option (a) Unique solution when $P = K$.
Since $K = -5$, this means $P = -5$.
If $P = -5$, then $\Delta = 7(-5) + 35 = -35 + 35 = 0$.
So, a unique solution is not possible.
This option is incorrect.
(When $P = K = -5$, $\Delta=0$.

To check for infinite or no solution, we can check $\Delta_x, \Delta_y, \Delta_z$.
It can be shown that for $P=-5$, $\Delta_x = \Delta_y = \Delta_z = 0$, implying infinite solutions.
Thus, "unique solution when $P=K$" is false).
Option (b) Infinite number of solutions when $P = l$.
Since $l = 4$, this means $P = 4$.
If $P = 4$, then $\Delta = 7(4) + 35 = 28 + 35 = 63$.
Since $\Delta \neq 0$, the system has a unique solution, not an infinite number of solutions.
This option is incorrect.

Option (c) No solution when $P = K + l$.
$K+l = -5 + 4 = -1$.
So, $P = -1$.
If $P = -1$, then $\Delta = 7(-1) + 35 = -7 + 35 = 28$.
Since $\Delta \neq 0$, the system has a unique solution, not no solution.
This option is incorrect.

Option (d) Unique solution when $P = l$.
Since $l = 4$, this means $P = 4$.
If $P = 4$, then $\Delta = 7(4) + 35 = 28 + 35 = 63$.
Since $\Delta \neq 0$, the system has a unique solution.
This option is correct.
\[ \boxed{\text{unique solution when } P = l} \]