2021-M2-08 – Solving Master

Ans: (a)

d = 3

{(3 - 32 t, 13 t - 1, t) | t \in R}

(b) No

For $(E)$ has infinitely many solutions, we have
$\begin{array}{rcl} | \begin{array}{c} 1 & d - 1 & d + 3 \\ 2 & d + 2 & - 1 \\ 3 & d + 4 & 5 \end{array} | & = & 0 \\ (1) | \begin{array}{c} d + 2 & - 1 \\ d + 4 & 5 \end{array} | - (2) | \begin{array}{c} d - 1 & d + 3 \\ d + 4 & 5 \end{array} | + (3) | \begin{array}{c} d - 1 & d + 3 \\ d + 2 & - 1 \end{array} | & = & 0 \\ 5 (d + 2) + (d + 4) - 2 [5 (d - 1) - (d + 3) (d + 4)] + 3 [- (d - 1) - (d + 3) (d + 2)] & = & 0 \\ 5 d + 10 + d + 4 - 10 d + 10 + 2 d^{2} + 14 d + 24 - 3 d + 3 - 3 d^{2} - 15 d - 18 & = & 0 \\ - d^{2} - 8 d + 33 & = & 0 \\ (d - 3) (d + 11) & = & 0 \end{array}$

$∴ d = 3$ or $d = - 11$ .

For $d = 3$ , we have

$\begin{array}{l} (\begin{array}{cccc} 1 & 2 & 6 & 1 \\ 2 & 5 & - 1 & 1 \\ 3 & 7 & 5 & 2 \end{array}) \\ \to_{R_{3} - 3 R_{1} \to R_{3}}^{R_{2} - 2 R_{1} \to R_{2}} (\begin{array}{cccc} 1 & 2 & 6 & 1 \\ 0 & 1 & - 13 & - 1 \\ 0 & 1 & - 13 & - 1 \end{array}) \\ \overset{R_{3} - R_{2} \to R_{3}}{\to} (\begin{array}{cccc} 1 & 2 & 6 & 1 \\ 0 & 1 & - 13 & - 1 \\ 0 & 0 & 0 & 0 \end{array}) \end{array}$

Let $z = t$ , where $t \in R$ .

$\begin{array}{rcl} y - 13 t & = & - 1 \\ y & = & 13 t - 1 \end{array}$

$\begin{array}{rcl} x + 2 (13 t - 1) + 6 t & = & 1 \\ x & = & 3 - 32 t \end{array}$

Therefore, the solution of $(E)$ is ${(3 - 32 t, 13 t - 1, t) | t \in R}$ .

For $d = - 11$ , we have

$\begin{array}{l} (\begin{array}{cccc} 1 & - 12 & - 8 & 15 \\ 2 & - 9 & - 1 & - 27 \\ 3 & - 7 & 5 & 2 \end{array}) \\ \to_{R_{3} - 3 R_{1} \to R_{3}}^{R_{2} - 2 R_{1} \to R_{2}} (\begin{array}{cccc} 1 & - 12 & - 8 & 15 \\ 0 & 15 & 15 & - 57 \\ 0 & 29 & 29 & - 43 \end{array}) \\ \to_{R_{3} - R_{2} \times \frac{29}{15} \to R_{3}}^{R_{2} \div 15 \to R_{2}} (\begin{array}{cccc} 1 & - 12 & - 8 & 15 \\ 0 & 1 & 1 & \frac{- 57}{15} \\ 0 & 0 & 0 & \frac{336}{5} \end{array}) \end{array}$

Therefore, $(E)$ has no solution if $d = - 11$ . Hence, $d \neq - 11$ .
By using the result of (a), sub. $x = 3 - 32 t$ , $y = 13 t - 1$ and $z = t$ into $x y + 2 x z = 3$ , we have
$\begin{array}{rcl} (3 - 32 t) (13 t - 1) + 2 (3 - 32 t) (t) & = & 3 \\ 39 t - 3 - 416 t^{2} + 32 t + 6 t - 64 t^{2} - 3 & = & 0 \\ 480 t^{2} - 77 t + 6 & = & 0 \\ t & = & \frac{- (- 77) \pm \sqrt{(- 77)^{2} - 4 (480) (6)}}{2 (480)} \\ t & = & \frac{77 \pm \sqrt{- 5591}}{960}, which is not a real number. \end{array}$

Hence, $(E)$ does not have a real solution $(x, y, z)$ satisfying $x y + 2 x z = 3$ .

Therefore, the claim is not correct.

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