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Solving Simultaneous Linear Equation Using Cramer's Rule

If sinβ≠cos...

Question

If $sin β \neq cos β$ and $x, y$ and $z$ satisfy the equations
$x cos α - y sin α + z = cos β + 1$
$x sin α + y cos α + z = - sin β + 1$
$x cos (α + β) - y sin (α + β) + z = 2$ ,
then $x^{2} + y^{2} + z^{2}$ equals

A

1

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B

2

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C

3

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D

5

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Solution

The correct option is B $2$
The determinant of the coefficient of the given set of equation is
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} cos α & - sin α & 1 sin α & cos α & 1 cos (α + β) & - sin (α + β) & 1 \end{matrix} ∣ ∣ ∣ ∣$
Applying $R_{3} \to R_{3} - cos β R_{1} + sin β R_{2}$
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} cos α & - sin α & 1 sin α & cos α & 1 0 & 0 & 1 + sin β - cos β \end{matrix} ∣ ∣ ∣ ∣$

$\Rightarrow Δ = (1 + sin β - cos β) ∣ ∣ ∣ \begin{matrix} cos α & - sin α sin α & cos α \end{matrix} ∣ ∣ ∣$

$\Rightarrow Δ = (1 + sin β - cos β) ({cos}^{2} α + {sin}^{2} α)$

$\Rightarrow Δ = (1 + sin β - cos β)$
To solve the equation by cramer's rule
Let
$Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos β + 1 & - sin α & 1 - sin β + 1 & cos α & 1 2 & - sin (α + β) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $C_{1} \to C_{1} - C_{3}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos β & - sin α & 1 - sin β & cos α & 1 1 & - sin (α + β) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
Applying $R_{3} \to R_{3} - cos β R_{1} + sin β R_{2}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos β & - sin α & 1 - sin β & cos α & 1 1 - {cos}^{2} β - {sin}^{2} β & 0 & 1 - cos β + sin β \end{matrix} ∣ ∣ ∣ ∣ ∣ Δ_{1} = (1 + sin β - cos β) cos (α + β)$
Similarly,
$Δ_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos α & cos β + 1 & 1 sin α & - sin β + 1 & 1 cos (α + β) & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$Δ_{2} = - (1 + sin β - cos β) sin (α + β)$
And
$Δ_{3} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos α & - sin α & cos β + 1 sin α & cos α & - sin β + 1 cos (α + β) & - sin (α + β) & 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ Δ_{3} = 1 + sin β - cos β$
Therefore from cramer's rule
$x = \frac{Δ_{1}}{Δ} = cos (α + β), y = \frac{Δ_{2}}{Δ} = - sin (α + β), z = \frac{Δ_{3}}{Δ} = 1$
Hence,
$x^{2} + y^{2} + z^{2} = {cos}^{2} (α + β) + {sin}^{2} (α + β) + 1 = 2$
Hence, option B.

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