Solveeit Logo
Login

Question

Question: If \(\left| \begin{matrix} y + z & x & y \\ z + x & z & x \\ x + y & y & z \end{matrix} \right| = k(...

If ∣y+zxyz+xzxx+yyz∣=k(x+y+z)(x−z)2\left| \begin{matrix} y + z & x & y \\ z + x & z & x \\ x + y & y & z \end{matrix} \right| = k(x + y + z)(x - z)^{2}, then k=k =

A

2xyz2xyz

B

1

C

xyzxyz

D

x2y2z2x^{2}y^{2}z^{2}

Answer

1

Explanation

Solution

∣y+zxyz+xzxx+yyz∣\left| \begin{matrix} y + z & x & y \\ z + x & z & x \\ x + y & y & z \end{matrix} \right|= (x+y+z)∣211z+xzxx+yyz∣(x + y + z)\left| \begin{matrix} 2 & 1 & 1 \\ z + x & z & x \\ x + y & y & z \end{matrix} \right|

by R1→R1+R2+R3R_{1} \rightarrow R_{1} + R_{2} + R_{3}

= (x+y+z)∣111xzxxyz∣(x + y + z)\left| \begin{matrix} 1 & 1 & 1 \\ x & z & x \\ x & y & z \end{matrix} \right| ; by C1→C1−C2C_{1} \rightarrow C_{1} - C_{2}

= (x+y+z).{(z2−xy)−(xz−x2)+(xy−xz)}(x + y + z).\{(z^{2} - xy) - (xz - x^{2}) + (xy - xz)\}

= (x+y+z)(x−z)2(x + y + z)(x - z)^{2} ⇒\Rightarrow k=1k = 1.

Trick : Put x=1,y=2x = 1,y = 2, z=3z = 3, then

\mathbf{5} & \mathbf{1} & \mathbf{2} \\ \mathbf{4} & \mathbf{3} & \mathbf{1} \\ \mathbf{3} & \mathbf{2} & \mathbf{3} \end{matrix} \right|\mathbf{= 5(7)}\mathbf{-}\mathbf{1(12}\mathbf{-}\mathbf{3) + 2(8}\mathbf{-}\mathbf{9)}$$ **=** $\mathbf{35 - 9 - 2 = 24}$& $(x + y + z)(x - z)^{2} = (6)( - 2)^{2} = 24$ $\mathbf{\therefore}\mathbf{k =}\frac{\mathbf{24}}{\mathbf{24}}\mathbf{= 1}$.