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Question: \[\left| \begin{matrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix} \right| =\]...

1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{matrix} \right| =$$
A

xyz(1+1x+1y+1z)xyz\left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)

B

xyzxyz

C

1+1x+1y+1z1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}

D

1x+1y+1z\frac{1}{x} + \frac{1}{y} + \frac{1}{z}

Answer

xyz(1+1x+1y+1z)xyz\left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)

Explanation

Solution

Δ=xyz∣1+1x1x1x1y1+1y1y1z1z1+1z∣\Delta = xyz\left| \begin{matrix} 1 + \frac{1}{x} & \frac{1}{x} & \frac{1}{x} \\ \frac{1}{y} & 1 + \frac{1}{y} & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & 1 + \frac{1}{z} \end{matrix} \right| =xyz(1+1x+1y+1z)= x y z \left( 1 + \frac { 1 } { x } + \frac { 1 } { y } + \frac { 1 } { z } \right) ∣1111y1+1y1y1z1z1+1z∣\left| \begin{matrix} 1 & 1 & 1 \\ \frac{1}{y} & 1 + \frac{1}{y} & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} & 1 + \frac{1}{z} \end{matrix} \right| (byR1→R1+R2+R3)(byR_{1} \rightarrow R_{1} + R_{2} + R_{3})

=xyz(1+1x+1y+1z)= x y z \left( 1 + \frac { 1 } { x } + \frac { 1 } { y } + \frac { 1 } { z } \right) ∣1001y101z01∣(byC2→C2−C1andC3→C3−C1)\left| \begin{matrix} 1 & 0 & 0 \\ \frac{1}{y} & 1 & 0 \\ \frac{1}{z} & 0 & 1 \end{matrix} \right|(byC_{2} \rightarrow C_{2} - C_{1}\text{and}C_{3} \rightarrow C_{3} - C_{1})

=xyz(1+1x+1y+1z)= x y z \left( 1 + \frac { 1 } { x } + \frac { 1 } { y } + \frac { 1 } { z } \right) ∣1001∣\left| \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right| =xyz(1+1x+1y+1z)= xyz\left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)

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

2 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 4 \end{matrix} \right| = 2(11) - 1(3) + 1(1 - 3) = 17$$ option (1) gives $1 \times 2 \times 3\left( 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \right) = 17$