Using properties of determinants, prove that 111+3x1+3y1111+3z1=9(3xy+xy+yz+zx)

From the question assume, L.H.S. =111+3x1+3y1111+3z1Taking x, y and z common from R1, R2 and R3 respectively, we get=xyz1x1x1x+31y+31y1y1z1z+31zOn applying R1→R1+R+2+R3, we get=xyz1x+1y+1z+31x+1y+1z+31x+1y+1z+31y+31y1y1z1z+31zTaking common 1x+1y+1z+3from R1, we get=xyz1x+1y+1z+31111y+31y1y1z1z+31zOn applying C2→C2−C1and C3→C3−C1, we get=xyz1x+1y+1z+31001y+3-3-31z30On expanding along R1, we get=xyz1x+1y+1z+31(0+9)=xyzyz+xz+xy+3xyzxyz(9)=9(3xyz+xy+yz+zx)=RHSTherefore, LHS=RHS has proven.

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Using properties of determinants, prove that $|\begin{array}{ccc} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{array}| = 9 (3 xy + xy + yz + zx)$

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Detailed Solution

From the question assume,
$L.H.S. = |\begin{array}{ccc} 1 & 1 & 1 + 3 x \\ 1 + 3 y & 1 & 1 \\ 1 & 1 + 3 z & 1 \end{array}|$
Taking x, y and z common from R₁, R₂ and R₃ respectively, we get
$= xyz |\begin{array}{ccc} \frac{1}{x} & \frac{1}{x} & \frac{1}{x} + 3 \\ \frac{1}{y} + 3 & \frac{1}{y} & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} + 3 & \frac{1}{z} \end{array}|$
On applying R₁ $\to$ R₁+R+₂+R₃, we get
$= xyz |\begin{array}{ccc} \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 3 & \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 3 & \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 3 \\ \frac{1}{y} + 3 & \frac{1}{y} & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} + 3 & \frac{1}{z} \end{array}|$
Taking common $(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 3)$ from R₁, we get
$= xyz (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 3) |\begin{array}{ccc} 1 & 1 & 1 \\ \frac{1}{y} + 3 & \frac{1}{y} & \frac{1}{y} \\ \frac{1}{z} & \frac{1}{z} + 3 & \frac{1}{z} \end{array}|$
On applying $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ , we get
$= xyz (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 3) |\begin{array}{ccc} 1 & 0 & 0 \\ \frac{1}{y} + 3 & - 3 & - 3 \\ \frac{1}{z} & 3 & 0 \end{array}|$
On expanding along R₁, we get
$\begin{array}{l} = xyz (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + 3) 1 (0 + 9) \\ = xy z (\frac{yz + xz + xy + 3 xyz}{xyz}) (9) \\ = 9 (3 xyz + xy + yz + zx) \end{array}$
=RHS
Therefore, LHS=RHS has proven.