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Using the properties of determinants in evaluate:
\left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|

Answers (1)

Let \: \: A=\left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|

Apply -  C_{1} \rightarrow C_{1} + C_{2} + C_{3}

\\\begin{aligned} &=\left|\begin{array}{ccc} 3 x-x+y-x+z & -x+y & -x+z \\ x-y+3 y+z-y & 3 y & z-y \\ x-z+y-z+3 z & y-z & 3 z \end{array}\right|\\ &=\left|\begin{array}{ccc} x+y+z & -x+y & -x+z \\ x+y+z & 3 y & z-y \\ x+y+z & y-z & 3 z \end{array}\right|\\ &\text { Take }(x+y+z) \text { common from } C_{1}\\ &=(x+y+z)\left|\begin{array}{ccc} 1 & -x+y & -x+z \\ 1 & 3 y & z-y \\ 1 & y-z & 3 z \end{array}\right| \end{aligned}

\\\begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1,} \text { you will get }\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 1-1 & 3 \mathrm{y}-(-\mathrm{x}+\mathrm{y}) & \mathrm{z}-\mathrm{y}-(-\mathrm{x}+\mathrm{z}) \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\\end{aligned}

\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & 3 \mathrm{y}+\mathrm{x}-\mathrm{y} & \mathrm{z}-\mathrm{y}+\mathrm{x}-\mathrm{z} \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 1 & \mathrm{y}-\mathrm{z} & 3 \mathrm{z} \end{array}\right|\\ &\text { Now apply, } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 1-1 & \mathrm{y}-\mathrm{z}-(-\mathrm{x}+\mathrm{y}) & 3 \mathrm{z}-(-\mathrm{x}+\mathrm{z}) \end{array}\right| \end{aligned}

 

\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{y}-\mathrm{z}+\mathrm{x}-\mathrm{y} & 3 \mathrm{z}+\mathrm{x}-\mathrm{z} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z} & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\ &\text { Apply, } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3},\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y}-(-\mathrm{x}+\mathrm{z}) & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y}-(\mathrm{x}-\mathrm{y}) & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z}-(2 \mathrm{z}+\mathrm{x}) & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\\end{aligned}

\\\\\begin{aligned} &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & -\mathrm{x}+\mathrm{y}+\mathrm{x}-\mathrm{z} & -\mathrm{x}+\mathrm{z} \\ 0 & \mathrm{x}+2 \mathrm{y}-\mathrm{x}+\mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & \mathrm{x}-\mathrm{z}-2 \mathrm{z}-\mathrm{x} & 2 \mathrm{z}+\mathrm{x} \end{array}\right|\\ &=(\mathrm{x}+\mathrm{y}+\mathrm{z})\left|\begin{array}{ccc} 1 & \mathrm{y}-\mathrm{z} & -\mathrm{x}+\mathrm{z} \\ 0 & 3 \mathrm{y} & \mathrm{x}-\mathrm{y} \\ 0 & -3 \mathrm{z} & 2 \mathrm{z}+\mathrm{x} \end{array}\right| \end{aligned}

Now, expand the determinant along Column 1

 

\begin{align*} &= (x + y + z) [1 \times \{ (3y)(2z + x) - (-3z)(x - y) \}] \\ &= (x + y + z) [6yz + 3yx + (3z)(x - y)] \\ &= (x + y + z) [6yz + 3yx + 3zx - 3zy] \\ &= (x + y + z) [3yz + 3zx + 3yx] \\ &= 3(x + y + z)(yz + zx + yx) \end{align*}

 

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