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प्रश्न
Using properties of determinants, prove that \[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\] .
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उत्तर
Solving L.H.S., we get : \[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix}\] \[C_1 \to C_1 + C_2 + C_3\]
= \[\begin{vmatrix}a + x + y + z & y & z \\ a + x + y + z & a + y & z \\ a + x + y + z & y & a + z\end{vmatrix}\]
Taking common a+x+y+z from C1, we get:
(a+x+y+z)
\[\begin{vmatrix}1 & y & z \\ 1 & a + y & z \\ 1 & y & a + z\end{vmatrix}\]
\[R_1 \to R_1 - R_2 , R_2 \to R_2 - R_3\]
=(a+x+y+z)
\[\begin{vmatrix}0 & - a & 0 \\ 0 & a & - a \\ 1 & y & a + z\end{vmatrix}\]
Expanding along C1, we get:
(a+x+y+z)
\[\left\{ 1 \times 1 \times \left( a^2 - 0 \right) \right\}\]
= \[a^2 \left( a + x + y + z \right)\]
L.H.S=R.H.S
Hence proved.
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