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MARCH 2024 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMARCH 20243 Marks
Question
Using the cofactor of elements of third column.
evaluate $\Delta=\left[\begin{array}{ccc}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right]$.
✓

Answer

$\Delta=\left|\begin{array}{lll}
1 & x & y z \\
1 & y & z x \\
1 & z & x y
\end{array}\right|$
Co-factor of the element $y z \quad A_{13}=(-1)^4\left|\begin{array}{ll}1 & y \\ 1 & z\end{array}\right|$
$\begin{array}{l}=(1)(z-y) \\ =(z-y)\end{array}$
Co-factor of the element $z x \quad A_{23}=(-1)^5\left|\begin{array}{ll}1 & x \\ 1 & z\end{array}\right|$
$\begin{array}{l}=(-1)(z-x) \\ =x-z\end{array}$
Co-factor of the element $x y \quad A_{33}=(-1)^6\left|\begin{array}{ll}1 & x \\ 1 & y\end{array}\right|$
$\begin{array}{l}=(1)(y-x) \\ =y-x\end{array}$
$\begin{aligned} \Delta & =a_{13} A_{13}+a_{23} A_{23}+a_{33} A_{33} \\ & =(y z)(z-y)+(z x)(x-z)+(x y)(y-x) \\ & =y z^2-y^2 z+z x^2-z^2 x+x y^2-x^2 y \\ & =z\left(x^2-y^2\right)+z^2(y-x)+x y(y-x) \\ & =z[(x-y)(x+y)]+z^2(y-x)+x y(y-x) \\ & =(y-x)\left(-z(x+y)+z^2+x y\right) \\ & =(y-x)\left(-z x-y z+z^2+x y\right) \\ & =(y-x)(z(z-x)-y(z-x)) \\ & =(y-x)(z-x)(z-y) \\ & =(x-y)(y-z)(z-x)\end{aligned}$

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