Question

Prove that
$\sf \left[\begin{array}{ccc}\sf a+x&\sf b&\sf c\\\sf a&\sf b+y&\sf c\\\sf a&\sf b&\sf c+z\end{array}\right] = xyz\bigg(1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z}\bigg)$
Note
No spam or copied answer

Answer 1

Appropriate Question :-

Prove that

$\rm \:\: \begin{gathered}\sf \left | \begin{array}{ccc}a + x&b& c\\a&b + y& c\\a& b&c + z\end{array}\right | \end{gathered} = xyz\bigg(1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z} \bigg)$

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\:\rm \:\: \begin{gathered}\sf \left | \begin{array}{ccc}a + x&b& c\\a&b + y& c\\a& b&c + z\end{array}\right | \end{gathered}$

Take x common from Column 1, y from Column 2 and z from Column 3, we get

$\rm \: = \:\:xyz \begin{gathered}\sf \left | \begin{array}{ccc}\dfrac{a}{x} + 1&\dfrac{b}{y}& \dfrac{c}{z} \\ \\\dfrac{a}{x}&\dfrac{b}{y} + 1& \dfrac{c}{z}\\ \\ \dfrac{a}{x}& \dfrac{b}{y}&\dfrac{c}{z} + 1\end{array}\right | \end{gathered}$

$\rm :\longmapsto\:OP \: C_1 \: \to \: C_1 + C_2 + C_3$

$\rm \: = \:\:xyz \begin{gathered}\sf \left | \begin{array}{ccc}1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z} &\dfrac{b}{y}& \dfrac{c}{z} \\ \\1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z}&\dfrac{b}{y} + 1& \dfrac{c}{z}\\ \\1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z}& \dfrac{b}{y}&\dfrac{c}{z} + 1\end{array}\right | \end{gathered}$

$\red{ \boxed{ \rm{ \: Take \: out \: 1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z} \: common \: from \: C_1}}}$

$\rm\:=xyz\bigg(1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z} \bigg) \begin{gathered}\sf \left | \begin{array}{ccc}1 &\dfrac{b}{y}& \dfrac{c}{z} \\ \\1&\dfrac{b}{y} + 1& \dfrac{c}{z}\\ \\1 & \dfrac{b}{y}&\dfrac{c}{z} + 1\end{array}\right | \end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \rm{ \: OP \: R_2 \: \to \: R_2 - R_1}}}$

$\rm\:=xyz\bigg(1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z} \bigg) \begin{gathered}\sf \left | \begin{array}{ccc}1 &\dfrac{b}{y}& \dfrac{c}{z} \\ \\0&1&0\\ \\1 & \dfrac{b}{y}&\dfrac{c}{z} + 1\end{array}\right | \end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{ \rm{ \: OP \: R_3 \: \to \: R_3 - R_1}}}$

$\rm\:=xyz\bigg(1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z} \bigg) \begin{gathered}\sf \left | \begin{array}{ccc}1 &\dfrac{b}{y}& \dfrac{c}{z} \\ \\0&1&0\\ \\0 &0&1\end{array}\right | \end{gathered}$

Expand along Column 1, we get

$\rm\:=xyz\bigg(1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z} \bigg)(1 - 0)$

$\rm\:=xyz\bigg(1 + \dfrac{a}{x} + \dfrac{b}{y} + \dfrac{c}{z} \bigg)$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

1. The value of determinant is 0, if every element of row or column is 0.

2. The value of determinant is 0, if any two rows or columns are identical.

3. The value of determinant is multiplied by (- 1) if successive rows or columns are interchanged.

4. The determinant value remains same if rows and columns are interchanged.

Answer 2

Bruhhhh

DC aaja ek baar...

insta hi aaja

@virhana_vcc_17