Value of $left| {begin{array}{*{20}{c}} {{{(b + c)}^2}}&{{a^2}}&{{a^2}}

English

Gujarati

Hindi

3 and 4 .Determinants and Matrices

normal

Value of $\left| {\begin{array}{*{20}{c}}
{{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\
{{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\
{{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|$ is equal to

$2abc(a + b + c)$

$2abc(a + b + c)^2$

$2abc(a + b + c)^3$

$abc$

Solution

Method- 1

Applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} ; \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$ then we get

$\Delta=(a+b+c)^{2}\left|\begin{array}{ccc}{(b+c)^{2}} & {a-(b+c)} & {a-(b+c)} \\ {b^{2}} & {(a+c)-b} & {0} \\ {c^{2}} & {0} & {(a+b)-c}\end{array}\right|$

Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\left(\mathrm{R}_{2}+\mathrm{R}_{3}\right)$

$\Delta=(a+b+c)^{2}\left|\begin{array}{ccc}{2 b c} & {-2 c} & {-2 b} \\ {b^{2}} & {a+c-b} & {0} \\ {c^{2}} & {0} & {a+b-c}\end{array}\right|$

Now applying $\mathrm{C}_{2} \rightarrow\left(\mathrm{C}_{2}+\frac{1}{\mathrm{b}} \mathrm{C}_{1}\right)$ and

${{{\rm{C}}_3} \to \left( {{{\rm{C}}_3} + \frac{1}{{\rm{c}}}{{\rm{C}}_1}} \right);{\rm{ we get }}}$

$\Delta = {\left( {a + b + c} \right)^2}\left| {\begin{array}{*{20}{c}}
{2bc}&0&0\\
{{b^2}}&{a + c}&{\frac{{{b^2}}}{c}}\\
{{c^2}}&{\frac{{{c^2}}}{b}}&{a + b}
\end{array}} \right|$

${ \Rightarrow {\rm{ So }}\Delta = 2abc{{(a + b + c)}^3}}$

Method-2 $\quad$ Put $a=1, b=1$ and $c=1$ and check from option

Standard 12

Mathematics

If $\omega $is a cube root of unity, then $\left| {\,\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\end{array}\,} \right| = $

easy

Prove that

$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$

medium

If the determinant $\left| {\begin{array}{*{20}{c}}{a\, + \,p}&{1\, + \,x}&{u\, + \,f}\\ {b\, + \,q}&{m\, + \,y}&{v\, + \,g}\\{c\, + \,r}&{n\, + \,z}&{w\, + \,h}\end{array}} \right|$ splits into exactly $K$ determinants of order $3$, each element of which contains only one term, then the value of $K$, is

normal

Without expanding the determinant, prove that

$\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

medium

Show that

$\Delta=\left|\begin{array}{ccc}
(y+z)^{2} & x y & z x \\
x y & (x+z)^{2} & y z \\
x z & y z & (x+y)^{2}
\end{array}\right|=2 x y z(x+y+z)^{3}$

hard

Value of $\left| {\begin{array}{*{20}{c}} {{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\ {{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\ {{c^2}}&{{c^2}}&{{{(a + b)}^2}} \end{array}} \right|$ is equal to

Solution

Similar Questions

If $\omega $is a cube root of unity, then $\left| {\,\begin{array}{*{20}{c}}{x + 1}&\omega &{{\omega ^2}}\\\omega &{x + {\omega ^2}}&1\\{{\omega ^2}}&1&{x + \omega }\end{array}\,} \right| = $

Prove that $\Delta=\left|\begin{array}{ccc} a+b x & c+d x & p+q x \\ a x+b & c x+d & p x+q \\ u & v & w \end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll} a & c & p \\ b & d & q \\ u & v & m \end{array}\right|$

Without expanding the determinant, prove that $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

Show that $\Delta=\left|\begin{array}{ccc} (y+z)^{2} & x y & z x \\ x y & (x+z)^{2} & y z \\ x z & y z & (x+y)^{2} \end{array}\right|=2 x y z(x+y+z)^{3}$

Start a Free Trial Now

Value of $\left| {\begin{array}{*{20}{c}}
{{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\
{{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\
{{c^2}}&{{c^2}}&{{{(a + b)}^2}}
\end{array}} \right|$ is equal to

Prove that

$\Delta=\left|\begin{array}{ccc}
a+b x & c+d x & p+q x \\
a x+b & c x+d & p x+q \\
u & v & w
\end{array}\right|=\left(1-x^{2}\right)\left|\begin{array}{lll}
a & c & p \\
b & d & q \\
u & v & m
\end{array}\right|$

Without expanding the determinant, prove that

$\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

Show that

$\Delta=\left|\begin{array}{ccc}
(y+z)^{2} & x y & z x \\
x y & (x+z)^{2} & y z \\
x z & y z & (x+y)^{2}
\end{array}\right|=2 x y z(x+y+z)^{3}$