Left| {,begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&amp

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3 and 4 .Determinants and Matrices

medium

$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right| = $

$4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$

$3\,\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$

$2\,\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$

None of these

Solution

(a) Apply ${R_2} – {R_3}$ and note that ${(x + y)^2} – {(x – y)^2} = 4xy$

$\therefore $ $\Delta = 4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\{{{(a – 1)}^2}}&{{{(b – 1)}^2}}&{{{(c – 1)}^2}}\end{array}\,} \right|$

= $4\,\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}\,} \right|$

{Applying ${R_3} – ({R_1} – 2{R_2}) $}

Standard 12

Mathematics

If $a, b, c$ are all different and $\left| {\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4}\, – \,1}\\b&{{b^3}}&{{b^4}\, – \,1}\\c&{{c^3}}&{{c^4}\, – \,1}\end{array}} \right|$ $= 0$ , then :

normal

If the system of equations $ax + y + z = 0$, $x + by + z = 0$ and $x + y + cz = 0 $, where $a,b,c \ne 1,$ has a non trivial solution, then the value of $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}}$is

medium

The determinant $\left| {\begin{array}{*{20}{c}}{{b_1}\, + \,\,{c_1}}&{{c_1}\, + \,\,{a_1}}&{{a_1}\, + \,\,{b_1}}\\{{b_2}\, + \,\,{c_2}}&{{c_2}\, + \,\,{a_2}}&{{a_2}\, + \,\,{b_2}}\\{{b_3}\, + \,\,{c_3}}&{{c_3}\, + \,\,{a_3}}&{{a_3}\, + \,\,{b_3}} \end{array}} \right|$ $=$

medium

If $x, y, z$ are different and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0,$ then show that $1+x y z=0$.

hard

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

$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right| = $

Solution

Similar Questions

If $a, b, c$ are all different and $\left| {\begin{array}{*{20}{c}}a&{{a^3}}&{{a^4}\, – \,1}\\b&{{b^3}}&{{b^4}\, – \,1}\\c&{{c^3}}&{{c^4}\, – \,1}\end{array}} \right|$ $= 0$ , then :

If the system of equations $ax + y + z = 0$, $x + by + z = 0$ and $x + y + cz = 0 $, where $a,b,c \ne 1,$ has a non trivial solution, then the value of $\frac{1}{{1 – a}} + \frac{1}{{1 – b}} + \frac{1}{{1 – c}}$is

The determinant $\left| {\begin{array}{*{20}{c}}{{b_1}\, + \,\,{c_1}}&{{c_1}\, + \,\,{a_1}}&{{a_1}\, + \,\,{b_1}}\\{{b_2}\, + \,\,{c_2}}&{{c_2}\, + \,\,{a_2}}&{{a_2}\, + \,\,{b_2}}\\{{b_3}\, + \,\,{c_3}}&{{c_3}\, + \,\,{a_3}}&{{a_3}\, + \,\,{b_3}} \end{array}} \right|$ $=$

If $x, y, z$ are different and $\Delta=\left|\begin{array}{lll}x & x^{2} & 1+x^{2} \\ y & y^{2} & 1+y^{2} \\ z & z^{2} & 1+z^{2}\end{array}\right|=0,$ then show that $1+x y z=0$.

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

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Show that

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