Number of positive integral solutions of equation left| {begin{array}{*{20}{c}}{{x^3} + 1}&amp

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3 and 4 .Determinants and Matrices

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The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right|$ $= 11$ is

$0$

$3$

$6$

$12$

Solution

$\left| {\,\begin{array}{*{20}{c}}{{x^4} + x}&{{x^3}y}&{{x^3}z}\\{x{y^3}}&{{y^4} +y}&{{y^3}z}\\{x{z^3}}&{y{z^3}}&{{z^4} + z}\end{array}\,} \right|$ $= 11$

$\frac{1}{{xyz}}\left| {\,\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^3}}&{{x^3}}\\ {{y^3}}&{{y^3} + 1}&{{y^3}}\\{{z^3}}&{{z^3}}&{{z^3} + 1}\end{array}\,} \right|$ $= 11$

use $R_1\rightarrow R_1 + R_2 + R_3$

$D =$ $(x^3 + y^3 + z^3 + 1)$ $\left| {\,\begin{array}{*{20}{c}}1&1&1\\{{y^3}}&{{y^3} + 1}&{{y^3}}\\{{z^3}}&{{z^3}}&{{z^3} + 1}\end{array}\,} \right|$ $= 11$

hence $x^3 + y^3 + z^3 = 10$

$(2, 1, 1) , (1, 2, 1) , (1, 1, 2)$

Standard 12

Mathematics

Evaluate $\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$

easy

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

$\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + y}&1\\1&1&{1 + z}\end{array}\,} \right| = $

hard

If ${a^2} + {b^2} + {c^2} = – 2$ and $f(x) = \left| {\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}} \right|$ then $f(x)$ is a polynomial of degree

hard

(AIEEE-2005)

The value of $\left| {\,\begin{array}{*{20}{c}}a&{a + b}&{a + 2b}\\{a + 2b}&a&{a + b}\\{a + b}&{a + 2b}&a\end{array}\,} \right|$ is equal to

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