If \[\left| {\begin{array}{*{20}{c}}
{{x^n}}&{{x^{n + 2}}}&{{x^{2n}}}\\
1&{{x^a}}&a\\
{{x^{n + 5}}}&{{x^{a + 6}}}&{{x^{2n + 5}}}
\end{array}} \right| = 0\,\,\forall \,\,x \in R,\] where $$n \in N$$ then value of $$'a'$$ is
A.
$$n$$
B.
$$n - 1$$
C.
$$n + 1$$
D.
None of these
Answer :
$$n + 1$$
Solution :
Taking $$x^5$$ common from last row, we get
\[{x^5}\left| {\begin{array}{*{20}{c}}
{{x^n}}&{{x^{n + 2}}}&{{x^{2n}}}\\
1&{{x^a}}&a\\
{{x^n}}&{{x^{a + 1}}}&{{x^{2n}}}
\end{array}} \right| = 0\,\,\forall \,\,x \in R,\]
⇒ $$a + 1 = n + 2$$
⇒ $$a = n + 1$$
(as it will make first and third row identical)
Releted MCQ Question on Algebra >> Matrices and Determinants
Releted Question 1
Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then
A.
$$C$$ is empty
B.
$$B$$ has as many elements as $$C$$
C.
$$A = B \cup C$$
D.
$$B$$ has twice as many elements as elements as $$C$$
Let $$a, b, c$$ be the real numbers. Then following system of equations in $$x, y$$ and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ has