If $$\left| {{A_{n \times n}}} \right| = 3$$ and $$\left| {adj\,A} \right| = 243,$$ what is the value of $$n \,?$$
A.
4
B.
5
C.
6
D.
7
Answer :
6
Solution :
As given : $$\left| {{A_{n \times n}}} \right| = 3{\text{ and }}\left| {adj\,A} \right| = 243$$
Dterminant of adjoint $$A$$ 13 given by : $$\left| {adj\,A} \right| = {\left| {{A_{n \times n}}} \right|^{n - 1}}$$
$$\eqalign{
& \Rightarrow 243 = {3^{n - 1}} \cr
& \Rightarrow {3^5} = {3^{n - 1}} \cr
& \Rightarrow n - 1 = 5 \cr
& \Rightarrow n = 6 \cr} $$
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