If $$\left| A \right| = 8,$$ where $$A$$ is square matrix of order 3, then what is $$\left| {adj\,A} \right|$$ equal to ?
A.
16
B.
24
C.
64
D.
512
Answer :
64
Solution :
Let $$\left| A \right| = 8$$ and $$A$$ is a square matrix of order 3.
We know that $$\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}.I$$ where $$'n'$$ is the order of the matrix $$A.$$
$$\therefore \left| {adj\,A} \right| = {8^{3 - 1}} = {8^2} = 64$$
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