If the matrix $$B$$ is the adjoint of the square matrix $$A$$ and $$\alpha $$ is the value of the determinant of $$A,$$ then what is $$AB$$ equal to ?
A.
$$\alpha $$
B.
$$\left( {\frac{1}{\alpha }} \right)I$$
C.
$$I$$
D.
$$\alpha I$$
Answer :
$$\alpha I$$
Solution :
Since, adjoint of the square matrix $$A$$ is $$B$$
$$\eqalign{
& \Rightarrow \,\frac{B}{{\left| A \right|}} = {A^{ - 1}} \cr
& \Rightarrow \,\frac{{AB}}{{\left| A \right|}} = A{A^{ - 1}} = I \cr
& \Rightarrow \,AB = \left| A \right|I \cr
& \Rightarrow \,AB = \alpha I \cr} $$
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