If a matrix $$A$$ is such that $$3{A^3} + 2{A^2} + 5A + I = 0.$$ Then what is $$A^{–1}$$ equal to ?
A.
$$ - \left( {3{A^2} + 2A + 5I} \right)$$
B.
$$3A^2 + 2A + 5I$$
C.
$$3A^2 – 2A – 5I$$
D.
$$ \left( {3{A^2} + 2A - 5I} \right)$$
Answer :
$$ - \left( {3{A^2} + 2A + 5I} \right)$$
Solution :
Let $$A$$ be a matrix such that $$3A^3 + 2A^2 + 5A + I = 0$$
Post multiply by $$A^{-1}$$ on both the sides, we get,
$$\eqalign{
& 3{A^3}{A^{ - 1}} + 2{A^2}{A^{ - 1}} + 5A{A^{ - 1}} + I{A^{ - 1}} = 0 \cr
& \Rightarrow 3{A^2} + 2A + 5I + {A^{ - 1}} = 0 \cr
& \Rightarrow {A^{ - 1}} = - \left( {3{A^2} + 2A + 5I} \right) \cr} $$
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