If \[B = \left[ {\begin{array}{*{20}{c}}
3&4\\
2&3
\end{array}} \right]\] and \[C = \left[ {\begin{array}{*{20}{c}}
3&{ - 4}\\
{ - 2}&3
\end{array}} \right]\] and $$X = BC,$$ find $$X^n$$
A.
$$0$$
B.
$$I$$
C.
$$2I$$
D.
None of these
Answer :
$$I$$
Solution :
\[\begin{array}{l}
X = BC = \left[ {\begin{array}{*{20}{c}}
3&4\\
2&3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
3&{ - 4}\\
{ - 2}&3
\end{array}} \right]\\
\Rightarrow X = BC = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right] = I
\end{array}\]
$${\text{So, }}{X^n} = {I^n} = I$$
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