Question

If \[A = \left[ {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right],\]   then $$A^{16}$$ is equal to :

A. \[\left[ {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}} 0&{1}\\ 1&0 \end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 0&1 \end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}} 1&{ 0}\\ 0&1 \end{array}} \right]\]  
Answer :   \[\left[ {\begin{array}{*{20}{c}} 1&{ 0}\\ 0&1 \end{array}} \right]\]
Solution :
\[\begin{array}{l} {\rm{We \,\,have, }}\,\,A = \left[ {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right]\\ {\rm{Now, }}\,\,{A^2} = A.A = \left( {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 1}&0\\ 0&{ - 1} \end{array}} \right) = - I\\ {\rm{where }}\,\,I = \left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right){\rm{ is\,\, identity\,\, matrix}} \end{array}\]
$$\eqalign{ & {\left( {{A^2}} \right)^8} = {\left( { - I} \right)^8} = I. \cr & {\text{Hence, }}{A^{16}} = I \cr} $$

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 $$C$$
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Releted Question 2

If $$\omega \left( { \ne 1} \right)$$  is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

A. 0
B. 1
C. $$i$$
D. $$\omega $$
View Answer
Releted Question 3

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

A. no solution
B. unique solution
C. infinitely many solutions
D. finitely many solutions
View Answer
Releted Question 4

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

A. $$A + B = B + A$$
B. $$A + B = A - B$$
C. $$A - B = B - A$$
D. $$AB=BA$$
View Answer

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Matrices and Determinants

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AlgebraSequences and SeriesQuadratic EquationComplex NumberMatrices and DeterminantsPermutation and CombinationBinomial TheoremMathematical InductionMathematical Reasoning
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