Question

The value of \[\left| {\begin{array}{*{20}{c}} {{i^m}}&{{i^{m + 1}}}&{{i^{m + 2}}} \\ {{i^{m + 5}}}&{{i^{m + 4}}}&{{i^{m + 3}}} \\ {{i^{m + 6}}}&{{i^{m + 7}}}&{{i^{m + 8}}} \end{array}} \right|,\]     where $$i = \sqrt { - 1} ,$$   is

A. $$1$$ if $$m$$ is a multiple of $$4$$
B. $$0$$ for all real $$m$$  
C. $$- i$$ if $$m$$ is a multiple of $$3$$
D. None of these
Answer :   $$0$$ for all real $$m$$
Solution :
\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} {{i^m}}&{{i^{m + 1}}}&{{i^{m + 2}}}\\ {{i^{m + 5}}}&{{i^{m + 4}}}&{{i^{m + 3}}}\\ {{i^{m + 6}}}&{{i^{m + 7}}}&{{i^{m + 8}}} \end{array}} \right|\\ = {i^m}{i^{m + 1}}{i^{m + 2}}\left| \begin{array}{l} 1\,\,\,\,\,\,\,1\,\,\,\,\,\,\,1\\ {i^5}\,\,\,\,{i^3}\,\,\,\,i\\ {i^6}\,\,\,\,{i^6}\,\,\,\,{i^6} \end{array} \right|\\ = {i^{3m + 3}}{i^6}\left| \begin{array}{l} 1\,\,\,\,\,1\,\,\,\,\,1\\ {i^5}\,\,\,{i^3}\,\,\,\,i\\ 1\,\,\,\,\,1\,\,\,\,\,1 \end{array} \right|\\ = 0 \end{array}\]
Hence, for all real $$m,$$ the value of given determinant is 0.

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 $$C$$
View Answer
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

Practice More Releted MCQ Question on
Matrices and Determinants

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AlgebraSequences and SeriesQuadratic EquationComplex NumberMatrices and DeterminantsPermutation and CombinationBinomial TheoremMathematical InductionMathematical Reasoning
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GeometryStraight LinesCircleParabolaEllipseHyperbolaLocus3D Geometry and VectorsThree Dimensional Geometry
TrigonometryTrigonometric Ratio and IdentitiesTrignometric EquationsProperties and Solutons of TriangleInverse Trigonometry Function