Question

Suppose \[\Delta = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|\]    and \[\Delta ' = \left| {\begin{array}{*{20}{c}} {{a_1} + p{b_1}}&{{b_1} + q{c_1}}&{{c_1} + r{a_1}}\\ {{a_2} + p{b_2}}&{{b_2} + q{c_2}}&{{c_2} + r{a_2}}\\ {{a_3} + p{b_3}}&{{b_3} + q{c_3}}&{{c_3} + r{a_3}} \end{array}} \right|.\]       Then

A. $$\Delta ' = \Delta $$
B. $$\Delta ' = \Delta \left( {1 - pqr} \right)$$
C. $$\Delta ' = \Delta \left( {1 + p + q + r} \right)$$
D. $$\Delta ' = \Delta \left( {1 + pqr} \right)$$  
Answer :   $$\Delta ' = \Delta \left( {1 + pqr} \right)$$
Solution :
\[\Delta ' = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {p{b_1}}&{q{c_1}}&{r{a_1}}\\ {p{b_2}}&{q{c_2}}&{r{a_2}}\\ {p{b_3}}&{q{c_3}}&{r{a_3}} \end{array}} \right|\]       [All other determinants vanish]
$$ = \Delta + pqr\Delta = \Delta \left( {1 + pqr} \right)$$

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
Statistics and ProbabilityStatisticsProbability
CalculusSets and RelationsFunctionLimitsContinuityDifferentiability and DifferentiationApplication of DerivativesDifferential EquationsIndefinite IntegrationDefinite IntegrationApplication of Integration
GeometryStraight LinesCircleParabolaEllipseHyperbolaLocus3D Geometry and VectorsThree Dimensional Geometry
TrigonometryTrigonometric Ratio and IdentitiesTrignometric EquationsProperties and Solutons of TriangleInverse Trigonometry Function