Question

Let $$\alpha \,{\text{and }}\beta $$   be the roots of the equation $${x^2} + x + 1 = 0.$$   Then for $$y \ne 0\,\,{\text{in }}R,$$   \[\left| {\begin{array}{*{20}{c}} {y + 1}&\alpha &\beta \\ \alpha &{y + \beta }&1\\ \beta &1&{y + \alpha } \end{array}} \right|\]     is equal to:

A. $$y\left( {{y^2} - 1} \right)$$
B. $$y\left( {{y^2} - 3} \right)$$
C. $${{y^3}}$$  
D. $${{y^3} - 1}$$
Answer :   $${{y^3}}$$
Solution :
$${\text{Let}}\,\alpha \, = \omega \,\,{\text{and}}\,\beta = {\omega ^2} = 2$$      are roots of $${x^2} + x + 1 = 0$$
\[\& \,{\text{Let}}\,\Delta = \left| {\begin{array}{*{20}{c}} {y + 1}&\omega &{{\omega ^2}}\\ \omega &{y + {\omega ^2}}&1\\ {{\omega ^2}}&1&{y + \omega } \end{array}} \right| = \Delta \]
$${\text{Applying}}\,{C_1} \to {C_1} + {C_2} + {C_3},{\text{we}}\,{\text{get}}$$
\[\Delta {\rm{ = }}\left| {\begin{array}{*{20}{c}} {y + 1 + \omega + {\omega ^2}}&\omega &{{\omega ^2}}\\ {y + 1 + \omega + {\omega ^2}}&{y + {\omega ^2}}&1\\ {1 + \omega + {\omega ^2} + y}&1&{y + \omega } \end{array}} \right|\]
\[\Delta = \left| {\begin{array}{*{20}{c}} y&\omega &{{\omega ^2}} \\ y&{y + {\omega ^2}}&1 \\ y&1&{y + \omega } \end{array}} \right|\left( {\because \,\,1 + \omega + {\omega ^2} = 0} \right)\]
\[\Delta = y\left| {\begin{array}{*{20}{c}} 1&\omega &{{\omega ^2}}\\ 1&{y + {\omega ^2}}&1\\ 1&1&{y + \omega } \end{array}} \right|\]
Applying $${R_2} \to {R_2} - {R_1}\,\,\& \,\,{R_3} \to {R_3} - {R_1},\,{\text{we get}}$$
\[\Delta = y\left| {\begin{array}{*{20}{c}} {y + {\omega ^2} - \omega }&{1 - {\omega ^2}}\\ {1 - \omega }&{y + \omega - {\omega ^2}} \end{array}} \right|\]
$$\eqalign{ & \Rightarrow \,\,\Delta = y \cr & \left[ {y - \left( {\omega - {\omega ^2}} \right)\left( {y + \left( {\omega - {\omega ^2}} \right)} \right) - \left( {1 - \omega } \right)\left( {1 - {\omega ^2}} \right)} \right] \cr & \Rightarrow \,\,\Delta = y\left[ {{y^2} - {{\left( {\omega - {\omega ^2}} \right)}^2} - 1 + {\omega ^2} + \omega - {\omega ^3}} \right] \cr & \Rightarrow \,\,\Delta = y\left[ {{y^2} - {\omega ^2} - {\omega ^4} + 2{\omega ^3} - 1 + {\omega ^2} + {\omega ^4} - {\omega ^3}} \right]\,\,\left( {\because \,\,{\omega ^4} = \omega } \right) \cr & \Rightarrow \,\,\Delta = y\left( {{y^2}} \right) = {y^3} \cr} $$

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 $$C$$
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. $$\omega $$
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
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. $$AB=BA$$

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