Question

The value of \[\left| {\begin{array}{*{20}{c}} {{a_1}x + {b_1}y}&{{a_2}x + {b_2}y}&{{a_3}x + {b_3}y} \\ {{b_1}x + {a_1}y}&{{b_2}x + {a_2}y}&{{b_3}x + {a_3}y} \\ {{b_1}x + {a_1}}&{{b_2}x + {a_2}}&{{b_3}x + {a_3}} \end{array}} \right|\]       is equal to

A. $${x^2} + {y^2}$$
B. $$0$$  
C. $${a_1}{a_2}{a_3}{x^2} + {b_1}{b_2}{b_3}{y^2}$$
D. None of these
Answer :   $$0$$
Solution :
\[\vartriangle = \left| {\begin{array}{*{20}{c}} {{a_1}x + {b_1}y}&{{a_2}x + {b_2}y}&{{a_3}x + {b_3}y} \\ {{a_1}\left( {y - 1} \right)}&{{a_2}\left( {y - 1} \right)}&{{a_3}\left( {y - 1} \right)} \\ {{b_1}x + {a_1}}&{{b_2}x + {a_2}}&{{b_3}x + {a_3}} \end{array}} \right|\]
$$\left( {{R_2} \to {R_2} - {R_3}} \right)$$
\[\vartriangle = \left( {y - 1} \right)\left| {\begin{array}{*{20}{c}} {{b_1}y}&{{b_2}y}&{{b_3}y} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}x + {a_1}}&{{b_2}x + {a_2}}&{{b_3}x + {a_3}} \end{array}} \right|\]
$$\left( {{R_1} \to {R_1} - x \times {R_2}} \right)$$
\[\vartriangle = \left( {y - 1} \right)y\left| {\begin{array}{*{20}{c}} {{b_1}}&{{b_2}}&{{b_3}} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ 0&0&0 \end{array}} \right|\left( {{\text{Using }}{R_3} \to {R_3} - x \times {R_1} - {R_2}} \right).\]

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 $$C$$
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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Matrices and Determinants


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