if 20c x y and y 2x and x y 20c 2 1 20c 3 2 222021892

If \[\left[ {\begin{array}{*{20}{c}} {x + y}&y \\ {2x}&{x - y} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2 \\ { - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right]\] then $$x \cdot y$$ is equal to

A. $$ - 5$$

B. $$5$$

C. $$4$$

D. $$6$$

Answer : $$ - 5$$

Solution :
\[\begin{array}{l} {\rm{Given,}}\\ \left[ {\begin{array}{*{20}{c}} {x + y}&y\\ {2x}&{x - y} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right]\\ \Rightarrow \left[ {\begin{array}{*{20}{c}} {2x + 2y}&{ - y}\\ {4x}&{ - x + y} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right]\\ \Rightarrow \left[ \begin{array}{l} 2x + y\\ 3x + y \end{array} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right] \end{array}\]
$$\eqalign{ & {\text{Comparing the corresponding elements}} \cr & 2x + y = 3......\left( {\text{i}} \right) \cr & 3x + y = 2......\left( {{\text{ii}}} \right) \cr & {\text{Subtracting, we get}} \cr & - x = 1 \cr & \Rightarrow x = - 1 \cr & {\text{Substituting the vale of }}x{\text{ in equation }}\left( {\text{i}} \right) \cr & 2\left( { - 1} \right) + y = 3 \cr & \Rightarrow - 2 + y = 3 \cr & \Rightarrow y = 3 + 2 = 5 \cr & {\text{Hence, }}x = - 1,\,y = 5 \cr & \therefore \,x \cdot y = - 5 \cr} $$

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 $$C$$

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

If \[\left[ {\begin{array}{{20}{c}} {x + y}&y \\ {2x}&{x - y} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 2 \\ { - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right]\] then $$x \cdot y$$ is equal to

Releted MCQ Question on
Algebra >> Matrices and Determinants

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

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

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

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Matrices and Determinants

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If \[\left[ {\begin{array}{*{20}{c}} {x + y}&y \\ {2x}&{x - y} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2 \\ { - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right]\] then $$x \cdot y$$ is equal to

Releted MCQ Question on Algebra >> Matrices and Determinants

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

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

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

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Practice More MCQ Question on Maths Section

If \[\left[ {\begin{array}{{20}{c}} {x + y}&y \\ {2x}&{x - y} \end{array}} \right]\left[ {\begin{array}{{20}{c}} 2 \\ { - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right]\] then $$x \cdot y$$ is equal to

Releted MCQ Question on
Algebra >> Matrices and Determinants

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

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Matrices and Determinants