if 20c 1 and 3 and 2 2 and 5 and 1 15 and 3 and 2 20c 1 2 x

If \[{\left[ {\begin{array}{*{20}{c}} 1&x&1 \end{array}} \right]_{1 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1&3&2\\ 2&5&1\\ {15}&3&2 \end{array}} \right]_{3 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ x \end{array}} \right]_{3 \times 1}} = 0,\] then $$x$$ is

A. $$2$$

B. $$- 2$$

C. $$14$$

D. None of these

Answer : $$- 2$$

Solution :
\[\begin{array}{l} {\rm{Given,\, }}{\left[ {\begin{array}{*{20}{c}} 1&x&1 \end{array}} \right]_{1 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1&3&2\\ 2&5&1\\ {15}&3&2 \end{array}} \right]_{3 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ x \end{array}} \right]_{3 \times 1}} = 0\\ \Rightarrow \left[ {1 + 2x + 15\,\,\,3 + 5x + 3\,\,\,2 + x + 2} \right]\left[ \begin{array}{l} 1\\ 2\\ x \end{array} \right] = 0\\ \Rightarrow \left[ {16 + 2x + 15\,\,\,3 + 5x + 3\,\,\,2 + x + 2} \right]\left[ \begin{array}{l} 1\\ 2\\ x \end{array} \right] = 0\\ \Rightarrow \left[ {\left( {16 + 2x} \right) \cdot 1 + \left( {6 + 5x} \right) \cdot 2 + \left( {4 + x} \right) \cdot x} \right] = 0\\ \Rightarrow \left( {16 + 2x} \right) + \left( {12 + 10x} \right) + \left( {4x + {x^2}} \right) = 0\\ \Rightarrow {x^2} + 16x + 28 = 0\\ \Rightarrow \left( {x + 14} \right)\left( {x + 2} \right) = 0\\ \Rightarrow x + 14 = 0{\rm{ \,\,or\,\, }}x + 2 = 0\\ {\rm{Hence, }}x = - 14{\rm{ }}\,{\rm{\,\,or\,\, }}x = - 2 \end{array}\]

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 \[{\left[ {\begin{array}{{20}{c}} 1&x&1 \end{array}} \right]_{1 \times 3}}{\left[ {\begin{array}{{20}{c}} 1&3&2\\ 2&5&1\\ {15}&3&2 \end{array}} \right]_{3 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ x \end{array}} \right]_{3 \times 1}} = 0,\] then $$x$$ is

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

Practice More MCQ Question on Maths Section

If \[{\left[ {\begin{array}{*{20}{c}} 1&x&1 \end{array}} \right]_{1 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1&3&2\\ 2&5&1\\ {15}&3&2 \end{array}} \right]_{3 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ x \end{array}} \right]_{3 \times 1}} = 0,\] then $$x$$ is

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 \[{\left[ {\begin{array}{{20}{c}} 1&x&1 \end{array}} \right]_{1 \times 3}}{\left[ {\begin{array}{{20}{c}} 1&3&2\\ 2&5&1\\ {15}&3&2 \end{array}} \right]_{3 \times 3}}{\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ x \end{array}} \right]_{3 \times 1}} = 0,\] then $$x$$ is

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