let a beginarray20c 1 and 1 and 1 2 and 1 and 3 1 and 1 and

Let \[A = \left( {\begin{array}{*{20}{c}} 1&{ - 1}&1\\ 2&1&{ - 3}\\ 1&1&1 \end{array}} \right)\] and 10 \[B = \left( {\begin{array}{*{20}{c}} 4&2&2\\ { - 5}&0&\alpha \\ 1&{ - 2}&3 \end{array}} \right).\] If $$B$$ is the inverse of matrix $$A,$$ then $$\alpha$$ is

A. $$5$$

B. $$- 1$$

C. $$2$$

D. $$ - 2$$

Answer : $$5$$

Solution :
Here,
$$ \Rightarrow $$ \[B = \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 4&2&2\\ { - 5}&0&\alpha \\ 1&{ - 2}&3 \end{array}} \right]\]
Also since, $$B = A^{- 1}$$
⇒ $$AB = I$$
\[\begin{array}{l} \Rightarrow \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&1\\ 2&1&{ - 3}\\ 1&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4&2&2\\ { - 5}&0&\alpha \\ 1&{ - 2}&3 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ \Rightarrow \frac{1}{{10}}\left[ {\begin{array}{*{20}{c}} {10}&0&{5 - 2}\\ 0&{10}&{ - 5 + \alpha }\\ 0&0&{5 + \alpha } \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] \end{array}\]
$$\eqalign{ & \Rightarrow \frac{{5 - \alpha }}{{10}} = 0 \cr & \Rightarrow \alpha = 5 \cr} $$

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

Let \[A = \left( {\begin{array}{{20}{c}} 1&{ - 1}&1\\ 2&1&{ - 3}\\ 1&1&1 \end{array}} \right)\] and 10 \[B = \left( {\begin{array}{{20}{c}} 4&2&2\\ { - 5}&0&\alpha \\ 1&{ - 2}&3 \end{array}} \right).\] If $$B$$ is the inverse of matrix $$A,$$ then $$\alpha$$ 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

Let \[A = \left( {\begin{array}{*{20}{c}} 1&{ - 1}&1\\ 2&1&{ - 3}\\ 1&1&1 \end{array}} \right)\] and 10 \[B = \left( {\begin{array}{*{20}{c}} 4&2&2\\ { - 5}&0&\alpha \\ 1&{ - 2}&3 \end{array}} \right).\] If $$B$$ is the inverse of matrix $$A,$$ then $$\alpha$$ 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

Let \[A = \left( {\begin{array}{{20}{c}} 1&{ - 1}&1\\ 2&1&{ - 3}\\ 1&1&1 \end{array}} \right)\] and 10 \[B = \left( {\begin{array}{{20}{c}} 4&2&2\\ { - 5}&0&\alpha \\ 1&{ - 2}&3 \end{array}} \right).\] If $$B$$ is the inverse of matrix $$A,$$ then $$\alpha$$ 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