Question

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

A. 5  
B. $$- 1$$
C. 2
D. $$- 2$$
Answer :   5
Solution :
\[{\rm{Given\,\, that\,\, 10}}\,B = \left[ \begin{array}{l} \,\,\,\,4\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,2\\ - 5\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\alpha \\ \,\,\,\,1\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,3 \end{array} \right]\]
\[ \Rightarrow \,B = \frac{1}{{10}}\left[ \begin{array}{l} \,\,\,4\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,2\\ - 5\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\alpha \\ \,\,\,1\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,3 \end{array} \right]\]
$${\text{Also since, }}B = {A^{ - 1}} \Rightarrow \,\,AB = I$$
\[ \Rightarrow \,\,\frac{1}{{10}}\left[ \begin{array}{l} 1\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,1\\ 2\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\, - 3\\ 1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,1 \end{array} \right]\left[ \begin{array}{l} \,\,\,\,4\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,2\\ - 5\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\alpha \\ \,\,\,\,\,1\,\,\,\, - 2\,\,\,\,\,\,\,\,3 \end{array} \right] = \left[ \begin{array}{l} 1\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,1\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,0\,\,\,\,\,\,\,1 \end{array} \right]\]
\[ \Rightarrow \,\,\frac{1}{{10}}\left[ \begin{array}{l} 10\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5 - 2\\ \,0\,\,\,\,\,\,10\,\,\,\,\,\,\, - 5 + \alpha \\ \,0\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5 + \alpha \end{array} \right] = \left[ \begin{array}{l} 1\,\,\,\,\,\,0\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,1\,\,\,\,\,\,\,0\\ 0\,\,\,\,\,\,0\,\,\,\,\,\,\,1 \end{array} \right]\]
$$\eqalign{ & \Rightarrow \,\,\frac{{5 - \alpha }}{{10}} = 0 \cr & \Rightarrow \,\,\alpha = 5 \cr} $$

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 $$C$$
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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

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Matrices and Determinants

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