Question

Consider the following in respect of the matrix \[A = \left( {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right):\]
$$\eqalign{ & 1.{A^2} = - A \cr & 2.{A^3} = 4A \cr} $$
Which of the above is/are correct ?

A. 1 only  
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2
Answer :   1 only
Solution :
\[\begin{array}{l} A = \left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\\ A.A = \left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 2&{ - 2}\\ { - 2}&2 \end{array}} \right] = - 2\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right] \end{array}\]
$$A^2 = - 2A$$
\[\begin{array}{l} {A^2}.A = - 2\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right]\\ = - 2\left[ {\begin{array}{*{20}{c}} 2&{ - 2}\\ { - 2}&2 \end{array}} \right] = 4\left[ {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right] \end{array}\]
$$A^3 = 4A$$
Hence, $${A^2} \ne - A,{A^3} = 4A.$$

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. $$AB=BA$$

Practice More Releted MCQ Question on
Matrices and Determinants


Practice More MCQ Question on Maths Section