If $$A$$ is an orthogonal matrix of order 3 and \[B = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
{ - 3}&0&2\\
2&5&0
\end{array}} \right],\] then which of the following is/are correct ?
$$\eqalign{
& 1.\left| {AB} \right| = \pm 47 \cr
& 2.AB = BA \cr} $$
Select the correct answer using the code given below :
A.
1 only
B.
2 only
C.
Both 1 and 2
D.
Neither 1 nor 2
Answer :
Both 1 and 2
Solution :
The determinent of a orthogonal matrix is always $$ \pm 1$$
$$\left| A \right| = \pm 1$$
\[B = \left[ {\begin{array}{*{20}{c}}
1&2&3\\
{ - 3}&0&2\\
2&5&0
\end{array}} \right]\]
$$\eqalign{
& \left| B \right| = - 10 - 2\left( { - 4} \right) + 3\left( { - 15} \right) \cr
& = - 47 \cr
& \left| {AB} \right| = \left| A \right|\left| B \right| \cr
& = \left( { \pm 1} \right)\left( { - 47} \right) \cr
& = \pm 47 \cr} $$
Taking $$A$$ as identity matrix we can prove $$AB = BA$$
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 $$C$$
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