Question

Let $$A$$ and $$B$$ be two symmetric matrices of order 3.
Statement - 1 : $$A(BA)$$  and $$(AB)A$$  are symmetric matrices.
Statement - 2 : $$AB$$  is symmetric matrix if matrix multiplication of $$A$$ with $$B$$ is commutative.

A. Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.  
B. Statement - 1 is true, Statement - 2 is false.
C. Statement - 1 is false, Statement - 2 is true.
D. Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
Answer :   Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
Solution :
$$\eqalign{ & \therefore \,\,A' = A,\,\,B' = B \cr & {\text{Now }}\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A' \cr & = \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right) \cr & {\text{Similarly}}\,\,\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A \cr} $$
So, $$A(BA)$$  and $$(AB)A$$  are symmetric matrices.
Again $$(AB)' = B'A’ = BA$$
Now if $$BA = AB,$$   then $$AB$$  is symmetric matrix.

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. $$AB=BA$$

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Matrices and Determinants


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