If $$A = {\left[ {{a_{ij}}} \right]_{n \times n}}$$ be a diagonal matrix with diagonal
element all different and $$B = {\left[ {{a_{ij}}} \right]_{n \times n}}$$ be some
another matrix. Let $$AB = {\left[ {{c_{ij}}} \right]_{n \times n}}$$ then $$c_{ij}$$ is equal to
A.
$${a_{jj}}{b_{ij}}$$
B.
$${a_{ii}}{b_{ij}}$$
C.
$${a_{ij}}{b_{ij}}$$
D.
$${a_{ij}}{b_{ji}}$$
Answer :
$${a_{ii}}{b_{ij}}$$
Solution :
$${c_{ij}} = \sum\limits_{k = 1}^n {{a_{ik}}{b_{kj}}} \,\,\,\left( {{\text{In general}}} \right)$$
and in a diagonal matrix non-diagonal elements are zero
i.e., \[{a_{ij}} = \left\{ \begin{gathered}
0,\,\,{\text{if }}i \ne j \hfill \\
{a_{ii}}{\text{, if }}i = j \hfill \\
\end{gathered} \right.\]
So, $${c_{ij}} = {a_{ii}}{b_{ij}}$$
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