If \[A = \left[ {\begin{array}{*{20}{c}}
0&1&3\\
1&2&3\\
3&a&1
\end{array}} \right]\] and \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}}&{ - \frac{1}{2}}&{\frac{1}{2}}\\
{ - 4}&3&c\\
{\frac{5}{2}}&{ - \frac{3}{2}}&{\frac{1}{2}}
\end{array}} \right],\] then the value of $$a + c$$ is equal to
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