If \[\left| {\begin{array}{*{20}{c}}
{{b^2} + {c^2}}&{ab}&{ac} \\
{ba}&{{c^2} + {a^2}}&{bc} \\
{ca}&{cb}&{{a^2} + {b^2}}
\end{array}} \right| = \] square of a determinant \[\vartriangle \] of the third order then \[\vartriangle \] is equal to
A.
\[\left| {\begin{array}{*{20}{c}}
0&c&b \\
c&0&a \\
b&a&0
\end{array}} \right|\]
B.
\[\left| {\begin{array}{*{20}{c}}
a&b&c \\
b&c&a \\
c&a&b
\end{array}} \right|\]
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