Let \[\left| {\begin{array}{*{20}{c}}
{{\lambda ^2} + 3\lambda }&{\lambda - 1}&{\lambda + 3} \\
{\lambda + 1}&{ - 2\lambda }&{\lambda - 4} \\
{\lambda - 3}&{\lambda + 4}&{3\lambda }
\end{array}} \right| = p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s\lambda + t\] be an identity in $$\lambda $$ where $$p, q, r, s, t$$ are independent of $$\lambda .$$ Then the value of $$t$$ is
A.
4
B.
0
C.
1
D.
None of these
Answer :
0
Solution :
Put $$\lambda = 0$$ on both sides. Then
\[\left| {\begin{array}{*{20}{c}}
0&{ - 1}&3 \\
1&0&{ - 4} \\
{ - 3}&4&0
\end{array}} \right| = t.\]
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