If $$f\left( x \right),g\left( x \right)$$ and $$h\left( x \right)$$ are three polynomials of degree 2 and \[\Delta \left( x \right) = \left| {\begin{array}{*{20}{c}}
{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\
{f'\left( x \right)}&{g'\left( x \right)}&{h'\left( x \right)}\\
{f''\left( x \right)}&{g''\left( x \right)}&{h''\left( x \right)}
\end{array}} \right|,\] then $$\Delta \left( x \right)$$ is a polynomial of degree
A.
2
B.
3
C.
at most 2
D.
at most 3
Answer :
at most 2
Solution :
Let, $$f\left( x \right) = {a_0}{x^2} + {a_1}x + {a_2}$$
$$\eqalign{
& g\left( x \right) = {b_0}{x^2} + {b_1}x + {b_2} \cr
& h\left( x \right) = {c_0}{x^2} + {c_1}x + {c_2} \cr} $$
Then, \[\Delta \left( x \right) = \left| {\begin{array}{*{20}{c}}
{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\
{2{a_0}x + {a_1}}&{2{b_0}x + {b_1}}&{2{c_0}x + {c_1}}\\
{2{a_0}}&{2{b_0}}&{2{c_0}}
\end{array}} \right|\]
\[ = x\left| {\begin{array}{*{20}{c}}
{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\
{2{a_0}}&{2{b_0}}&{2{c_0}}\\
{2{a_0}}&{2{b_0}}&{2{c_0}}
\end{array}} \right| + \left| {\begin{array}{*{20}{c}}
{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\
{{a_1}}&{{b_1}}&{{c_1}}\\
{2{a_0}}&{2{b_0}}&{2{c_0}}
\end{array}} \right|\]
\[ = 0 + 2\left| {\begin{array}{*{20}{c}}
{f\left( x \right)}&{g\left( x \right)}&{h\left( x \right)}\\
{{a_1}}&{{b_1}}&{{c_1}}\\
{{a_0}}&{{b_0}}&{{c_0}}
\end{array}} \right|\]
$$ = 2\left[ {\left( {{b_1}{c_0} - {b_0}{c_1}} \right]f\left( x \right) - \left( {{a_1}{c_0} - {a_0}{c_1}} \right)g\left( x \right) + \left( {{a_1}{b_0} - {a_0}{b_1}} \right)h\left( x \right)} \right]$$
Hence degree of $$\Delta \left( x \right) \leqslant 2.$$
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