\[A = \left| {\begin{array}{*{20}{c}}
{2a}&{3r}&x\\
{4b}&{6s}&{2y}\\
{ - 2c}&{ - 3t}&{ - z}
\end{array}} \right| = \lambda \left| {\begin{array}{*{20}{c}}
a&r&x\\
b&s&y\\
c&t&z
\end{array}} \right|,\] then what is the value of $$\lambda \,?$$
A.
$$12$$
B.
$$ - 12$$
C.
$$7$$
D.
$$ - 7$$
Answer :
$$ - 12$$
Solution :
Given, \[\left| {\begin{array}{*{20}{c}}
{2a}&{3r}&x\\
{4b}&{6s}&{2y}\\
{ - 2c}&{ - 3t}&{ - z}
\end{array}} \right| = \lambda \left| {\begin{array}{*{20}{c}}
a&r&x\\
b&s&y\\
c&t&z
\end{array}} \right|\]
Taking 2 common from $$C_1$$ and 3 from $$C_2$$ in L.H.S.
\[\therefore 2 \times 3\left| {\begin{array}{*{20}{c}}
a&r&x\\
{2b}&{2s}&{2y}\\
{ - c}&{ - t}&{ - z}
\end{array}} \right| = \lambda \left| {\begin{array}{*{20}{c}}
a&r&x\\
b&s&y\\
c&t&z
\end{array}} \right|\]
Taking 2 common from $$R_2$$ and $$- 1$$ from $$R_3$$ in L.H.S.
\[\therefore \, - 12\left| {\begin{array}{*{20}{c}}
a&r&x\\
b&s&y\\
c&t&z
\end{array}} \right| = \lambda \left| {\begin{array}{*{20}{c}}
a&r&x\\
b&s&y\\
c&t&z
\end{array}} \right|\]
$$ \Rightarrow \,\lambda = - 12$$
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