If \[A = \left[ {\begin{array}{*{20}{c}}
0&{ - 4}&1 \\
2&\lambda &{ - 3} \\
1&2&{ - 1}
\end{array}} \right]\] then $${A^{ - 1}}$$ exists (i.e., $$A$$ is invertible) if
A.
$$\lambda \ne 4$$
B.
$$\lambda \ne 8$$
C.
$$\lambda = 4$$
D.
None of these
Answer :
$$\lambda \ne 8$$
Solution :
$$A$$ is invertible if \[\left| A \right| \ne 0,\,{\text{i}}{\text{.e}}{\text{.,}}\left| {\begin{array}{*{20}{c}}
0&{ - 4}&1 \\
2&\lambda &{ - 3} \\
1&2&{ - 1}
\end{array}} \right| \ne 0.\]
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