If the system of equations $$\lambda {x_1} + {x_2} + {x_3} = 1,{x_1} + \lambda {x_2} + {x_3} = 1,{x_1} + {x_2} + \lambda {x_3} = 1$$ is consistent, then $$\lambda$$ can be
A.
$$5$$
B.
$$ - \frac{2}{3}$$
C.
$$ - 3$$
D.
None of these
Answer :
None of these
Solution :
Let \[\Delta = \left| {\begin{array}{*{20}{c}}
\lambda &1&1 \\
1&\lambda &1 \\
1&1&\lambda
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{\lambda + 2}&1&1 \\
{\lambda + 2}&\lambda &1 \\
{\lambda + 2}&1&\lambda
\end{array}} \right|\left[ {{C_1} \to {C_1} + {C_2} + {C_3}} \right]\]
\[ = \left( {\lambda + 2} \right)\left| {\begin{array}{*{20}{c}}
1&1&1 \\
1&\lambda &1 \\
1&1&\lambda
\end{array}} \right| = \left( {\lambda + 2} \right)\left| {\begin{array}{*{20}{c}}
1&0&0 \\
1&{\lambda - 1}&0 \\
1&0&{\lambda - 1}
\end{array}} \right|\]
$$ = \left( {\lambda + 2} \right){\left( {\lambda - 1} \right)^2}$$ [ using $${{C_2} \to {C_2} - {C_1}}$$ and $${{C_3} \to {C_3} - {C_1}}$$ ]
If $$\Delta = 0,$$ then $$\lambda = - 2{\text{ or }}\lambda = 1.$$
But when $$\lambda = 1,$$ the system of equation becomes $${x_1} + {x_2} + {x_3} = 1$$ which has infinite number of solutions. When $$\lambda = - 2,$$ by adding three equations, we obtain $$0 = 3$$ and thus, the system of equations is inconsistent.
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