Question:
Let \(\lambda\in R\). The system of linear equations
\(2x_1-4x_2+\lambda x_3=1\)
\(x_1-6x_2+x_3=2\)
\(\lambda x_1-10x_2+4x_3=3\)
is inconsistent for:
Let \(\lambda\in R\). The system of linear equations
\(2x_1-4x_2+\lambda x_3=1\)
\(x_1-6x_2+x_3=2\)
\(\lambda x_1-10x_2+4x_3=3\)
is inconsistent for:
\(2x_1-4x_2+\lambda x_3=1\)
\(x_1-6x_2+x_3=2\)
\(\lambda x_1-10x_2+4x_3=3\)
is inconsistent for:
Show Hint
Find the \(\lambda\) values for which the coefficient determinant vanishes, then test the consistency of each case separately.
Updated On: Nov 5, 2026
- exactly one positive value of \(\lambda\)
- exactly one negative value of \(\lambda\)
- exactly two values of \(\lambda\)
- every value of \(\lambda\)
Show Solution
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The Correct Option is B
Solution and Explanation
Coefficient determinant: \(\Delta=\begin{vmatrix}2&-4&\lambda\\1&-6&1\\\lambda&-10&4\end{vmatrix}=2(-24+10)+4(4-\lambda)+\lambda(-10+6\lambda)=6\lambda^2-14\lambda-12\).
\(\Delta=2(3\lambda+2)(\lambda-3)=0\Rightarrow\lambda=3\) or \(\lambda=-\frac23\). For other \(\lambda\) the system has a unique solution (consistent).
Case \(\lambda=3\): rows of coefficients \(R_1=(2,-4,3)\), \(R_2=(1,-6,1)\), \(R_3=(3,-10,4)\). Here \(R_1+R_2=R_3\) and RHS: \(1+2=3\). Dependent and consistent (infinitely many solutions).
Case \(\lambda=-\frac23\): find \(a,b,c\) with \(aR_1+bR_2+cR_3=0\): \(a=21,\ b=-34,\ c=12\) works for the coefficient columns. Apply it to the RHS: \(21(1)-34(2)+12(3)=-11\neq0\). Inconsistent.
So the system is inconsistent for exactly one value, \(\lambda=-\frac23\) (negative).
Answer: (2) exactly one negative value of \(\lambda\)
\(\Delta=2(3\lambda+2)(\lambda-3)=0\Rightarrow\lambda=3\) or \(\lambda=-\frac23\). For other \(\lambda\) the system has a unique solution (consistent).
Case \(\lambda=3\): rows of coefficients \(R_1=(2,-4,3)\), \(R_2=(1,-6,1)\), \(R_3=(3,-10,4)\). Here \(R_1+R_2=R_3\) and RHS: \(1+2=3\). Dependent and consistent (infinitely many solutions).
Case \(\lambda=-\frac23\): find \(a,b,c\) with \(aR_1+bR_2+cR_3=0\): \(a=21,\ b=-34,\ c=12\) works for the coefficient columns. Apply it to the RHS: \(21(1)-34(2)+12(3)=-11\neq0\). Inconsistent.
So the system is inconsistent for exactly one value, \(\lambda=-\frac23\) (negative).
Answer: (2) exactly one negative value of \(\lambda\)
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