Write the system in matrix form
The equations are
{ 2 x 1 − 4 x 2 + λ x 3 = 1 x 1 − 6 x 2 + x 3 = 2 λ x 1 − 10 x 2 + 4 x 3 = 3 \begin{cases}
2x_1-4x_2+\lambda x_3=1 \\
x_1-6x_2+x_3=2 \\
\lambda x_1-10x_2+4x_3=3
\end{cases} ⎩ ⎨ ⎧ 2 x 1 − 4 x 2 + λ x 3 = 1 x 1 − 6 x 2 + x 3 = 2 λ x 1 − 10 x 2 + 4 x 3 = 3 Coefficient matrix:
A = ( 2 − 4 λ 1 − 6 1 λ − 10 4 ) A=\begin{pmatrix}
2 & -4 & \lambda\\
1 & -6 & 1\\
\lambda & -10 & 4
\end{pmatrix} A = 2 1 λ − 4 − 6 − 10 λ 1 4 Augmented matrix:
[ A ∣ B ] = ( 2 − 4 λ 1 1 − 6 1 2 λ − 10 4 3 ) [A|B]=\begin{pmatrix}
2 & -4 & \lambda & 1\\
1 & -6 & 1 & 2\\
\lambda & -10 & 4 & 3
\end{pmatrix} [ A ∣ B ] = 2 1 λ − 4 − 6 − 10 λ 1 4 1 2 3 For inconsistency, we need
rank ( A ) < rank ( [ A ∣ B ] ) . \operatorname{rank}(A) < \operatorname{rank}([A|B]). rank ( A ) < rank ([ A ∣ B ]) . This can happen only when det ( A ) = 0 \det(A)=0 det ( A ) = 0
Compute det ( A ) \det(A) det ( A )
det ( A ) = ∣ 2 − 4 λ 1 − 6 1 λ − 10 4 ∣ \det(A)=\begin{vmatrix}
2 & -4 & \lambda\\
1 & -6 & 1\\
\lambda & -10 & 4
\end{vmatrix} det ( A ) = 2 1 λ − 4 − 6 − 10 λ 1 4 Expand along the first row:
det ( A ) = 2 ∣ − 6 1 − 10 4 ∣ − ( − 4 ) ∣ 1 1 λ 4 ∣ + λ ∣ 1 − 6 λ − 10 ∣ \det(A)=2\begin{vmatrix}-6 & 1\\-10 & 4\end{vmatrix}-(-4)\begin{vmatrix}1 & 1\\ \lambda & 4\end{vmatrix}+\lambda\begin{vmatrix}1 & -6\\ \lambda & -10\end{vmatrix} det ( A ) = 2 − 6 − 10 1 4 − ( − 4 ) 1 λ 1 4 + λ 1 λ − 6 − 10 = 2 [ ( − 6 ) ( 4 ) − 1 ( − 10 ) ] + 4 [ ( 1 ) ( 4 ) − λ ] + λ [ ( 1 ) ( − 10 ) − ( − 6 ) λ ] =2[(-6)(4)-1(-10)]+4[(1)(4)-\lambda]+\lambda[(1)(-10)-(-6)\lambda] = 2 [( − 6 ) ( 4 ) − 1 ( − 10 )] + 4 [( 1 ) ( 4 ) − λ ] + λ [( 1 ) ( − 10 ) − ( − 6 ) λ ] = 2 ( − 24 + 10 ) + 4 ( 4 − λ ) + λ ( − 10 + 6 λ ) =2(-24+10)+4(4-\lambda)+\lambda(-10+6\lambda) = 2 ( − 24 + 10 ) + 4 ( 4 − λ ) + λ ( − 10 + 6 λ ) = − 28 + 16 − 4 λ − 10 λ + 6 λ 2 =-28+16-4\lambda-10\lambda+6\lambda^2 = − 28 + 16 − 4 λ − 10 λ + 6 λ 2 = 6 λ 2 − 14 λ − 12 =6\lambda^2-14\lambda-12 = 6 λ 2 − 14 λ − 12 = 2 ( 3 λ 2 − 7 λ − 6 ) = 2 ( 3 λ + 2 ) ( λ − 3 ) =2(3\lambda^2-7\lambda-6)=2(3\lambda+2)(\lambda-3) = 2 ( 3 λ 2 − 7 λ − 6 ) = 2 ( 3 λ + 2 ) ( λ − 3 ) So,
det ( A ) = 0 ⟺ λ = 3 or λ = − 2 3 . \det(A)=0 \iff \lambda=3 \quad \text{or} \quad \lambda=-\frac23. det ( A ) = 0 ⟺ λ = 3 or λ = − 3 2 .
Check consistency for each value
Case 1: λ = 3 \lambda=3 λ = 3
System becomes
{ 2 x 1 − 4 x 2 + 3 x 3 = 1 x 1 − 6 x 2 + x 3 = 2 3 x 1 − 10 x 2 + 4 x 3 = 3 \begin{cases}
2x_1-4x_2+3x_3=1 \\
x_1-6x_2+x_3=2 \\
3x_1-10x_2+4x_3=3
\end{cases} ⎩ ⎨ ⎧ 2 x 1 − 4 x 2 + 3 x 3 = 1 x 1 − 6 x 2 + x 3 = 2 3 x 1 − 10 x 2 + 4 x 3 = 3 Notice that
(Eq. 3) = (Eq. 1) + (Eq. 2) \text{(Eq. 3)} = \text{(Eq. 1)} + \text{(Eq. 2)} (Eq. 3) = (Eq. 1) + (Eq. 2) for the left-hand sides, since
( 2 , − 4 , 3 ) + ( 1 , − 6 , 1 ) = ( 3 , − 10 , 4 ) . (2,-4,3)+(1,-6,1)=(3,-10,4). ( 2 , − 4 , 3 ) + ( 1 , − 6 , 1 ) = ( 3 , − 10 , 4 ) . Also on the right-hand side,
1 + 2 = 3. 1+2=3. 1 + 2 = 3. So the third equation is dependent and consistent. Hence the system is consistent for λ = 3 \lambda=3 λ = 3
Case 2: λ = − 2 3 \lambda=-\frac23 λ = − 3 2
System becomes
{ 2 x 1 − 4 x 2 − 2 3 x 3 = 1 x 1 − 6 x 2 + x 3 = 2 − 2 3 x 1 − 10 x 2 + 4 x 3 = 3 \begin{cases}
2x_1-4x_2-\frac23 x_3=1 \\
x_1-6x_2+x_3=2 \\
-\frac23 x_1-10x_2+4x_3=3
\end{cases} ⎩ ⎨ ⎧ 2 x 1 − 4 x 2 − 3 2 x 3 = 1 x 1 − 6 x 2 + x 3 = 2 − 3 2 x 1 − 10 x 2 + 4 x 3 = 3 Let us check whether the third row is a linear combination of the first two in coefficients.
Suppose
R 3 = a R 1 + b R 2 R_3=aR_1+bR_2 R 3 = a R 1 + b R 2 for coefficient rows. Then from the first two entries,
2 a + b = − 2 3 , 2a+b=-\frac23, 2 a + b = − 3 2 , − 4 a − 6 b = − 10. -4a-6b=-10. − 4 a − 6 b = − 10. Solving:
From the first,
b = − 2 3 − 2 a . b=-\frac23-2a. b = − 3 2 − 2 a . Substitute into second:
− 4 a − 6 ( − 2 3 − 2 a ) = − 10 -4a-6\left(-\frac23-2a\right)=-10 − 4 a − 6 ( − 3 2 − 2 a ) = − 10 − 4 a + 4 + 12 a = − 10 -4a+4+12a=-10 − 4 a + 4 + 12 a = − 10 8 a = − 14 8a=-14 8 a = − 14 a = − 7 4 . a=-\frac74. a = − 4 7 . Then
b = − 2 3 + 7 2 = 17 6 . b=-\frac23+\frac72=\frac{17}{6}. b = − 3 2 + 2 7 = 6 17 . Now check the third coefficient:
a λ + b = ( − 7 4 ) ( − 2 3 ) + 17 6 = 7 6 + 17 6 = 4 , a\lambda+b = \left(-\frac74\right)\left(-\frac23\right)+\frac{17}{6}
=\frac76+\frac{17}{6}=4, aλ + b = ( − 4 7 ) ( − 3 2 ) + 6 17 = 6 7 + 6 17 = 4 , which matches the third entry of row 3. So coefficient-wise,
R 3 = − 7 4 R 1 + 17 6 R 2 . R_3=-\frac74 R_1+\frac{17}{6}R_2. R 3 = − 4 7 R 1 + 6 17 R 2 . But for consistency, the constants must satisfy the same relation:
− 7 4 ( 1 ) + 17 6 ( 2 ) = − 7 4 + 17 3 = − 21 + 68 12 = 47 12 ≠ 3. -\frac74(1)+\frac{17}{6}(2)= -\frac74+\frac{17}{3}
=\frac{-21+68}{12}=\frac{47}{12}\neq 3. − 4 7 ( 1 ) + 6 17 ( 2 ) = − 4 7 + 3 17 = 12 − 21 + 68 = 12 47 = 3. Hence the augmented row does not satisfy the same relation.
Therefore,
rank ( A ) < rank ( [ A ∣ B ] ) , \operatorname{rank}(A)<\operatorname{rank}([A|B]), rank ( A ) < rank ([ A ∣ B ]) , so the system is inconsistent for
λ = − 2 3 . \lambda=-\frac23. λ = − 3 2 .
Conclusion
The system is inconsistent for exactly one value of λ \lambda λ , namely
λ = − 2 3 , \lambda=-\frac23, λ = − 3 2 , which is negative .
So the correct option is:
B: exactly one negative value of λ \boxed{\text{B: exactly one negative value of }\lambda} B: exactly one negative value of λ