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Exercise 1.6 · Q3

Q.Investigate the values of λ\lambda and μ\mu so that the system of linear equations 2x+3y+5z=92x+3y+5z=9, 7x+3y−5z=87x+3y-5z=8, 2x+3y+λz=μ2x+3y+\lambda z=\mu have

(i) no solution
(ii) a unique solution
(iii) an infinite number of solutions.
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Step 1. Compute the coefficient determinant.

∣A∣=∣23573−523λ∣.|A|=\begin{vmatrix} 2 & 3 & 5\\ 7 & 3 & -5\\ 2 & 3 & \lambda\end{vmatrix}.

Rows 1 and 3 already agree in the first two columns, so subtract: R3→R3−R1R_3\to R_3-R_1 (this leaves the determinant unchanged) gives ∣23573−500λ−5∣\begin{vmatrix} 2 & 3 & 5\\ 7 & 3 & -5\\ 0 & 0 & \lambda-5\end{vmatrix}. Expanding along the new third row:

∣A∣=(λ−5)∣2373∣=(λ−5)(6−21)=−15(λ−5)=75−15λ.|A|=(\lambda-5)\begin{vmatrix}2 & 3\\ 7 & 3\end{vmatrix}=(\lambda-5)(6-21)=-15(\lambda-5)=75-15\lambda.

This vanishes exactly at λ=5\lambda=5.

Step 2. λ≠5\lambda\ne5: unique solution. Here ∣A∣≠0⇒ρ(A)=ρ([A∣B])=3=n|A|\ne0\Rightarrow\rho(A)=\rho([A|B])=3=n for every value of μ\mu, so the system has a unique solution whenever λ≠5\lambda\ne5.

Step 3. λ=5\lambda=5: reduce and test consistency. With λ=5\lambda=5 the third equation is 2x+3y+5z=μ2x+3y+5z=\mu, which has the same left-hand side as the first equation 2x+3y+5z=92x+3y+5z=9. Row-reduce [A∣B][A|B]:

[235973−58235μ]→R3→R3−R1[235973−58000μ−9].\left[\begin{array}{ccc|c} 2 & 3 & 5 & 9\\ 7 & 3 & -5 & 8\\ 2 & 3 & 5 & \mu\end{array}\right]\xrightarrow{R_3\to R_3-R_1}\left[\begin{array}{ccc|c} 2 & 3 & 5 & 9\\ 7 & 3 & -5 & 8\\ 0 & 0 & 0 & \mu-9\end{array}\right].

Rows 1 and 2 are independent (their 2×22\times2 minor ∣2373∣=−15≠0\begin{vmatrix}2&3\\7&3\end{vmatrix}=-15\ne0), so ρ(A)=2\rho(A)=2 regardless of μ\mu. …

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