Equations x + y + z = 6, x + 2y + 3z = 10 and x + 2y + λz = μ have infinite number of solutions if

Concept:

A = [aij]m × n = coefficient of matrix, X = Column matrix of the variables

B = column matrix of constants

The system AX = B has

1) A unique solution If Rank of A = Rank [A|B] and is equal to the number of variables.

2) Infinitely many solutions, If Rank of A = Rank of [A|B] < number of variables

3) no solution, If Rank of A ≠ Rank of [A|B], i.e. Rank of A < Rank of [A|B].

Calculation:

Let 

\(let\;A = \left[ {\begin{array}{*{20}{c}} 1&1&{ 1}\\ 1&2&3\\ 1&{ 2}&{ λ} \end{array}} \right]\)

We can see that, at λ = 3, R₂ = R₃ and hence, |A| = 0.

For λ = 3 either infinite solutions exist or no solution exists.

The coefficient matrix of the given linear equation is

\(\left[ {\begin{array}{*{20}{c}} 1&1&{ 1}\\ 1&2&3\\ 1&{ 2}&{ λ} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} x\\ y\end{array}}\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} { 6}\\ { 10} \end{array}}\\ \mu\end{array}} \right]\;\)

R3 → R3 - R2

⇒ \(\left[ {\begin{array}{*{20}{c}} 1&1&{ 1}\\ 1&2&3\\ 0&{ 0}&{ λ\ -\ 3} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} x\\ y\end{array}}\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} { 6}\\ { 10} \end{array}}\\ \mu\ -\ 3\end{array}} \right]\;\)

Let the augmented matrix be

\(X = \;\left[ {\begin{array}{*{20}{c}} 1&1&{ 1}\\ 1&2&3\\ 0&{ 0}&{ λ \ -\ 3} \end{array}\left| {\begin{array}{*{20}{c}} { 6}\\ { 10}\\ \mu\ -\ 10 \end{array}} \right.} \right]\)

For no solution,  μ = 10 

Hence, λ = 3, μ = 10 is the correct answer.

Shortcut Trickx + y + z = 6

x + 2y + 3z = 10

x + 2y + λz = μ 

Hence, if we put λ = 3, μ = 10, two-equation will get coincide, which result in an infinite number of solution.