Question

Consider the system of equations: x + y = 2 and 2x + 2y = 5. This system has

1.

One solution<<br>2. No solution
3. Infinite solution
4. Four solutions

Answer 1

Correct Answer - Option 2 :

Concept:

Consider the system of 'm' linear equations with 'n' unknown.

a11 x1 + a12 x2 + … + a1n xn = b₁

 

a21 x1 + a22 x2 + … + a2n xn = b₂

…

am1 x1 + am2 x2 + … + amn xn = b_m

To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrix.

\(C = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \ldots &{{a_{1n}}}&{b_1}\\ {{a_{21}}}&{{a_{22}}}& \ldots &{{a_{2n}}}&{b_2}\\ \ldots & \ldots & \ldots & \ldots \\ {{a_{m1}}}&{{a_{m2}}}& \ldots &{{a_{mn}}}&{b_m} \end{array}} \right]\)

C is the augmented matrix of the given system of equations.

Rank of A = Rank of C = no of unknown (consistent and unique solution)

Rank of A = Rank of C < no of unknown (inconsistent and infinite solution)

Rank of A < Rank of C (inconsistent and no solution)

Calculation:

Given:

x + y = 2

2x + 2y = 5

\(C=\begin{bmatrix} 1 & 1 &2 \\ 2& 2 &5 \end{bmatrix}\)

R₂' → R₂ - 2R1

\(C=\begin{bmatrix} 1 & 1 &2 \\ 0& 0 &1\end{bmatrix}\)

rank of C = 2 and rank of A = 1

i.e. Rank of A < Rank of C

∴ the given matrix is inconsistent and has no solution.