| 1. |
Consider the system of equations: x + y = 2 and 2x + 2y = 5. This system has1.One solution<<br>2. No solution3. Infinite solution4. Four solutions |
|
Answer» Correct Answer - Option 2 :
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 = bm 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. |
|