Correct Answer - Option 1 : 1
Concept:
For at least one solution solve the given equation for the unique value.
Given:
x + y = 2 ------ (1)
kx + y = 4 ------ (2)
x + ky = 5 ------ (3)
Calculation:
After adding equation (2) & (3),
kx + y + x + ky = 4 + 5
k(x + y) + (x + y) = 9
(k + 1)(x + y) = 9 ------ (4)
And, We have
x + y = 2
From equation (4)
(k + 1) × 2 = 9
k = 3.5
Then,
For the unique solution K have only one value.
Let us consider a system of equations in three variables:
a1 × x + b1 × y + c1 × z = d1
a2 × x + b2 × y + c2 × z = d2
a3 × x + b3 × y + c3 × z = d3
Then, \({\rm{\Delta }} = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,\;\;{{\rm{\Delta }}_1} = \left| {\begin{array}{*{20}{c}} {{d_1}}&{{b_1}}&{{c_1}}\\ {{d_2}}&{{b_2}}&{{c_2}}\\ {{d_3}}&{{b_3}}&{{c_3}} \end{array}} \right|,\;{{\rm{\Delta }}_2} = \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{d_1}}&{{c_1}}\\ {{a_2}}&{{d_2}}&{{c_2}}\\ {{a_3}}&{{d_3}}&{{c_3}} \end{array}} \right|\;and\;{{\rm{\Delta }}_3} = \;\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{d_1}}\\ {{a_2}}&{{b_2}}&{{d_2}}\\ {{a_3}}&{{b_3}}&{{d_3}} \end{array}} \right|\)
By cramer’s rule:
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If Δ ≠ 0, then the system of the equation has a unique solution and it is given by: \(x = \frac{{{{\rm{\Delta }}_1}}}{{\rm{\Delta }}},\;y = \frac{{{{\rm{\Delta }}_2}}}{{\rm{\Delta }}}\;and\;z = \frac{{{{\rm{\Delta }}_3}}}{{\rm{\Delta }}}\)
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If Δ = 0 and at least one of the determinants Δ, Δ1, Δ2, and Δ3 is non-zero, then the given system is inconsistent.
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If Δ = 0 and Δ1 = Δ2 = Δ3 = 0, then the system is consistent and has infinitely many solutions.
