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If the following equations are consistent, then value of k is
4x – 3y + 1 = 0
kx – 8y + 10 = 0
x + y – 5 = 0
1. -6
2. 7
3. 6
4. -7

1 Answer

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Best answer
Correct Answer - Option 2 : 7

Concept: 

If the system of equation a1x + b1y + c1 = 0,  a2x + b2y + c2 = 0, a3x + b3y + c3 = 0 be consistent, then

\(\begin{vmatrix} a_1 &b_1 & c_1\\ a_2 &b_2 & c_2 \\ a_3 &b_3 & c_3 \end{vmatrix} = 0\)

Calculation:

Given:

\(\begin{vmatrix} 4 &-3 & 1\\ k&-8 & 10 \\ 1 &1 & -5 \end{vmatrix} = 0\)

⇒ 4(40 - 10) - (-3)(-5k - 10) + 1(k + 8) = 0

⇒ 120 - 15k - 30 + k + 8 = 0

⇒ 98 - 14k = 0

∴ k = 7

Consider the system of m linear equations

a11 x1 + a12 x2 + … + a1n xn = b1

a21 x1 + a22 x2 + … + a2n xn = b2

am1 x1 + am2 x2 + … + amn xn = bm

The above equations contain the n unknowns x1, x2, …, xn. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrices.

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

A is the coefficient matrix and [A|B] is called an augmented matrix of the given system of equations.

We can find the consistency of the given system of equations as follows:

(i) If the rank of matrix A is equal to rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.

The rank of A = Rank of augmented matrix = n

(ii) If the rank of matrix A is equal to rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.

The rank of A = Rank of augmented matrix < n

|A| = 0

(iii) If the rank of matrix A is not equal to rank of the augmented matrix, then the system is inconsistent, and it has no solution.

The rank of A ≠ Rank of an augmented matrix

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