For what value of k is the system of equations KX 3y K 2 12x Ky K inconsistent?

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The system of equations:



is 

(1)  consistent and independent if 

AE ≠ BD 

(2)  consistent and dependent if 

AE = BD and AF = CD



Here A=k, B=3, C=k-2, D=12, E=k, F=k

Since we want them to be inconsistent, we must rule out 
both case (1) and case (2).

To rule out case (1) we require AE ≠ BD to be false, so we must
require:

  AE = BD

k(k) = 3(12)
  k� = 36
   k = �6

But we also must check to see that k = �6 also rules out 
case (2)

That is we must be sure that AF ≠ CD


    AF ≠ CD
  k(k) ≠ (k-2)(12)

�6(�6) ≠ (�6-2)(12)

Taking the + we have

   36 ≠ (6-2)(12)

   36 ≠ (4)(12)

   36 ≠ 48

which is true.

Taking the - we have

   36 ≠ (-6-2)(12)

   36 ≠ (-8)(12)

   36 ≠ -96

which is also true.

So case (2) is ruled out be taking k = �6

It was necessary to rule out case (2), however.

Edwin

SOLUTION

The given system of equations can be written as follows:

kx + 3y – (k – 2) = 0 and 12x + ky – k = 0

The given equations are of the following form:

a1x + b1x + c1 = 0 and a2x + b2y + c2 = 0

Here, a1 = k, b1 = 3, c1 = –(k – 2) and a2 = 12, b2 = k and c2 = –k

∴ 𝑎1/ 𝑎2 = 𝑘 /12 , 𝑏1/ 𝑏2 = 3 /𝑘 and 𝑐1 /𝑐2 = −(𝑘−2) /−𝑘 = (𝑘−2)/ 𝑘

For inconsistency, we must have:

𝑎1/ 𝑎2 = 𝑏1 /𝑏2 ≠ 𝑐1/ 𝑐2 ⇒ 𝑘 /12 = 3/ 𝑘 ≠ (𝑘−2)/ 𝑘

⇒  k2 = (3 × 12) = 36

⇒  k = √36 = ±6

Hence, the pair of equations is inconsistent if k = ±6

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Updated On: 27-06-2022

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