Question

The value of k for which the system of linear
equations 3x-2y = 4; kx + y=3, has a unique
solution is:
a) K = 3/2
(b)k = -3/2 (c)k = - 2/3 (d) k is not equals to -3/2

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Answer 1

Step-by-step explanation:

For unique solution a1/a2.not equal to b1/b2

For unique solution a1/a2.not equal to b1/b2So a1=3,a2=k

For unique solution a1/a2.not equal to b1/b2So a1=3,a2=k B1=-2. B2=1

For unique solution a1/a2.not equal to b1/b2So a1=3,a2=k B1=-2. B2=1 So, 3/k not equal - 2/1

For unique solution a1/a2.not equal to b1/b2So a1=3,a2=k B1=-2. B2=1 So, 3/k not equal - 2/1 Valuw of k not equal - 3/2

For unique solution a1/a2.not equal to b1/b2So a1=3,a2=k B1=-2. B2=1 So, 3/k not equal - 2/1 Valuw of k not equal - 3/2Answer d is correct

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Answer 2

$\fontsize{18}{10}{\textup{\textbf{Correct option is (d). }}}-\dfrac{3}{2}.$

Step-by-step explanation:   We know that a system of non homogeneous linear equations Ax = B has a unique solution if the determinant of matrix A is not equal to 0.

The given system is

$3x-2y=4,\\\\kx+y=3.$

Writing in matrix form, Ax = B, we have

$A=\left[\begin{array}{ccc}3&-2\\4&5\end{array}\right] ,~~~~~x=\left[\begin{array}{ccc}x\\y\end{array}\right],~~~~~B=\left[\begin{array}{ccc}4\\3\end{array}\right].$

Therefore, the system will have unique solution for

$|A|\neq0\\\\\\\Rightarrow \left[\begin{array}{ccc}3&-2\\k&1\end{array}\right] \neq0\\\\\\\Rightarrow 3+2k\neq0\\\\\Rightarrow 2k\neq-3\\\\\Rightarrow k\neq-\dfrac{3}{2}.$

Thus, the required value of k is $-\dfrac{3}{2}.$

Option (d) is CORRECT.