Question

for what value of k system of linear equations x-3y+kz=0, 2x+y+3z=-1, 3x-2y+2z=1 has a unique solution​

Answer 1

Given : system of linear equations

x-3y+kz  =  0

2x+y+3z =  -1

3x-2y+2z =  1

To Find :   value of k for unique solution

Solution:

x-3y+kz  =  0

2x+y+3z =  -1

3x-2y+2z =  1

$\left[\begin{array}{ccc}1&-3&k\\2&1&3\\3&-2&2\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}0\\-1\\1\end{array}\right]$

AX = B

X = A⁻¹B

unique solution would exist if inverse of A exists

Hence | A | ≠ 0

1(2 + 6) + 3(4 - 9) + k (-4 - 3)   ≠ 0

=> 8 - 15  - 7k ≠ 0

=> -7k  ≠ 7

=>  k  ≠  -  1

value of k system of linear equations x-3y+kz=0, 2x+y+3z=-1, 3x-2y+2z=1 has a unique solution​  is  k  ≠  -  1

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