find the value of k for which the following equation system has a unique solution:
kx+2y=5,3x+y=1
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kx+2y=5,3x+y=1
a1 = k b1 = 2 and c1 = -5
a2=3 b2 = 1 c2 = -1
after using matrix method we get
x= (b1c2-b2c1)/a1b2-a2b1)
x= -2-5/k-6
for y = (c2a1 -c1a2) / (a1b2-a2b1)
y= -k+15/k-6
put the values of x and y in second eaquation
3x + y = 1
3(-7/k-6) + (15-k)/k-6 = 1
-21 + 15 - k = k - 6
2k =0 k =0 ans i hope it is valid
a1 = k b1 = 2 and c1 = -5
a2=3 b2 = 1 c2 = -1
after using matrix method we get
x= (b1c2-b2c1)/a1b2-a2b1)
x= -2-5/k-6
for y = (c2a1 -c1a2) / (a1b2-a2b1)
y= -k+15/k-6
put the values of x and y in second eaquation
3x + y = 1
3(-7/k-6) + (15-k)/k-6 = 1
-21 + 15 - k = k - 6
2k =0 k =0 ans i hope it is valid
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