I can explain for you in another way: If k=1 or 2, then the solution of the system will give infinitely many solutions. Otherwise, the solution of the system will be no solution. So, there is no k that will make this system a unique solution (one solution)!
@@Mulkek But when you put k=1 in no solution you get -6 as B value which satisfy it (000/-6) and same goes if you put k=2 you get -12 (000/-12). So K can equal 1,2. Or am I doing the math wrong.
@@gracedavay9932 See, if you want to substitute k=1, you have to substitute this in the last augmented matrix to check which solution you will get. For example, If k=1, you have to substitute this with the last row in the last augmented matrix which we have reached [0 0 0|-k^2 + 3k -2]. So, you get [0 0 0|-(1)^2 + 3(1) -2] = [0 0 0|-1+3-2] = [0 0 0|0] and this is infinitely many solution case. If k=2, again you have to substitute this with the last row in the last augmented matrix which we have reached [0 0 0|-k^2 + 3k -2]. So, you get [0 0 0|-(2)^2 + 3(2) -2] = [0 0 0|-4+6-2] = [0 0 0|0] and this is again infinitely many solution case. If you need any more illustrations let me know 😇
Wonderful. 👍👍
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you saved me you have no idea
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Why k can't be equal to 1 and 2 in No solution..?
I can explain for you in another way:
If k=1 or 2, then the solution of the system will give infinitely many solutions.
Otherwise, the solution of the system will be no solution.
So, there is no k that will make this system a unique solution (one solution)!
@@Mulkek But when you put k=1 in no solution you get -6 as B value which satisfy it (000/-6) and same goes if you put k=2 you get -12 (000/-12). So K can equal 1,2. Or am I doing the math wrong.
@@gracedavay9932 See, if you want to substitute k=1, you have to substitute this in the last augmented matrix to check which solution you will get. For example,
If k=1, you have to substitute this with the last row in the last augmented matrix which we have reached [0 0 0|-k^2 + 3k -2].
So, you get [0 0 0|-(1)^2 + 3(1) -2] = [0 0 0|-1+3-2] = [0 0 0|0] and this is infinitely many solution case.
If k=2, again you have to substitute this with the last row in the last augmented matrix which we have reached [0 0 0|-k^2 + 3k -2].
So, you get [0 0 0|-(2)^2 + 3(2) -2] = [0 0 0|-4+6-2] = [0 0 0|0] and this is again infinitely many solution case.
If you need any more illustrations let me know 😇
@@Mulkek Thank you, this is going to save me on my test later on today.
@@gracedavay9932 Glad it helped, and good luck 👍