Find the values of k which give, one Solution, no Solution, or infinitely Many Solutions

❖ In this problem, we have tried to find the values of unknown constant k, if any, will give one solution, no solution infinitely, many solutions to the system of equations.
This video presents linear algebra in the School of Mathematics and Statistics which is in the Faculty of Science in most universities.
The link to this playlist (Linear Algebra):
• Linear Algebra
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[@danielnyamuche2639]

Wonderful. 👍👍

[@Mulkek]

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[@maryam1777]

Legend 💯

[@Mulkek]

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[@bonganipeter9433]

Thanks man.

[@Mulkek]

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[@justicetom5768]

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[@Jakxb_S]

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[@aymoonnah4913]

So useful ❤

[@Mulkek]

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[@jacksmith1664]

doing gods work

[@Mulkek]

Wishing you the best & if you're struggling with the material here, I'm happy to help

[@theesanshaik393]

thanks alot!!! :)

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[@naitikrathore3317]

Why k can't be equal to 1 and 2 in No solution..?

[@Mulkek]

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)!

[@gracedavay9932]

@@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.

[@Mulkek]

​@@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 😇

[@gracedavay9932]

@@Mulkek Thank you, this is going to save me on my test later on today.

[@Mulkek]

@@gracedavay9932 Glad it helped, and good luck 👍