Video: Finding the Unknown Coefficient in a Set of Given Equations So That They Have a Solution Other Than Zero

Find the value of π‘˜ that would make the equations 4π‘₯ + 9𝑦 + 5𝑧 = 0, 16π‘₯ + 36𝑦 + π‘˜π‘§ = 0, and 9π‘₯ βˆ’ 8𝑦 βˆ’ 3𝑧 = 0 have a solution other than zero.

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Video Transcript

Find the value of π‘˜ that would make the equations four π‘₯ plus nine 𝑦 plus five 𝑧 equals zero, 16π‘₯ plus 36𝑦 plus π‘˜π‘§ equals zero, and nine π‘₯ minus eight 𝑦 minus three 𝑧 equals zero have a solution other than zero.

So, in this question, what we need to do is find the value of π‘˜ that means that our three equations have a solution other than zero. So, there’re three typical scenarios that could occur with the solutions of our equations. Now there are infinite solutions if one equation can be deduced from another. If two of our equations conflict, then there’re gonna been no solutions. And finally, if any cannot be deduced from each other, and there are no conflicts, then there will be a unique solution.

Well, we want there to be a solution. So therefore, scenario two is not what we want because we don’t want any conflicts. Because if there’re two conflicts, then there’re no solutions. And if we’re wondering what a conflict is, here’s an example of a conflict. If we had one equation that said that π‘Ž plus 𝑏 was equal to four, but another equation said that π‘Ž plus 𝑏 would be equal to seven, then there cannot be any solutions because this cannot be true. Because these two are conflicting.

So, now if you want to see if there any of our equations can be deduced from each other, we can take a look at the first and second equations. Because from inspection, we can see that they probably can be deduced from each other. And that’s because we can see that if we multiply the first equation by four, we can get the first two terms of the second equation. For example, four multiplied by four gives us 16, so we get from four π‘₯ to 16π‘₯. Nine multiplied by four is 36, so we get from nine 𝑦 to 36𝑦. So therefore, let’s have a look at what happens if we rearrange our equations.

We’ve now got four π‘₯ plus nine 𝑦 equals negative five 𝑧. And then, we have 16π‘₯ plus 36𝑦 equals negative π‘˜π‘§. Now we’ve already said that the first equation multiplied by four gives us the second equation. So therefore, if we divide the second equation by four, this would give us the first equation. So, if we do that, we get four π‘₯ plus nine 𝑦, as we wanted, but then is equal to negative π‘˜ over four 𝑧. That’s because we had 16π‘₯ plus 36𝑦 equals negative π‘˜π‘§. And we’ve divided each of the terms by four.

So therefore, the corresponding values in the right-hand side are negative five and negative π‘˜ over four. So, we can say that negative five must be equal to negative π‘˜ over four, to avoid a conflict. Because if these weren’t equal, then we’d have four π‘₯ plus nine 𝑦 being equal to different things. So therefore, we’d have our conflict. And therefore, there’ll be no solutions.

So, now what we have is an equation that we can solve to find π‘˜. Cause we have negative five is equal to negative π‘˜ over four. So then, what we do is multiply each side of the equation by four. So, we get negative 20 is equal to negative π‘˜. So then, if we divide each side of the equation by negative one, and we just flip it round the other side to have the π‘˜ on the left-hand side, we get π‘˜ is equal to 20.

So therefore, we can say that it, the value of π‘˜ that would make the equations four π‘₯ plus nine 𝑦 plus five 𝑧 equal zero, 16π‘₯ plus 36𝑦 plus π‘˜π‘§ equal zero, and nine π‘₯ minus eight 𝑦 minus three 𝑧 equal zero have a solution other than zero is π‘˜ is equal to 20.

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