### Video Transcript

In this video, weβre going to work with matrices and check out a claimed proof that the matrix one, one, zero, zero is equal to the matrix one, zero, zero, one.

So letβs start off by defining matrix π΄ to be one, one, zero, zero. And if I multiply π΄ by itself, first I multiply one by one, and then I add one times zero to get this term here. Then I multiply one by one and add one times zero to get this term here. Then zero times one and zero times zero to get this term here. And lastly zero times one and zero times zero to get this term here.

Well one times one is one. One times zero is zero. One and zero is one. And one times one is one. One times zero is zero. One and zero is one. Zero times one is zero. Zero times zero is zero. Zero plus zero is zero. And so again, zero times one is zero. Zero times zero is zero. Zero plus zero is zero.

So that means if I multiply matrix π΄ by matrix π΄, by itself in other words, I get the result which is matrix π΄. Now I can premultiply both sides of my equation by the inverse matrix of π΄. And because matrix multiplication is associative, it doesnβt matter whether I multiply π΄ inverse by the result of π΄ times π΄ or whether I multiply the result of π΄ inverse π΄ by matrix π΄. In either case, weβll get the same result.

Now the definition of the inverse matrix of π΄ is that when I pre multiply π΄ by its inverse, I get the identity matrix one, zero, zero, one. And Iβm doing that here and over here. So the identity matrix times π΄ is equal to the identity matrix. And because premultiplying by the identity matrix is a bit like multiplying a number by one, it gives a result thatβs equal to the original matrix, that means that π΄ is equal to πΌ.

And through a series of logical steps, weβve proved that the matrix one, one, zero, zero is equal to the matrix one, zero, zero, one. Well that was all very plausible, but I can assure you that the matrix one, one, zero, zero is not equal to the matrix one, zero, zero, one. So why donβt you pause the video now. Have a look through. And see if you can work out whereabouts our logic broke down. Right! First of all, Iβm just gonna tidy this up a little bit to make some space for us to write some more things.

Okay, so our first line says that π΄ is defined as being the matrix one, one, zero, zero. Well thatβs a perfectly fine matrix so thereβs no problems there. And in the next line we multiply matrix π΄ by itself and get the result one, one, zero, zero. Well the method for multiplying two two-by-two matrices together is to take this term and multiply by this term and then add this term multiplied by this term. And this gives us this term in the answer, which is π times π plus π times π.

Then to get this term over here, I do π times π plus π times β. Then to get this term here, I do π times π plus π times π. Then to get this term here, I do π times π plus π times β. And indeed when I apply that logic to one, one, zero, zero times one, one, zero, zero, I do get one times one plus one times zero, one times one plus one times zero, zero times one plus zero times zero, and zero times one plus zero times zero, which is indeed one, one, zero, zero. So that second line is also correct.

Then just representing the matrices by their letters, we can see that π΄π΄ or π΄ times π΄ does in fact equal π΄. Now in the next line we start talking about π΄ inverse, π΄ to the power of minus one, the inverse matrix of π΄. And the meaning of the word inverse here is the matrix that I need to multiply π΄ by in order to get the identity matrix one, zero, zero, one. So if our matrix π΄ starting off as being π, π, π, π, the inverse of π΄ is the inverse of this matrix. And this turns out to be one over ππ minus ππ times the matrix π, negative π, negative π, π.

Now ππ minus ππ is a value that we called the determinant of the matrix. And thatβs quite easy to calculate for a two-by-two matrix. Itβs a bit more complicated for bigger matrices. But weβll just stick to a two-by-two matrix for now. And then within the matrix itself, you can see that weβve swapped the π and the π over here. And weβve taken the negative of π and π. So when we define the inverse matrix in that way, we find that π΄ inverse times π΄ gives us the identity matrix one, zero, zero, one.

Letβs run through and see it in action. With π΄ being defined as matrix π, π, π, π and the inverse of π΄ being one over the determinant times π, negative π, negative π, π, weβve got π΄ inverse π΄ is equal to all that lot. Now this is just a constant times every term in the resultant matrix. So letβs multiply these two matrices together using the method that we talked about earlier.

Weβre gonna have to do π times π add negative π times π for this term up here. Then itβs going to be π times π add negative π times π for this term over here. Then negative π times π add π times π for this term down here. And finally, negative π times π plus π times π down here. Now π times π is the same as π times π. And then if weβre adding negative π times π, thatβs the same as taking away π times π. So I can rewrite that term as ππ minus ππ.

Then ππ, adding the negative of π times π, well thatβs the same as π times π take away π times π, which is nothing or zero. And then negative ππ plus ππ, well π times π is the same as π times π. So this is negative ππ plus ππ. That again is zero. And then down here, negative ππ is the same as negative ππ. And if I write these two in a different order, I get ππ minus ππ. And now, I have to multiply each of the terms in that matrix by the constant term outside.

And of course zero divided by ππ minus ππ is just zero. And ππ minus ππ divided by ππ minus ππ is just one. So in general, yes the inverse of π΄ times π΄ does give us this identity matrix, πΌ: one, zero, zero, one. But letβs take a look at the specific case when the matrix π΄ is one, one, zero, zero. Then the inverse of π΄ is going to be one over the determinant times β well I need to swap over the zero and the one here. And I need to take the negative of the one and the zero here.

Well of course, negative zero is just zero. So Iβve just gonna write that in here and then I can work out what the determinant is. Well itβs one times zero minus one times zero. And one times zero is zero. So that becomes one over zero minus zero. And obviously zero minus zero is zero. And weβve just set off the dividing-by-zero alarm! One divided by zero is undefined. So weβre trying to multiply this matrix by an undefined number.

It turns out that the inverse of matrix π΄ is undefined. We call π΄ a singular or a degenerate matrix. And thatβs just a square matrix that doesnβt have an inverse where the determinant is zero. So going back to our so-called proof, when multiplying by π΄ inverse, weβre multiplying by an undefined number. And from then on in, everything is complete nonsense. Because the inverse of π΄ is undefined, we canβt do further calculations with undefined numbers. We havenβt proved that one, one, zero, zero is equal to one, zero, zero, one after all. It was a fake proof.