WEBVTT
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Calculate the eigenvalues of the matrix π΄ equals negative one, zero, zero, negative one.
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Well, first when we take a look at our matrix, we can see that itβs a two-by-two matrix.
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Well, we know that an π by π matrix will always have π eigenvalues.
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So therefore, we know that with our matrix being a two-by-two matrix, weβre gonna have two eigenvalues.
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Okay, great, so we know thatβs what weβre looking for.
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But what can we do to work out the eigenvalues?
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So to help us solve this problem and find our eigenvalues, we have a relationship.
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And that one is that the determinant of the matrix π΄ minus ππΌ, where πΌ is the identity matrix β β well, this is equal to zero.
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And Itβs the π in our relationship thatβs gonna help us calculate our eigenvalue.
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So therefore, using our relationship, we can say that the determinant of the matrix π΄, which is negative one, zero, zero, negative one, minus π multiplied by the identity matrix, which is the matrix one, zero, zero, one, is just gonna be equal to zero.
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And as we said, the identity matrix is one thatβs in this form.
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But we have these terms as one.
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So the diagonal running from the top left to bottom right is one.
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So therefore, if we multiplied by the scaler, which was our π , weβre gonna get negative one, zero, zero, negative one minus π, zero, zero, π.
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And we want to find the determinant of this.
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So once we carried out the subtraction of our matrices, weβre gonna get the determinant of the matrix.
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Negative one minus π, zero, zero, negative one minus π is equal to zero.
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So when we look at a two-by-two matrix and we want to find the determinant of that, if you remind ourselves what we do, well, we multiply diagonally and then subtract.
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So we have ππ minus ππ.
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So if we have the matrix π, π, π, π the determinant of that is ππ minus ππ.
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So therefore, if we do that, weβre gonna get negative one minus π multiplied by negative one minus π minus zero equals zero.
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So then, if we distribute the first parentheses over the second parentheses, weβre gonna get β well, first of all, weβre gonna get one plus π.
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Thatβs cause negative one multiplied by negative one is one and negative one multiplied by negative π is just π.
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So itβs one plus π.
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Then weβve got plus another π and then plus π squared.
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Then this is equal to zero.
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So then if we tidy this up, we get π squared plus two π plus one equals zero.
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So now what we need to do is solve to find π cause itβs gonna give us our eigenvalues.
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Well now to solve this, what we can do is factor because we can factor our expression because weβve got π squared plus two π plus one.
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Well, we need two values.
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Theyβre gonna multiply together to give us one and add together to give us two.
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So therefore, weβre gonna get π plus one multiplied by π plus one is equal to zero.
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And thatβs because if we have one multiplied by one is one; one add one is two.
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So they satisfy this.
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So now what we need to do is solve this to find our π value.
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Well, what weβre gonna do is set one of our parentheses equal to zero.
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And thatβs because to get a result of zero, we need to have zero multiplied by something to get zero.
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So Iβm gonna have π plus one equals zero.
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So therefore, π is gonna be equal to negative one.
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And actually, as you can see in this one, weβve got π plus one multiplied by π plus one.
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This is same as π plus one squared.
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So weβve got repeated roots.
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However, we were expecting two eigenvalues cause we showed that earlier.
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There should be two eigenvalues for two-by-two matrix.
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But weβve got a repeated eigenvalue, so repeated root.
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So therefore, what we say is that our eigenvalue, or eigenvalues, is negative one.
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But negative one is degenerate.
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And itβs degenerate because both of our eigenvalues are identical because weβve got, as we said, repeated roots.