# Video: Using the Cayley-Hamilton Theorem to Find the Inverse of Three-by-Three Matrices

Using the Cayley-Hamilton theorem, find 𝐴⁻¹, if possible, for the matrix 𝐴 = [2, 3, 4 and −5, 5, 6 and 7, 8, 9].

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### Video Transcript

Using the Cayley–Hamilton theorem, find the inverse of 𝐴, if possible, for the matrix 𝐴 equals two, three, four, negative five, five, six, seven, eight, nine.

So if we want to use the Cayley–Hamilton theorem to find inverse for our matrix, then the first thing we need to do is find the characteristic polynomial. So if we want to find the characteristic polynomial of our matrix, then this is gonna be equal to the determinant of 𝐴 minus 𝜆𝐼, where 𝐼 is the identity matrix. And the identity matrix is a matrix whose top left to bottom right terms diagonally are all one. Then, all the other values are zero. So therefore, we can say that the characteristic polynomial is gonna be equal to the determinant of the matrix two minus 𝜆 three, four, negative five, five minus 𝜆, six, seven, eight, nine minus 𝜆.

And we have those terms in our diagonal because originally we could see that the characteristic polynomial is equal to the determinant of 𝐴 minus 𝜆𝐼. And 𝜆𝐼 is gonna be 𝜆 multiplied by our identity matrix. And our identity matrix only has terms of any value across the diagonal. So therefore, you’re gonna have 𝜆 multiplied by one for each of those terms. And that just gives us 𝜆, which we subtract from the original terms. So now, what we need to do is find the determinant of our three-by-three matrix. And we’re gonna use our first row as coefficients to do this. And what we’re gonna do is multiplying out using those.

It’s worth noting at this point that each column has a different sign. So first column positive, second column negative, third column positive. And what this means in practice is that the sign of the coefficient in the positive columns will remain the same. But in the negative column, it will change. So first of all, what we’re gonna have is two minus 𝜆. So that’s our first value in the first row multiplied by the determinant of the submatrix. And that’s a two-by-two submatrix, which is formed if we delete the column in the row that two minus 𝜆 are in. And if we do that, we get the submatrix five minus 𝜆, six, eight, nine minus 𝜆.

So then, what we’re gonna have is minus three multiplied by the determinant of the submatrix negative five six seven nine minus 𝜆. And again, we got this by deleting the row in column that the three were in. It’s also worth noting the sign is negative three because it’s our second column. And then, finally, we have plus four multiplied by the determinant of the submatrix negative five, five minus 𝜆, seven, eight. Again, we found this exact the same way because we deleted the row in column that the four were in. And we can also see here that the four has a positive sign. Well, great. Well, now, what’s the next stage? What we need to do now?

So now, what we need to do is find the determinants of our two-by-two submatrices. And to do that, what we do is if we have a matrix 𝑎, 𝑏, 𝑐, 𝑑 and we want to find the determinant, then we multiply the diagonals, so 𝑎𝑑 and 𝑏𝑐. And what we do is we subtract 𝑏𝑐 from 𝑎𝑑. So first of all, we’re gonna have this two minus 𝜆 multiplied by. Then, we’ve got five minus 𝜆 and multiplied by nine minus 𝜆. So that’s like our 𝑎 and 𝑑 and then minus 48. And that’s because six multiplied by eight is 48. So then, next, we’re gonna have minus three multiplied by and we got negative five multiplied by nine minus 𝜆 and then minus 42. Again, we get that from a cross multiplying. So then, we’ve got finally positive four multiplied by. Then, we’ve got negative 40 which is our negative five multiplied by eight minus seven multiplied by five minus 𝜆.

Okay, great. So we’ve got to this stage. We can also see that we’ve been careful with our coefficients when they’re positive, negative, and positive throughout. So now, what we need to do is work this out by distributing across our parentheses. And the first thing we need to do is distribute across the parentheses five minus 𝜆 and nine minus 𝜆. So what I’m gonna do is distribute five minus 𝜆 across that. And what we’re gonna do here is multiply each term in the left-hand parentheses by each term in the right-hand parentheses.

And when we do that, we’re gonna get 45. So that’s five multiplied by nine minus five 𝜆, which is five multiplied by negative 𝜆 minus nine 𝜆 plus 𝜆 squared. Be careful here. You’ve got negative 𝜆 multiplied by negative 𝜆. That gives us positive 𝜆, so positive 𝜆 squared. So now, what we’re gonna do is quickly simplify. And when I do that, I get 45 minus 14𝜆 plus 𝜆 squared. And then, what I’m gonna do is subtract 48 from this. And if we subtract 48 from 45, we get negative three. So therefore, when we have done this, if I rearrange, we’re gonna have 𝜆 squared minus 14𝜆 minus three.

So now, what we need to do is to find a part for this term which is multiply this through by two minus 𝜆. So what I’m gonna do is distribute this across the parentheses. So if I do this, first of all, we’re gonna get two 𝜆 squared minus 28𝜆 minus six then minus 𝜆 cubed plus 14𝜆 squared plus three 𝜆. And if we simplify this, we’re gonna get negative 𝜆 cubed plus 16𝜆 squared minus 25𝜆 minus six. So now, we can move on to our next part.

So then, first of all, for this, what we need to do is distribute negative five across the parentheses nine minus 𝜆. And when we do this, we’re gonna get negative three multiplied by. And then, we’ve got negative 45 and that’s because negative five multiplied by nine is negative 45 plus five 𝜆 and that’s because we got a negative multiplied by a negative minus 42. And then, if we simplify this, we get negative three multiplied by five 𝜆 minus 87. And then, if we multiply through by negative three, we’re gonna get minus 15𝜆 plus 261. Okay, great. Let’s move on to the final part.

So for the final part, what we need to do is distribute negative seven across the parentheses first. And when I do this, I get four multiplied by. Then, we’ve got negative 40 minus 35 plus seven 𝜆. Then, if I tidied it up, we get four multiplied by negative 75 plus seven 𝜆. Then if we distribute the four across the parentheses, we’re gonna get negative 300 plus 28𝜆. Okay, great. So we’ve now finished that stage.

So now, what we can do is simplify the full expression. So first of all, we’ve got negative 𝜆 cubed plus 16𝜆 squared. And that’s because they’re the only cubed and squared terms. Then, we’ve got negative 25𝜆 minus 15𝜆 plus 28𝜆, which is gonna give us negative 12𝜆. And then, finally, we’ve got negative six plus 261 minus 300, which will give us negative 45. So great. Now what we found is our characteristic polynomial. So now, what we’re gonna do is use the Cayley–Hamilton theorem. And we can see that our characteristic polynomial with the matrix 𝐴 substituted in for 𝜆 is gonna be equal to zero. So therefore, we’re gonna get negative 𝐴 cubed plus 16𝐴 squared minus 12𝐴 minus 45𝐼 equals zero, where 𝐴 is our matrix 𝐴 and 𝐼 is our identity matrix.

So now, what we do is we take 𝐴 out as a factor and add 45𝐼 to each side. When we do that, we get 𝐴 multiplied by negative 𝐴 squared plus 16𝐴 minus 12𝐼 equals 45𝐼. We have the 𝐼 with the 12 because we’ve got a constant term. So we need to include our identity matrix at that point. So then, if we divide through by 45, we get 𝐴 multiplied by one over 45 multiplied by negative 𝐴 squared plus 16𝐴 minus 12𝐼 is all equal to 𝐼. So therefore, we can say that the inverse of our matrix 𝐴 is equal to one over 45 multiplied by negative 𝐴 squared plus 16𝐴 minus 12𝐼.

So now, we’re at a stage where we can work out the inverse of our matrix. Well, to work out the inverse for our matrix, what we need to do is work out each part separately. So we’re gonna start with negative 𝐴 squared. And It’s gonna be equal to negative. And then, we’ve got the matrix two, three, four, negative five, five, six, seven, eight, nine multiplied by the matrix two, three, four, negative five, five, six, seven, eight, nine. So now, to work out the first term of our matrix, what we’re gonna do is multiply the terms in the first row of our first matrix by the corresponding terms in the first column of our second matrix.

So for the first term, we’d have two multiplied by two plus three multiplied by negative five plus four multiplied by seven. So then, to find the second term, we’d have two multiplied by three plus three multiplied by five plus four multiplied by eight. And that’s because we moved across the second column. So then, we continue this method until we form the matrix 17, 53, 62, seven, 58, 64, 37, 133, 57. Now, we need to remember that we’ve got a negative in front of this because it needs to be negative of this matrix. So what we’re gonna do is multiple each term by negative one. So this returns us to the same matrix, just with every term being negative.

So then, we’ve got 16𝐴. And 16𝐴 is gonna be equal to 16 multiplied by our matrix 𝐴. So that’s gonna give us the matrix 32, 48, 64, negative 80, 80, 96, 112, 128, 144. And We got that by multiplying each term by 16. And then, finally, we need to find 12𝐼, which is just 12 multiplied by our identity matrix, which is gonna give us 12, zero, zero, zero, 12, zero, zero, zero, 12 as our matrix.

So great, we have all our components. And what we need to do here is put it together. So once, we put them all together, we can calculate to find the inverse for our matrix 𝐴. So then, what we’re gonna do is take each of the corresponding terms from our matrices and then use the operations that we’ve been given. And so, we’re gonna do that to form the top left term, for example. So in this case, we’ve got negative 17 plus 32 minus 12 which will give us three. So then, we carry this on in the same way to find the other missing terms.

So we can say that using the Cayley–Hamilton theorem, the inverse of matrix 𝐴 if matrix 𝐴 is two, three, four, negative five, five, six, seven, eight, nine is one over 45 multiplied by the matrix three, negative five, two, negative 87, 10, 32, 75, negative five, negative 25.