### Video Transcript

Find the inverse of the three-by-three matrix π to the power of π‘, cos π‘, sin π‘, π to the power of π‘, negative sin π‘, cos π‘, π to the power of π‘, negative cos π‘, negative sin π‘.

In this question, weβre given a three-by-three matrix and weβre asked to find the inverse of this matrix. And whenever weβre asked to find the inverse of a matrix, this means the multiplicative inverse of our matrix. We have a few different methods of doing this. In this video, weβre going to use the adjoint method. So letβs start by recalling the steps we need to find the inverse of a matrix by using the adjoint method. And to make things easier, weβre going to call the matrix given to us in the question the matrix π΄.

We recall there are five steps to finding the inverse of a matrix by using the adjoint method. First, we need to check that the determinant of our matrix is not equal to zero. And this is because if the determinant is equal to zero, then we know our matrix is not invertible. And also, weβre going to need to use it later to find the inverse. The second step is to find the determinants of all of the matrix minors in our matrix π΄. These are represented by π΄ sub ππ. And remember, the matrix minor π΄ ππ is the matrix we get by removing row π and column π from our matrix π΄. The third step is to construct the cofactor matrix of matrix π΄.

The element in row π and column π of our cofactor matrix of π΄ is given by negative one to the power of π plus π multiplied by the determinant of our cofactor matrix π΄ sub ππ. The fourth step is to construct the adjoint matrix of matrix π΄. And we get this just by taking the transpose of our cofactor matrix. And remember, to find the transpose of a matrix, we need to switch the rows and columns around. Finally, we can construct the inverse matrix of matrix π΄ by multiplying the adjoint matrix of π΄ by one divided by the determinant of π΄.

So to find our inverse, letβs take this one step at a time. Letβs start by finding the determinant of matrix π΄. We need to find the determinant of our three-by-three matrix π΄. We can see this is a very complicated-looking matrix, and thereβs a few different options we could use. If we wanted to do this by expanding over our row or our column, we would usually do this by finding the row or column with the most number of zeros. However, this matrix doesnβt have any. Instead, we need to notice that all three entries in our first column are the same. So if we expand over this column, weβll have a shared factor that we can take out. So weβll find this determinant by expanding over the first column.

Letβs start with the first element in the first column. Weβre going to need to multiply this by the determinant of the matrix minor we get by removing the first row and the first column. And donβt forget we also get a coefficient of negative one raised to the power of the row number plus the column number of the entry weβre expanding over. In this case, weβre expanding over the entry in row one, column one. So we get a coefficient of negative one to the power of one plus one. We then want to do exactly the same for the second entry in our column. We want to multiply this by the determinant of the matrix minor we get by removing the first column and the second row.

And donβt forget we also want to multiply this by negative one raised to the power of the row number plus the column number of the entry weβre expanding over. In this case, thatβs the first column in the second row. And we need to do this one more time for the last entry in our column. We need to multiply this by the determinant of the matrix minor we get by removing the first column in the third row. And of course, we still need to multiply this by negative one raised to the power of the column number plus the row number of the entry weβre expanding over. In this case, thatβs the entry in column one, row three. This gives us the following expression for the determinant of matrix π΄.

And we can start by simplifying the sine of each of our three terms. We get one, negative one, and one, respectively. This gives us the following expression. Next, we see we can take out the shared factor of π to the power of π‘. And this gives us the following expression. Now, all we need to do is evaluate all three of our two-by-two determinants. And to do this, we recall the determinant of the two-by-two matrix, π, π, π, π, is equal to ππ minus ππ. We find the difference between the products of the diagonals. And we can start by applying this to our first term. We get negative the sin of π‘ multiplied by negative the sin of π‘ minus the cos of π‘ multiplied by negative the cos of π‘. And instead of carrying on, we can simplify this expression now. It simplifies to give us the sin squared of π‘ plus the cos squared of π‘.

However, we actually know that this is an identity. Itβs exactly equal to one for all values of π‘. So the determinant of the matrix in our first term gave us one. Letβs now calculate the determinant of the matrix in our second term. Using our formula and remembering that we need to subtract this value, we get the cos of π‘ multiplied by negative the sin of π‘ minus the sin of π‘ multiplied by negative the cos of π‘. And just like we did before, we could simplify this now. It simplifies to give us negative sin π‘ times the cos of π‘ plus the sin of π‘ multiplied by the cos of π‘. Of course, this is just equal to zero.

Finally, all we need to do is calculate the determinants of the matrix in our third term. We find the difference in the products of the diagonals and simplify. This gives us the cos squared of π‘ plus the sin squared of π‘. And once again, by using the exact same identity, we know that this is equivalent to one for all values of π‘. So we can replace this entire expression with one, giving us the determinant of matrix π΄ is equal to π to the power of π‘ multiplied by one minus zero plus one. And we can calculate this expression. Itβs equal to two times π to the power of π‘. And thereβs something very important to realize about this. We know itβs positive for all values of π‘. Remember, our matrix will not be invertible if the determinant of our matrix was equal to zero.

So because weβve shown our determinant is positive for all values of π‘, weβve also proven that our matrix will be invertible for all values of π‘. However, this is only the first step in finding the inverse of this matrix, so letβs clear some space for the rest of our working and keep in mind the determinant of matrix π΄ is two π to the power of π‘. The second step we need to do is find the determinants of all of the matrix minors of our matrix π΄. Letβs start by finding the matrix minor π΄ one, one. Remember, the matrix minor π΄ one, one is the matrix we get by removing the first row and first column from matrix π΄. And we need to find the determinant of this. Thatβs the determinant of the two-by-two matrix negative sin π‘, cos π‘, negative cos π‘, negative sin π‘.

And just like we did before, to find the determinant of this matrix, we need to find the difference in the product of the diagonals. Simplifying this expression, we get sin squared π‘ plus cos squared π‘, which once again is an identity. Itβs equal to one for all values of π‘. But remember, we need to do this for all of our matrix minors. Letβs now find the determinant of matrix minor π΄ sub one two. This is the matrix minor we get by removing the first row and second column from our matrix π΄. So we need to find the determinant of the two-by-two matrix π to the power of π‘, cos π‘, π to the power of π‘, negative sin π‘.

We do this in the same way we did before. We need to find the difference in the products of the diagonals. Simplifying this, we get negative π to the power of π‘ times sin π‘ minus π to the power π‘ cos π‘. And we can simplify this by taking out the shared factor of negative π to the power of π‘. This gives us the determinant of the matrix minor π΄ sub one two is equal to negative π to the power of π‘ times the sin of π‘ plus the cos of π‘. Weβre going to need to do this for all nine of our matrix minors. So letβs clear some space and rewrite down the determinants of the two matrix minors weβve found so far. And by doing the exact same method and simplifying all of our expressions, we can calculate the determinants of the remaining seven matrix minors. We get the following expressions.

So thatβs our second step done. We found the determinants of all nine matrix minors in matrix π΄. And remember, we needed all of these to construct the cofactor matrix of matrix π΄. Each entry in our cofactor matrix is going to be the determinant of one of these matrix minors. But then we multiply this by negative one to the power of the row number plus the column number. So in row one, column one of our cofactor matrix, weβre going to get negative one to the power of one plus one because this is row one and column one. And we need to multiply this by the determinant of the matrix minor π΄ sub one one, which weβve shown is equal to one. This gives us negative one to the power of one plus one times one, which of course is just equal to one.

So we found the entry in row one, column one of our cofactor matrix. Letβs now find the entry in row one, column two. This time, we get negative one to the power of one plus two multiplied by the determinant of the matrix minor π΄ sub one two, which is negative π to the power π‘ times the sin of π‘ plus the cos of π‘. And we can simplify this. Negative one to the power of one plus two is equal to negative one. And if we multiply this by our other factor of negative one, we just get one. This gives us the entry in row one, column two of our cofactor matrix is π to the power of π‘ times the sin of π‘ plus the cos of π‘.

Weβre now going to want to do exactly the same for the entry in row one, column three. This time, we get negative one to the power of one plus three multiplied by the determinant of the matrix minor π΄ sub one three, which is π to the power of π‘ times the sin of π‘ minus the cos of π‘. And of course, negative one to the power of one plus three is just equal to one. So this just simplifies to give us π to the power π‘ times the sin of π‘ minus the cos of π‘. And weβll repeat this process to find all of the entries in our cofactor matrix. This gives us the following three-by-three matrix for our cofactor matrix πΆ. And by finding the cofactor matrix of matrix π΄, weβve now successfully done the third step, which means we now need to move on to the fourth step to find the inverse of matrix π΄.

We now need to construct the adjoint matrix of π΄, and we do this by finding the transpose of matrix πΆ. Remember, to find the transpose of a matrix, we need to switch the rows and columns around. So letβs find the transpose of our cofactor matrix πΆ. Weβll start by writing the first column as the first row. We need to rewrite our first column of one, zero, one as the first row in our new matrix, one, zero, one. We then want to do the same with the second column of our matrix. We need to write this as the second row in our transpose. This gives us the second row of our adjoint matrix is π to the power of π‘ times sin π‘ plus cos of π‘, negative two π to the power of π‘ times sin π‘, π to the power of π‘ multiplied by the sin of π‘ minus the cos of π‘.

Finally, weβre going to do exactly the same with the last column of our matrix πΆ. And this gives us the following matrix, which is the transpose of our cofactor matrix. And remember, this is also the adjoint matrix of our matrix π΄. And remember, to find the inverse of matrix π΄, all we needed to do was find the adjoint matrix of matrix π΄ and the determinants of matrix π΄. And weβve done both of these. So now we can just find an expression for the inverse of matrix π΄. We get the multiplicative inverse of matrix π΄ is going to be equal to one over the determinant of π΄ multiplied by the adjoint of matrix π΄. And we showed the determinant of matrix π΄ is two π to the power of π‘.

And we found the adjoint matrix of π΄. However, itβs not necessary to write this matrix out in full again. Instead, we need to notice weβre multiplying every single entry of our adjoint matrix by one divided by two π to the power of π‘. This means all we need to do is divide all of the entries of our adjoint matrix by two π to the power of π‘. Weβll start with the entry in row one, column one. We need to divide one by two π to the power of π‘. This just gives us one over two π to the power of π‘. But weβll rewrite this as one-half times π to the power of negative π‘. We could do the same to find the entry in row one, column two. We would need to divide zero by two by two π to the power π‘. However, of course, this is just equal to zero.

The entry in row one, column three is exactly the same as the entry in row one, column one. So weβll also end up with one-half times π to the power of negative π‘. Next, weβll find the entry in row two, column one of our inverse matrix. We need to divide π to the power of π‘ times the sin of π‘ plus the cos of π‘ by two π to the power of π‘. And weβll simplify this to be one-half multiplied by the sin of π‘ plus the cos of π‘. And by following the exact same process and simplifying, we can find all the remaining entries of an inverse. And this gives us our final answer. And it is worth pointing out here we could check our answer by multiplying it by our matrix π΄ and checking that we get the three-by-three identity matrix.

Therefore, we were able to find the multiplicative inverse of the matrix π΄ given to us in the question. And we also showed that this will be true for any value of π‘. We were able to show that π΄ inverse is equal to the three-by-three matrix one-half π to the power of negative π‘, zero, one-half π to the power of negative π‘, one-half times sin π‘ plus cos π‘, negative sin π‘, one-half multiplied by sin of π‘ minus cos of π‘, one-half sin π‘ minus cos π‘, cos π‘, negative one-half times sin π‘ plus cos π‘.