# Question Video: Evaluating an Expression Using the Properties of the Determinant Mathematics

If det (π΄) = 2, det (π΅) = 3, and the size of π΄ and that of π΅ is 2 Γ 2, find the value of det (2π΄) + det (3π΅) + det (π΄π΅) using the properties of determinants.

03:09

### Video Transcript

If the determinant of π΄ is equal to two, the determinant of π΅ is equal to three, and the size of π΄ and that of π΅ is two by two, find the value of the determinant of two π΄ plus the determinant of three π΅ plus the determinant of π΄ times π΅ using the properties of determinants.

In this question, weβre given some information about two matrices π΄ and π΅. Weβre told that both of these matrices are square matrices of the same order; theyβre both two-by-two matrices. Next weβre given the determinant of each matrix. The determinant of matrix π΄ is two and the determinant of matrix π΅ is three. We need to use this information to find the value of the determinant of two π΄ plus the determinant of three π΅ plus the determinant of matrix π΄ times matrix π΅.

We could start by trying to do this by using the definition of the determinant. However, weβre told to do this by using the property of determinants. And in fact, this is a good idea because it will make the process easier. To use the properties of determinants, letβs start by looking at each term of the expression weβre given to evaluate.

We can see that the first two terms are the determinants of a scalar multiplied by a matrix. And we can recall we can evaluate determinants like this by using one of our properties of the determinant. In particular, we can recall if π is a square matrix of order π by π, then for any scalar πΎ the determinant of πΎπ is equal to πΎ to the πth power multiplied by the determinant of π. We can use this to evaluate the first two terms of this expression.

First, the determinant of two times π΄ is equal to two squared times the determinant of π΄ because weβre told that π΄ is a two-by-two matrix. And we can keep evaluating. Remember, weβre told in the question the determinant of matrix π΄ is two. This gives us two squared times two, which is two cubed, which we can evaluate is equal to eight.

We can then do the same to evaluate the second term, the determinant of three times π΅. π΅ is also a two-by-two matrix. Therefore, the determinant of three π΅ is three squared multiplied by the determinant of π΅. And weβre told in the question the determinant of π΅ is three. So we get three squared times three, which is equal to three cubed, which we can evaluate is 27.

Now, we just need to evaluate the third term in this expression, the determinant of π΄ multiplied by π΅. And we can do this by using one of the properties of determinants. If π sub one and π sub two are square matrices of the same order, then the determinant of π sub one times π sub two is equal to the determinant of π sub one multiplied by the determinant of π sub two. And of course, weβre told in the question that both matrices π΄ and π΅ are square matrices of order two by two. So, we can rewrite this as the determinant of matrix π΄ times the determinant of matrix π΅. And weβre told in the question the determinant of π΄ is two and the determinant of π΅ is three. So, this is equal to two times three, which we can evaluate is six.

Now, all we need to do is substitute the three values we found into this expression. We get the determinant of two π΄ plus the determinant of three π΅ plus the determinant of π΄π΅ is equal to eight plus 27 plus six, which if we evaluate is equal to 41. Therefore, by using the properties of determinants, we were able to find the value of the expression given to us in the question; its value was 41.