Question

If A and B are square matrices of order 3 such that \[\left| A \right| = - 1\] , \[\left| B \right| = 3\] , then the determinant of 3AB is
A. -9
B. -27
C. -81
D. 81

Answer 1

Hint: For any square matrix of order n, we know that if we multiply it with any constant K then \[\left| {KA} \right| = {K^n}\left| A \right|\] . So we will use this property of matrix and determinants to solve this question.

Complete step by step answer:
For this Question we will use the property of determinants that is \[\left| {KA} \right| = {K^n}\left| A \right|\] where n is the order of matrix
So we are given that \[\left| A \right| = - 1,\left| B \right| = 3\] and we are told to find the value of \[\left| {3AB} \right|\]
We can break \[\left| {3AB} \right|\] as \[\left| {3A} \right| \times \left| B \right|\]
Now we know that both A and B are square matrices of order 3 which means if i want to take 3 out from \[\left| {3A} \right|\] It will come out as \[{3^3}\left| A \right|\]
Now we are left with \[\left| {3A} \right| \times \left| B \right| = {3^3} \times \left| A \right| \times \left| B \right|\]
Now we know that \[\left| A \right| = - 1\& \left| B \right| = 3\]
So putting the values of \[\left| A \right|\& \left| B \right|\] we will get
\[\begin{array}{l}
 \Rightarrow \left| {3AB} \right| = 27 \times ( - 1) \times 3\\
 \Rightarrow \left| {3AB} \right| = 27 \times ( - 3)\\
 \Rightarrow \left| {3AB} \right| = - 81
\end{array}\]

So, the correct answer is “Option C”.

Note: The thing is to remember that \[\left| {KA} \right| = {K^n}\left| A \right|\] , a lot of students usually forget this relation and can’t solve the question. Also remember that \[\left| {AB} \right| = \left| A \right| \times \left| B \right|\] .