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# 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. -9B. -27C. -81D. 81  Verified
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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.

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
$\begin{array}{l} \Rightarrow \left| {3AB} \right| = 27 \times ( - 1) \times 3\\ \Rightarrow \left| {3AB} \right| = 27 \times ( - 3)\\ \Rightarrow \left| {3AB} \right| = - 81 \end{array}$
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 $\left| {AB} \right| = \left| A \right| \times \left| B \right|$ .