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Question: If \(A\) and \(B\) are square matrices of order \(3\) such that \(\left| A \right|=-1\), \(\left| B ...

If AA and BB are square matrices of order 33 such that ∣A∣=−1\left| A \right|=-1, ∣B∣=3\left| B \right|=3, then ∣3AB∣\left| 3AB \right| is equal to
A. −9-9
B. −81-81
C. −27-27
D. 81

Explanation

Solution

We first explain the relation between determinants of multiplied matrices with their orders. We follow the rules of ∣aXY∣=a∣X∣m∣Y∣n\left| aXY \right|=a{{\left| X \right|}^{m}}{{\left| Y \right|}^{n}} where aa is constant. We put the values of ∣A∣=−1\left| A \right|=-1, ∣B∣=3\left| B \right|=3, m=n=3m=n=3 to find the solution.

Complete step by step answer:
If the given matrices are square matrices, then the value for the multiplication is dependent on the order of the matrices. Let X and Y are two square matrices with order respectively m and n, then ∣XY∣=∣X∣m∣Y∣n\left| XY \right|={{\left| X \right|}^{m}}{{\left| Y \right|}^{n}}.

Any constant multiplication will work following the formula of ∣aXY∣=a∣XY∣\left| aXY \right|=a\left| XY \right|.
In the given case AA and BB are square matrices of order 3 such that ∣A∣=−1\left| A \right|=-1, ∣B∣=3\left| B \right|=3.
We need to find the value of ∣3AB∣\left| 3AB \right|.
So, ∣3AB∣=3∣AB∣\left| 3AB \right|=3\left| AB \right|.
For the matrices we put the value of m=n=3m=n=3.
Therefore, 3∣AB∣=3∣A∣3∣B∣33\left| AB \right|=3{{\left| A \right|}^{3}}{{\left| B \right|}^{3}}.
We put the values ∣A∣=−1\left| A \right|=-1, ∣B∣=3\left| B \right|=3.So,
∣3AB∣=3×(−1)3×(3)3 ∴∣3AB∣=−81\left| 3AB \right|=3\times {{\left( -1 \right)}^{3}}\times {{\left( 3 \right)}^{3}} \\\ \therefore \left| 3AB \right| =-81.

Hence, the correct option is B.

Note: The matrices and the formulas won’t work in matrix division as we first have to convert it to multiplication form by converting the matrix to its inverse matrix. We can also consider the constant as the multiplication of an identity matrix and use the order form.