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If A and B are square matrices of order 3 such that \[\left| A \right|=-2\] and \[\left| B \right|=3\], find the value of $\left| -5AB \right|$.

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Last updated date: 13th Jun 2024
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Answer
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Hint: We find out the relation between matrix and determinant multiplication with a constant. We find the usage of the order of the matrix for the constant part. The multiplication difference between matrix and determinant is judged by putting the values.

Complete step-by-step solution:
We are going to use the matrix multiplication rules to find out the determinant value of $\left| -5AB \right|$.
As both A and B are square matrices of order 3, the determinant of those two matrices follows a simple multiplication rule.
So, $\left| AB \right|=\left| A \right|\left| B \right|$. Now it’s given that \[\left| A \right|=-2\] and \[\left| B \right|=3\].
So, $\left| AB \right|=\left| A \right|\left| B \right|=\left( -2 \right)\left( 3 \right)=-6$.
This is only possible because the determinants are square matrices and they are invertible.
Now for the constant value, we know that for a determinant when we multiply a constant, it is multiplied with the only row or one column. It doesn’t multiply every element in the determinant itself.
But in the case of a matrix, the constant gets multiplied with every element of the matrix.
So, to mathematically express it we can say that for a matrix P of order $n\times n$, if $P=\left[ {{p}_{ij}} \right]$ where i and j defines the rows and columns of the matrix then $aP=\left[ a\times {{p}_{ij}} \right]$. Here a is constant.
But in case of determinant $\left| Q \right|=\left| {{q}_{ij}} \right|$,if we multiply with ‘a’ which is a constant then $a\left| Q \right|=a\left| {{q}_{ij}} \right|$.
Now in our case, we are trying to find the determinant value of a matrix. The constant (-5) is being multiplied with the matrix and then we are trying to find its determinant value.
So, let’s find determinant value of aP where $aP=\left[ a\times {{p}_{ij}} \right]\Rightarrow \left| aP \right|=\left| a\times {{p}_{ij}} \right|$.
The constant ‘a’ is multiplied with every row (or column, consider only one case) of the determinant. So, we can take out the constant. We take the common constant out from every row or column. So, the number of times the constant can be taken common is similar to the number of rows (or columns). P is of order $n\times n$.
So, $\left| aP \right|=\left| a{{p}_{ij}} \right|={{a}^{n}}\left| {{p}_{ij}} \right|$
In our problem, the determinant is of the square matrix of order 3.
So, we take the cubic value of the constant (-5).
The determinant value becomes $\left| -5AB \right|={{\left( -5 \right)}^{3}}\left| A \right|\left| B \right|$.
Putting the values, we get $\left| -5AB \right|=\left( -125 \right)\left( -6 \right)=750$.
The value of $\left| -5AB \right|$ is 750.

Note: We need to remember that the determinant value is dependent on the matrix. The determinant is element dependent also. The relation between them determines the solution of the problem. The matrix has to be a square matrix for the inverse and multiplication formula to work, otherwise, it won’t work for general matrices.