Question

If A is an invertible matrix of order 3$\times$3 such that $\left| \text{A} \right|=5$, find the value of $\left| {{A}^{-1}} \right|$.

Answer 1

To solve the question, we should know the relation between the determinants of the matrices A and A-1. The relation is

& {{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right) \\\ & \left| {{A}^{-1}} \right|=\left| \dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right) \right| \\\ \end{aligned}$$ We know the property that $\left| kA \right|={{k}^{n}}\left| A \right|$ where n is the order of the matrix. $$\left| {{A}^{-1}} \right|=\dfrac{1}{{{\left| A \right|}^{n}}}\left| \left( adj\left( A \right) \right) \right|$$ We know the relation that $\left| \left( adjA \right) \right|={{\left| A \right|}^{n-1}}$ Using the property, we get that $\left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}$, we get the answer. **Complete step-by-step answer:** In the question, we are given the value of $\left| \text{A} \right|=5$ and we are asked to find the value of $\left| {{A}^{-1}} \right|$. To do that, we have to find the relation between the two values $\left| A \right|$ and $\left| {{A}^{-1}} \right|$. From the fundamental properties of inverse of the matrix, we can write that $$\begin{aligned} & {{A}^{-1}}=\dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right) \\\ & \left| {{A}^{-1}} \right|=\left| \dfrac{1}{\left| A \right|}\left( adj\left( A \right) \right) \right| \\\ \end{aligned}$$ We also know the property of the determinant of the matrix when a constant is multiplied to it. Here, $\left| A \right|$ is a constant, but not a matrix. The relation is If n is the order of the matrix and determinant of kA is written as $\left| kA \right|={{k}^{n}}\left| A \right|$. As the order of matrix A is n, the order of matrix which is adjoint of A is also n. Applying it in the above equation, we get $$\left| {{A}^{-1}} \right|=\dfrac{1}{{{\left| A \right|}^{n}}}\left| \left( adj\left( A \right) \right) \right|$$ The relation between the determinants of A and adjoint of A is given by $\left| \left( adjA \right) \right|={{\left| A \right|}^{n-1}}$ Using this in the above relation, we get $$\begin{aligned} & \left| {{A}^{-1}} \right|=\dfrac{{{\left| A \right|}^{n-1}}}{{{\left| A \right|}^{n}}} \\\ & \left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}\to \left( 1 \right) \\\ \end{aligned}$$ In the question, it is given that $\left| \text{A} \right|=5$. Substituting the value of determinant of A in equation-1, we get $$\left| {{A}^{-1}} \right|=\dfrac{1}{5}$$ $\therefore $The required value is $$\left| {{A}^{-1}} \right|=\dfrac{1}{5}$$. **Note:** This problem can be done in an alternative way. We know the relation between the product of matrices and its inverse. That is, for a square invertible matrix A of order n$\times $n $A\times {{A}^{-1}}=I$ Applying determinant on both sides, we get $\left| A\times {{A}^{-1}} \right|=\left| I \right|$ We know that the determinant of an identity matrix is 1. $\left| A\times {{A}^{-1}} \right|=1$ From the relation$\left| A\times B \right|=\left| A \right|\times \left| B \right|$, we get $\left| A \right|\times {{\left| A \right|}^{-1}}=1$ From the above relation, we can write $$\left| {{A}^{-1}} \right|=\dfrac{1}{\left| A \right|}$$ Using the value of $\left| \text{A} \right|=5$, we get $$\left| {{A}^{-1}} \right|=\dfrac{1}{5}$$ which is the required answer.