Question

In Exercises 31–36, mention an appropriate theorem in your explanation.

32. Suppose that A is a square matrix such that \(det{\rm{ }}{A^3} = 0\). Explain why A cannot be invertible.

Short answer

It is proved that A cannot be invertible.

Step-by-step solution

Write the multiplicative property

According totheorem 6,if A andB aresquare matrices, then thedeterminant of the product matrix AB is equal to the product of the determinant of A and the determinant of B.

\(\det AB = \left( {\det A} \right)\left( {\det B} \right)\)

If matrices A and B are the same, then the general form is

\(\det {A^n} = {\left( {\det A} \right)^n}\).

Based on theorem 4,matrix A of the order\(n \times n\)(square matrix) isinvertible if itsdeterminant is not 0, (\(\det \left( A \right) \ne 0\)).

Prove the statement

It is given that\(\det {A^3} = 0\).

By using theorem, it can be written as shown below:

\(\begin{aligned}{}\det {A^3} &= {\left( {\det A} \right)^3}\\{\left( {\det A} \right)^3} &= 0\\\det A &= 0\end{aligned}\)

Since \(\det A = 0\), matrix A cannot be invertible.

Hence, proved.