Question

If an $\mathit{n}\mathbf{\times }\mathit{n}$ matrixAis invertible, then there must be an $\left(n-1\right)\mathbf{\times }\left(n-1\right)$sub matrix of(obtained by deleting a row and a column of) that is invertible as well.

Short answer

Therefore, the sub matrix is invertible. So, the given statement is true.

Step-by-step solution

Matrix Definition

Matrix is aset of numbers arranged in rows and columns so as to form a rectangular array.

The numbers are called the elements, or entries, of the matrix.

If there are m rows and n columns, the matrix is said to be an "m by n” matrix, written “m$\times n$.”

To check whether the given condition is true or false

If A is an invertible matrix, then$detA\ne 0$ .

Thus, if we use the Laplace expansion along the first row, we can see that at least one of the addends must be different than 0.

Therefore, this must also apply to at least one $\left(n-1\right)\times \left(n-1\right)$sub matrix.

So that sub matrix is invertible.