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Expert-verified Found in: Page 309 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # For every nonzero ${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrix A there exists a ${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrix B such that ${\mathbit{d}}{\mathbit{e}}{\mathbit{t}}{\mathbf{\left(}}{\mathbit{A}}{\mathbf{+}}{\mathbit{B}}{\mathbf{\right)}}{\mathbf{\ne }}{\mathbit{d}}{\mathbit{e}}{\mathbit{t}}{\mathbf{}}{\mathbit{A}}{\mathbf{+}}{\mathbit{d}}{\mathbit{e}}{\mathbit{t}}{\mathbf{}}{\mathbit{B}}$.

Therefore, $det\left(A+B\right)\ne detA+detB.$ So, the given statement is true.

See the step by step solution

## Step 1: Matrix Definition

Matrix is a set 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 “$m×n$ .”

## Step 2: To check whether the given condition is true or false

For any invertible matrix A , we can choose

$B=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Indeed,

$det\left(A+B\right)=detA\phantom{\rule{0ex}{0ex}}det\left(A+B\right)=detA+0$

Therefore,

$det\left(A+B\right)\ne detA+detB$ .

So, the given statement is true. ### Want to see more solutions like these? 