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Q5E

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Found in: Page 289

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Use Gaussian elimination to find the determinant of the matrices A in Exercises 1 through 10.5. ${\mathbf{\left[}}\begin{array}{cccc}0& 2& 3& 4\\ 0& 0& 0& 4\\ 1& 2& 3& 4\\ 0& 0& 3& 4\end{array}{\mathbf{\right]}}$

Therefore, the determinant of given matrix is given by,

$detA=-24$

See the step by step solution

## Step 1: Definition

Gaussian elimination method is used to solve a system of linear equations.

Gaussian elimination provides a relatively efficient way of constructing the inverse to a matrix.

Interchanging two rows. Multiplying a row by a constant (any constant which is not zero).

## Step 2: Given

Given Matrix,

${A}{=}{\left[}\begin{array}{cccc}0& 2& 3& 4\\ 0& 0& 0& 4\\ 1& 2& 3& 4\\ 0& 0& 3& 4\end{array}{\right]}$

## Step 3: To find determinant by using Gaussian Eliminations

First, we switch the first and third row.

We get,

${A}_{1}=\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 0& 0& 4\\ 0& 2& 3& 4\\ 0& 0& 3& 4\end{array}\right]$

Then, we switch the second and third row.

We get,

${A}_{2}=\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 2& 3& 4\\ 0& 0& 0& 4\\ 0& 0& 3& 4\end{array}\right]$

Lastly, we switch the third and fourth row.

We get

${A}_{3}=\left[\begin{array}{cccc}1& 2& 3& 4\\ 0& 2& 3& 4\\ 0& 0& 3& 4\\ 0& 0& 3& 4\end{array}\right]$

We had three row interchanges, so

$\begin{array}{l}detA={\left(-1\right)}^{3}.1.2.3.4\\ detA=-24.\end{array}$