• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q3E

Expert-verified
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 10.3. $\left[\begin{array}{cccc}1& 3& 2& 4\\ 1& 6& 4& 8\\ 1& 3& 0& 0\\ 2& 6& 4& 12\end{array}\right]$

Therefore, the determinant of given matrix is given by,

$\mathrm{det}A=-24$

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,

$A=\left[\begin{array}{cccc}1& 3& 2& 4\\ 1& 6& 4& 8\\ 1& 3& 0& 0\\ 2& 6& 4& 12\end{array}\right]$

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

First, we multiply the first row by -1 and add it to the second and third row. Then, we multiply the first row by -2 and add it to the 4th row.

We get,

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

We had zero row swaps, so

$\begin{array}{l}detA={\left(-1\right)}^{0}.1.3.-2.4\\ detA=-24\end{array}$