Compute the indicated 3 x 3 determinant. ab 0 0 a b a 0 b
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Compute the indicated \(3 \times 3\) determinant:
\[
\begin{vmatrix}
a & b & 0 \\
0 & a & b \\
a & 0 & b \\
\end{vmatrix}
\]
**Steps to Compute the Determinant:**
To find the determinant of a \(3 \times 3\) matrix, use the rule of Sarrus or the cofactor expansion method. Here, we’ll use the cofactor expansion along the first row:
1. The matrix is:
\[
\begin{pmatrix}
a & b & 0 \\
0 & a & b \\
a & 0 & b \\
\end{pmatrix}
\]
2. The determinant can be computed as:
\[
\text{det}(M) = a \begin{vmatrix} a & b \\ 0 & b \end{vmatrix} - b \begin{vmatrix} 0 & b \\ a & b \end{vmatrix} + 0 \begin{vmatrix} 0 & a \\ a & 0 \end{vmatrix}
\]
3. Calculate each of the \(2 \times 2\) determinants:
- \(\begin{vmatrix} a & b \\ 0 & b \end{vmatrix} = ab - 0 = ab\)
- \(\begin{vmatrix} 0 & b \\ a & b \end{vmatrix} = 0 \cdot b - a \cdot b = -ab\)
4. Substitute back into the original equation:
\[
\text{det}(M) = a(ab) - b(-ab) = a^2b + ab^2
\]
5. The final result is:
\[
\text{det}(M) = ab(a + b)
\]
This determinant evaluates to the product of \(ab\) multiplied by the sum of \(a\) and \(b\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F992f5037-f1cc-4483-8aa3-731b184958ad%2Fe45992ea-c7c6-4c38-ae88-a645fd914898%2F1mfjvoc_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Compute the indicated \(3 \times 3\) determinant:
\[
\begin{vmatrix}
a & b & 0 \\
0 & a & b \\
a & 0 & b \\
\end{vmatrix}
\]
**Steps to Compute the Determinant:**
To find the determinant of a \(3 \times 3\) matrix, use the rule of Sarrus or the cofactor expansion method. Here, we’ll use the cofactor expansion along the first row:
1. The matrix is:
\[
\begin{pmatrix}
a & b & 0 \\
0 & a & b \\
a & 0 & b \\
\end{pmatrix}
\]
2. The determinant can be computed as:
\[
\text{det}(M) = a \begin{vmatrix} a & b \\ 0 & b \end{vmatrix} - b \begin{vmatrix} 0 & b \\ a & b \end{vmatrix} + 0 \begin{vmatrix} 0 & a \\ a & 0 \end{vmatrix}
\]
3. Calculate each of the \(2 \times 2\) determinants:
- \(\begin{vmatrix} a & b \\ 0 & b \end{vmatrix} = ab - 0 = ab\)
- \(\begin{vmatrix} 0 & b \\ a & b \end{vmatrix} = 0 \cdot b - a \cdot b = -ab\)
4. Substitute back into the original equation:
\[
\text{det}(M) = a(ab) - b(-ab) = a^2b + ab^2
\]
5. The final result is:
\[
\text{det}(M) = ab(a + b)
\]
This determinant evaluates to the product of \(ab\) multiplied by the sum of \(a\) and \(b\).
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