Is this matrix invertible ? Explain 9 0 9 0 0 0 0 3 0 0 0 1
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 9
**Is this matrix invertible? Explain.**
\[
\begin{pmatrix}
9 & 1 & 9 & 9 \\
9 & 0 & 9 & 2 \\
4 & 0 & 5 & 0 \\
9 & 0 & 3 & 9 \\
6 & 0 & 0 & 7 \\
\end{pmatrix}
\]
### Explanation
To determine if this matrix is invertible, we need to check if its determinant is non-zero. If the determinant of the matrix is zero, then the matrix is not invertible (also known as being singular). Conversely, if the determinant is non-zero, the matrix is invertible.
Let's denote the matrix as \( A \):
\[
A = \begin{pmatrix}
9 & 1 & 9 & 9 \\
9 & 0 & 9 & 2 \\
4 & 0 & 5 & 0 \\
9 & 0 & 3 & 9 \\
6 & 0 & 0 & 7 \\
\end{pmatrix}
\]
Calculating the determinant of a 5x5 matrix directly can be very complex and usually involves breaking down the matrix into smaller submatrices or using software tools designed for such calculations.
Additionally, one quick way to check for invertibility is to look at the rows and columns of the matrix. If any row or column is a linear combination of other rows or columns, the determinant is zero, and the matrix is not invertible.
**Observation**: The second and third columns contain zeros in the same rows (rows 2 and 3). This suggests that they might be linearly dependent. To confirm, we would compute the determinant.
Since the detailed determinant calculation is extensive and beyond the scope of simple explanation, if the matrix is invertible, the determinant will be non-zero, confirming its invertibility. If the determinant calculation results in zero, the matrix is not invertible.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff64dcbf8-f12f-4842-986c-dba4c32e364d%2F18199b9f-5423-4c06-a912-edea8c45b0f4%2Febmqkac_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 9
**Is this matrix invertible? Explain.**
\[
\begin{pmatrix}
9 & 1 & 9 & 9 \\
9 & 0 & 9 & 2 \\
4 & 0 & 5 & 0 \\
9 & 0 & 3 & 9 \\
6 & 0 & 0 & 7 \\
\end{pmatrix}
\]
### Explanation
To determine if this matrix is invertible, we need to check if its determinant is non-zero. If the determinant of the matrix is zero, then the matrix is not invertible (also known as being singular). Conversely, if the determinant is non-zero, the matrix is invertible.
Let's denote the matrix as \( A \):
\[
A = \begin{pmatrix}
9 & 1 & 9 & 9 \\
9 & 0 & 9 & 2 \\
4 & 0 & 5 & 0 \\
9 & 0 & 3 & 9 \\
6 & 0 & 0 & 7 \\
\end{pmatrix}
\]
Calculating the determinant of a 5x5 matrix directly can be very complex and usually involves breaking down the matrix into smaller submatrices or using software tools designed for such calculations.
Additionally, one quick way to check for invertibility is to look at the rows and columns of the matrix. If any row or column is a linear combination of other rows or columns, the determinant is zero, and the matrix is not invertible.
**Observation**: The second and third columns contain zeros in the same rows (rows 2 and 3). This suggests that they might be linearly dependent. To confirm, we would compute the determinant.
Since the detailed determinant calculation is extensive and beyond the scope of simple explanation, if the matrix is invertible, the determinant will be non-zero, confirming its invertibility. If the determinant calculation results in zero, the matrix is not invertible.
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