3.29. Let A = 1 -2 0 21 4 0 If A is invertible find A-¹. Otherwise show that it is not invertible. How many solutions does the system of equation Ax = 0 have?
3.29. Let A = 1 -2 0 21 4 0 If A is invertible find A-¹. Otherwise show that it is not invertible. How many solutions does the system of equation Ax = 0 have?
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 3.29
**Given:**
\[ A = \begin{pmatrix}
1 & 1 & -1 \\
-2 & 0 & 4 \\
2 & 1 & 0
\end{pmatrix} \]
If \( A \) is invertible, find \( A^{-1} \). Otherwise, show that it is not invertible. How many solutions does the system of equations \( A\mathbf{x} = 0 \) have?
---
**Solution Guide:**
1. **Determine if \( A \) is invertible:**
- Calculate the determinant of matrix \( A \).
- If the determinant is non-zero, then \( A \) is invertible, and we can proceed to find \( A^{-1} \).
- If the determinant is zero, then \( A \) is not invertible.
2. **Calculate the determinant:**
\[
\text{det}(A) = 1 \left( \begin{vmatrix}
0 & 4 \\
1 & 0
\end{vmatrix} \right) - 1 \left( \begin{vmatrix}
-2 & 4 \\
2 & 0
\end{vmatrix} \right) + (-1) \left( \begin{vmatrix}
-2 & 0 \\
2 & 1
\end{vmatrix} \right)
\]
Simplify the determinants:
\[
= 1 (0 \cdot 0 - 4 \cdot 1) - 1 ((-2) \cdot 0 - 4 \cdot 2) + (-1)((-2) \cdot 1 - 0 \cdot 2)
\]
\[
= 1 ( -4 ) - 1 ( - 8 ) + (-1) ( - 2)
\]
\[
= - 4 + 8 - 2
\]
\[
= 2
\]
Since the determinant is non-zero (2), matrix \( A \) is invertible.
3. **Find \( A^{-1} \):**
- Utilize the formula for the inverse of a 3x3 matrix by using co](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa2cfa344-ed79-4b65-9fc9-d1da7eeb8dd1%2F62c82984-e126-4aeb-b682-9100426f8d97%2Fsxvjjiuj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 3.29
**Given:**
\[ A = \begin{pmatrix}
1 & 1 & -1 \\
-2 & 0 & 4 \\
2 & 1 & 0
\end{pmatrix} \]
If \( A \) is invertible, find \( A^{-1} \). Otherwise, show that it is not invertible. How many solutions does the system of equations \( A\mathbf{x} = 0 \) have?
---
**Solution Guide:**
1. **Determine if \( A \) is invertible:**
- Calculate the determinant of matrix \( A \).
- If the determinant is non-zero, then \( A \) is invertible, and we can proceed to find \( A^{-1} \).
- If the determinant is zero, then \( A \) is not invertible.
2. **Calculate the determinant:**
\[
\text{det}(A) = 1 \left( \begin{vmatrix}
0 & 4 \\
1 & 0
\end{vmatrix} \right) - 1 \left( \begin{vmatrix}
-2 & 4 \\
2 & 0
\end{vmatrix} \right) + (-1) \left( \begin{vmatrix}
-2 & 0 \\
2 & 1
\end{vmatrix} \right)
\]
Simplify the determinants:
\[
= 1 (0 \cdot 0 - 4 \cdot 1) - 1 ((-2) \cdot 0 - 4 \cdot 2) + (-1)((-2) \cdot 1 - 0 \cdot 2)
\]
\[
= 1 ( -4 ) - 1 ( - 8 ) + (-1) ( - 2)
\]
\[
= - 4 + 8 - 2
\]
\[
= 2
\]
Since the determinant is non-zero (2), matrix \( A \) is invertible.
3. **Find \( A^{-1} \):**
- Utilize the formula for the inverse of a 3x3 matrix by using co
![Sure, here is the transcription for the educational website:
---
### Linear Algebra Equation:
Consider the following matrix equation presented in Example 2.29:
\[
\begin{bmatrix}
1 & 0 & 0 & -\frac{1}{5} \\
0 & 1 & 0 & -\frac{2}{5} \\
0 & 0 & 1 & 2
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
= t
\begin{bmatrix}
1 \\
2 \\
-10 \\
5
\end{bmatrix}
\]
In this setup, we have a 3x4 coefficient matrix on the left-hand side, multiplying a 4x1 column vector of variables \(x_1, x_2, x_3, x_4\). The resulting product is scaled by a scalar \(t\) and equated to a 4x1 column vector.
Each element of the equation represents a distinct linear relationship among the variables \(x_1, x_2, x_3, x_4\) and the scalar \(t\).
---
Make sure to understand the structure of the matrix and the dimensions of the involved vectors when solving or analyzing such systems. For further information, refer to the topic of "Systems of Linear Equations" in Linear Algebra.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa2cfa344-ed79-4b65-9fc9-d1da7eeb8dd1%2F62c82984-e126-4aeb-b682-9100426f8d97%2F3hbz7zb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Sure, here is the transcription for the educational website:
---
### Linear Algebra Equation:
Consider the following matrix equation presented in Example 2.29:
\[
\begin{bmatrix}
1 & 0 & 0 & -\frac{1}{5} \\
0 & 1 & 0 & -\frac{2}{5} \\
0 & 0 & 1 & 2
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
= t
\begin{bmatrix}
1 \\
2 \\
-10 \\
5
\end{bmatrix}
\]
In this setup, we have a 3x4 coefficient matrix on the left-hand side, multiplying a 4x1 column vector of variables \(x_1, x_2, x_3, x_4\). The resulting product is scaled by a scalar \(t\) and equated to a 4x1 column vector.
Each element of the equation represents a distinct linear relationship among the variables \(x_1, x_2, x_3, x_4\) and the scalar \(t\).
---
Make sure to understand the structure of the matrix and the dimensions of the involved vectors when solving or analyzing such systems. For further information, refer to the topic of "Systems of Linear Equations" in Linear Algebra.
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