Exercises 37 and 38 should be solved without performing row
operations. [Hint: Write Ax D 0 as a vector equation.]
37. Given A = [2 3 5, -5 1 -4, -3 -1 -4, 1 0 1] 
, observe that the third column
is the sum of the first two columns. Find a nontrivial solution
of Ax D 0.




To solve the system \( A\mathbf{x} = 0 \) without performing row operations, we can follow the hint and express the matrix equation as a vector equation. 

The matrix \( A \) is:

\[
A = \begin{bmatrix}
2 & 3 & 5 \\
-5 & 1 & -4 \\
-3 & -1 & -4 \\
1 & 0 & 1
\end{bmatrix}
\]

We are told that the third column is the sum of the first two columns. Let's denote the columns of \( A \) as \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \):

\[
\mathbf{v}_1 = \begin{bmatrix} 2 \\ -5 \\ -3 \\ 1 \end{bmatrix}, \quad
\mathbf{v}_2 = \begin{bmatrix} 3 \\ 1 \\ -1 \\ 0 \end{bmatrix}, \quad
\mathbf{v}_3 = \mathbf{v}_1 + \mathbf{v}_2 = \begin{bmatrix} 5 \\ -4 \\ -4 \\ 1 \end{bmatrix}
\]

Now, the vector equation \( A\mathbf{x} = 0 \) can be written as:

\[
x_1 \mathbf{v}_1 + x_2 \mathbf{v}_2 + x_3 \mathbf{v}_3 = \mathbf{0}
\]

Using the fact that \( \mathbf{v}_3 = \mathbf{v}_1 + \mathbf{v}_2 \), we can rewrite this as:

\[
x_1 \mathbf{v}_1 + x_2 \mathbf{v}_2 + x_3 (\mathbf{v}_1 + \mathbf{v}_2) = \mathbf{0}
\]

This simplifies to:

\[
(x_1 + x_3) \mathbf{v}_1 + (x_2 + x_3) \mathbf{v}_2 = \mathbf{0}
\]

Since \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are linearly independent (i.e., neither is a scalar multiple of the other), the only way this equation holds is if the coefficients of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are both zero. This gives us the system of equations:

\[
x_1 + x_3 = 0 \quad \text{(1)}
\]
\[
x_2 + x_3 = 0 \quad \text{(2)}
\]

From equation (1), we get \( x_1 = -x_3 \), and from equation (2), we get \( x_2 = -x_3 \).

Thus, the general solution for \( \mathbf{x} \) is:

\[
\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -x_3 \\ -x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}
\]

A nontrivial solution occurs when \( x_3 \neq 0 \), so we can take \( x_3 = 1 \). Therefore, one nontrivial solution is:

\[
\mathbf{x} = \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix}
\]

This is the required nontrivial solution of \( A\mathbf{x} = 0 \).

---

Would you like more details on any part of this solution? Here are 5 additional related questions for practice:

1. What is the general form of the solution to \( A\mathbf{x} = 0 \)?
2. How can you verify that \( \mathbf{x} = \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} \) satisfies \( A\mathbf{x} = 0 \)?
3. Why does the fact that the third column is the sum of the first two columns imply that the matrix is not of full rank?
4. What does it mean for vectors to be linearly dependent, and how does it apply in this context?
5. What would the solution be if the matrix had independent columns?

**Tip:** In homogeneous systems like \( A\mathbf{x} = 0 \), if the columns of \( A \) are linearly dependent, then there always exists a nontrivial solution.

Exercises 37 and 38 should be solved without performing row operations. [Hint: Write Ax D 0 as a vector equation.] 37. Given A = [2 3 5, -5 1 -4, -3 -1 -4, 1 0 1], observe that the third column is the sum of the first two columns. Find a nontrivial solution of Ax D 0.

Solve the matrix equation Ax = 0 where A = [2 3 5, -5 1 -4, -3 -1 -4, 1 0 1], given that the third column is the sum of the first two columns. Explore linear dependence in vector spaces and the concept of nontrivial solutions in homogeneous systems. Suitable for undergraduate students studying Linear Algebra.

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