Find a vector x such that T(x) = Ax, where the matrix A and the vector b are given, and solve for x.
Solution
The image you provided involves finding a vector x such that when transformed by a matrix A, the result equals a given vector. Specifically, it asks to find the vector x whose image under the transformation T(x)=Ax results in a given vector.
Problem Breakdown:
The matrix A is given as:
6 & -2 & -3 \\
5 & -4 & 2 \\
-42 & 11 & 6
\end{pmatrix}$$
- The transformation is described by $$T(\mathbf{x}) = A\mathbf{x}$$, where the result is:
$$\mathbf{b} = \begin{pmatrix} 0 \\ 5 \\ 39 \end{pmatrix}$$
- We need to solve for the vector $$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$ such that:
$$A\mathbf{x} = \mathbf{b}$$
### Solution Steps:
1. Set up the system of equations based on $$A\mathbf{x} = \mathbf{b}$$.
2. Solve the system of equations to find the vector $$\mathbf{x}$$.
Let me calculate that for you.The solution to the system $$A\mathbf{x} = \mathbf{b}$$ is the vector:
$$\mathbf{x} = \begin{pmatrix} -2.505 \\ -5.165 \\ -1.567 \end{pmatrix}$$
Thus, the vector whose image under the transformation is $$\mathbf{b} = \begin{pmatrix} 0 \\ 5 \\ 39 \end{pmatrix}$$ is approximately:
$$\mathbf{x} = \begin{pmatrix} -2.505 \\ -5.165 \\ -1.567 \end{pmatrix}$$
Additionally, based on the system of equations and the matrix, this solution is unique, as the matrix $$A$$ is invertible (its determinant is non-zero).
Would you like any additional details or explanations on the steps?
Here are five related questions you might find helpful:
1. How can we determine if a matrix is invertible?
2. What does it mean for a transformation to be one-to-one or onto?
3. How do you compute the determinant of a 3x3 matrix?
4. Can we always solve $$A\mathbf{x} = \mathbf{b}$$ for any matrix $$A$$?
5. How can eigenvalues be related to solving systems of linear equations?
**Tip:** If a system of linear equations has a unique solution, the matrix of coefficients must be invertible, meaning its determinant is non-zero.