Pregunta
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Find the bases for \( \operatorname{Col} \operatorname{A} \) and \( \operatorname{Nul} \mathrm{A} \), and then state the dimension of these subspaces for the matrix \( A \) and an echelon form of \( A \) below. \[ A=\left[\begin{array}{rrrrr} 1 & 4 & -2 & -1 & 3 \\ 5 & 20 & 1 & -27 & 8 \\ 4 & 16 & -1 & -18 & 7 \\ -2 & -8 & 1 & 8 & -6 \end{array}\right] \sim\left[\begin{array}{rrrrr} 1 & 4 & -2 & -1 & 3 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & -5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \] A basis for Col A is given by \( \square \) B. (Use a comma to separate vectors as needed.) The dimension of Col A is \( \square \) \( \square \). A basis for Nul A is given by \( \square \) B. (Use a comma to separate vectors as needed.) The dimension of Nul A is \( \square \) .

Ask by Wagner Wilson. in the United States
Mar 11,2025

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A basis for Col A is given by \(\begin{pmatrix} 1 \\ 5 \\ 4 \\ -2 \end{pmatrix}, \begin{pmatrix} -2 \\ 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 8 \\ 7 \\ -6 \end{pmatrix}\). The dimension of Col A is \(3\). A basis for Nul A is given by \(\begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 5 \\ 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}\). The dimension of Nul A is \(2\).

Solución

The echelon form shows pivots in columns 1, 3, and 5. This means: 1. The basis for \(\operatorname{Col} A\) is given by the corresponding columns of the original matrix: \[ \begin{pmatrix} 1 \\ 5 \\ 4 \\ -2 \end{pmatrix}, \quad \begin{pmatrix} -2 \\ 1 \\ -1 \\ 1 \end{pmatrix}, \quad \begin{pmatrix} 3 \\ 8 \\ 7 \\ -6 \end{pmatrix} \] The dimension of \(\operatorname{Col} A\) is \(3\). 2. To find the basis for \(\operatorname{Nul} A\), we solve \(A\mathbf{x}=\mathbf{0}\). Since the pivots are in columns 1, 3, and 5, let the free variables be: \[ x_2 = s \quad \text{and} \quad x_4 = t. \] From the echelon form we have the following equations: - **Row 1:** \[ x_1 + 4x_2 - 2x_3 - x_4 + 3x_5 = 0 \quad \Longrightarrow \quad x_1 = -4x_2 + 2x_3 + x_4 - 3x_5. \] - **Row 2:** \[ x_3 - 2x_4 = 0 \quad \Longrightarrow \quad x_3 = 2x_4 = 2t. \] - **Row 3:** \[ -5x_5 = 0 \quad \Longrightarrow \quad x_5 = 0. \] Substitute \(x_3 = 2t\) and \(x_5 = 0\) into the equation for \(x_1\): \[ x_1 = -4s + 2(2t) + t = -4s + 4t + t = -4s + 5t. \] So the general solution is: \[ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} = \begin{pmatrix} -4s+5t \\ s \\ 2t \\ t \\ 0 \end{pmatrix} = s\begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} 5 \\ 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}. \] Thus, a basis for \(\operatorname{Nul} A\) is: \[ \begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} 5 \\ 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}. \] The dimension of \(\operatorname{Nul} A\) is \(2\). Summary: A basis for Col A is given by \(\begin{pmatrix} 1 \\ 5 \\ 4 \\ -2 \end{pmatrix}, \begin{pmatrix} -2 \\ 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 8 \\ 7 \\ -6 \end{pmatrix}\). The dimension of Col A is \(3\). A basis for Nul A is given by \(\begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 5 \\ 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}\). The dimension of Nul A is \(2\).

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The Deep Dive

To find a basis for the column space \(\operatorname{Col} A\) and the null space \(\operatorname{Nul} A\), let's start with the echelon form of matrix \(A\). The echelon form of \(A\) is: \[ \left[\begin{array}{rrrrr} 1 & 4 & -2 & -1 & 3 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & -5 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \] ### For \(\operatorname{Col} A\): The pivot columns in the echelon form correspond to the columns in the original matrix \(A\) that form a basis for the column space. The pivot columns are the first, third, and fifth columns, leading us to: A basis for \(\operatorname{Col} A\) is given by: \[ \begin{pmatrix} 1 \\ 5 \\ 4 \\ -2 \end{pmatrix}, \begin{pmatrix} -2 \\ 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 8 \\ 7 \\ -6 \end{pmatrix} \] The dimension of \(\operatorname{Col} A\) is \(3\). ### For \(\operatorname{Nul} A\): To find the basis for the null space, we need to solve the equation \(A\mathbf{x} = \mathbf{0}\). From the given echelon form, we can express the solutions: Let \(x_2 = t\), \(x_4 = s\), and \(x_5 = 0\). Then from the second row, we can write \(x_3\) in terms of \(t\): \[ x_3 = 2s \] From the first row: \[ x_1 + 4t - 2(2s) - s + 3(0) = 0 \implies x_1 = -4t + 4s \] Thus, we can represent the solutions in vector form: \[ \mathbf{x} = t\begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + s\begin{pmatrix} 4 \\ 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}. \] This yields the basis for the null space: \[ \begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 4 \\ 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}. \] The dimension of \(\operatorname{Nul} A\) is \(2\). Thus, the final answer is: A basis for Col A is given by: \(\begin{pmatrix} 1 \\ 5 \\ 4 \\ -2 \end{pmatrix}, \begin{pmatrix} -2 \\ 1 \\ -1 \\ 1 \end{pmatrix}, \begin{pmatrix} 3 \\ 8 \\ 7 \\ -6 \end{pmatrix}\). The dimension of Col A is \(3\). A basis for Nul A is given by: \(\begin{pmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 4 \\ 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}\). The dimension of Nul A is \(2\).

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