ALG: Ejercicios de repaso

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

\[{{x}^{4}}-14 {{x}^{3}}+68 {{x}^{2}}-136 x+96\]

\[52.79\]

\[\operatorname{ }\begin{bmatrix}-1 & 4 & 3\\1 & -2 & 0\\1 & -2 & 2\\1 & 0 & 1\end{bmatrix}\]

\[\]\[\mbox{tiene pseudoinversa por la izquierda}\]\[\begin{bmatrix}-\frac{1}{12} & \frac{1}{4} & -\frac{5}{12} & \frac{13}{12}\\\frac{1}{12} & 0 & -\frac{1}{3} & \frac{5}{12}\\\frac{1}{6} & 0 & \frac{1}{3} & -\frac{1}{6}\end{bmatrix}» «\]

\[-2\]

Sean \(\mathbf{a}_1\) y \(\mathbf{a}_2\) las columnas de \(A\):
\[ \mathbf{a}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad \mathbf{a}_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

Paso 1: Aplicar el proceso de Gram-Schmidt para obtener las columnas ortogonales \(\mathbf{v}_1\) y \(\mathbf{v}_2\).

1. Vector \(\mathbf{v}_1\):
\[ \mathbf{v}_1 = \mathbf{a}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

2. Vector \(\mathbf{v}_2\):
\[ \mathbf{v}_2 = \mathbf{a}_2 – \text{proy}_{\mathbf{v}_1} \mathbf{a}_2 = \mathbf{a}_2 – \frac{\mathbf{a}_2 \cdot \mathbf{v}_1}{\|\mathbf{v}_1\|^2} \mathbf{v}_1 \]
Calculamos los términos:

• \(\mathbf{a}_2 \cdot \mathbf{v}_1 = (1)(1) + (0)(1) = 1\)
• \(\|\mathbf{v}_1\|^2 = 1^2 + 1^2 = 2\)

Sustituimos:
\[ \mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} – \frac{1}{2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 – 1/2 \\ 0 – 1/2 \end{bmatrix} = \begin{bmatrix} 1/2 \\ -1/2 \end{bmatrix} \]

Paso 2: Normalizar los vectores \(\mathbf{v}_1\) y \(\mathbf{v}_2\) para formar la matriz \(Q\).

1. Vector \(\mathbf{q}_1\):
\[ \|\mathbf{v}_1\| = \sqrt{2} \]
\[ \mathbf{q}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix} \]

2. Vector \(\mathbf{q}_2\):
\[ \|\mathbf{v}_2\| = \sqrt{(1/2)^2 + (-1/2)^2} = \sqrt{1/4 + 1/4} = \sqrt{1/2} = \frac{1}{\sqrt{2}} \]
\[ \mathbf{q}_2 = \frac{\mathbf{v}_2}{\|\mathbf{v}_2\|} = \sqrt{2} \begin{bmatrix} 1/2 \\ -1/2 \end{bmatrix} = \begin{bmatrix} \sqrt{2}/2 \\ -\sqrt{2}/2 \end{bmatrix} = \begin{bmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{bmatrix} \]

La matriz ortogonal \(Q\) es:
\[ Q = \begin{bmatrix} \mathbf{q}_1 & \mathbf{q}_2 \end{bmatrix} = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{bmatrix} \]

Paso 3: Calcular la matriz triangular superior \(R\).

La matriz \(R\) se calcula como \(R = Q^T A\):
\[ R = Q^T A = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \]

Los elementos de \(R\) son:

• \(r_{11} = \mathbf{q}_1 \cdot \mathbf{a}_1 = 1/\sqrt{2} + 1/\sqrt{2} = 2/\sqrt{2} = \sqrt{2}\)
• \(r_{12} = \mathbf{q}_1 \cdot \mathbf{a}_2 = 1/\sqrt{2} + 0 = 1/\sqrt{2}\)
• \(r_{21} = \mathbf{q}_2 \cdot \mathbf{a}_1 = 1/\sqrt{2} – 1/\sqrt{2} = 0\)
• \(r_{22} = \mathbf{q}_2 \cdot \mathbf{a}_2 = 1/\sqrt{2} – 0 = 1/\sqrt{2}\)

La matriz triangular superior \(R\) es:
\[ R = \begin{bmatrix} \sqrt{2} & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} \end{bmatrix} \]

Para terminar
\[[2,1]\begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{bmatrix}.[2,1]^t=\frac{7}{\sqrt{2}}\approx 4.9497\]

C.)

\[\begin{bmatrix}-4 & 0\\-4 & -7\end{bmatrix}\]

\[\left( x+4\right) \, \left( x+7\right) \]

\[\]\[solve: dependent equations eliminated: (2)\]\[solve: dependent equations eliminated: (1)\]

\[\begin{bmatrix}-7 & 0\\0 & -4\end{bmatrix}\]

\[\begin{bmatrix}0 & -\frac{3}{4}\\1 & 1\end{bmatrix}\]

\[\begin{bmatrix}-262144 & 0\\-53455284 & -40353607\end{bmatrix}\]

\[6.697\cdot {{10}^{7}}\]

\[18.01\]