91Ó°ÊÓ

Solve the equation for \(X\), given that \(A=\) \(\left[\begin{array}{ll}1 & 2 \\\ 3 & 4\end{array}\right]\) and \(B=\left[\begin{array}{rr}-1 & 0 \\ 1 & 1\end{array}\right]\). $$X-2 A+3 B=O$$

Let \(P=\left[\begin{array}{cc}0.5 & 0.3 \\ 0.5 & 0.7\end{array}\right]\) be the transition matrix for a Markov chain with two states. Let \(\mathbf{x}_{0}=\left[\begin{array}{l}0.5 0.5\end{array}\right]\) be the initial state vector for the population. Compute \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2}\).

Solve the system \(A \mathbf{x}=\mathbf{b}\) using the given \(L U\) factorization of \(A\). $$A=\left[\begin{array}{rr} -2 & 1 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ -1 & 1 \end{array}\right]\left[\begin{array}{rr} -2 & 1 \\ 0 & 6 \end{array}\right], \mathbf{b}=\left[\begin{array}{l} 5 \\ 1 \end{array}\right]$$

Let \(T_{A}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be the matrix transformation corre- sponding to \(A=\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right] .\) Find \(T_{A}(\mathbf{u})\) and \(T_{A}(\mathbf{v})\) where \(\mathbf{u}=\left[\begin{array}{l}1 \\ 2\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{r}3 \\ -2\end{array}\right]\).

Let \(S\) be the collection of vectors\(\left[\begin{array}{l}x \\\ y\end{array}\right]\) in \(\mathbb{R}^{2}\) that satisfy the given property. In each case, either prove that S forms a subspace of \(\mathbb{R}^{2}\) or give a counterexample to show that it does not. $$x=0$$

Let \(P=\left[\begin{array}{cc}0.5 & 0.3 \\ 0.5 & 0.7\end{array}\right]\) be the transition matrix for a Markov chain with two states. Let \(\mathbf{x}_{0}=\left[\begin{array}{l}0.5 0.5\end{array}\right]\) be the initial state vector for the population. What proportion of the state 1 population will be in state 2 after two steps?

Solve the equation for \(X\), given that \(A=\) \(\left[\begin{array}{ll}1 & 2 \\\ 3 & 4\end{array}\right]\) and \(B=\left[\begin{array}{rr}-1 & 0 \\ 1 & 1\end{array}\right]\). $$2 X=A-B$$

Solve the system \(A \mathbf{x}=\mathbf{b}\) using the given \(L U\) factorization of \(A\). $$A=\left[\begin{array}{rr} 4 & -2 \\ 2 & 3 \end{array}\right]=\left[\begin{array}{rr} 1 & 0 \\ \frac{1}{2} & 1 \end{array}\right]\left[\begin{array}{rr} 4 & -2 \\ 0 & 4 \end{array}\right], \mathbf{b}=\left[\begin{array}{l} 0 \\ 8 \end{array}\right]$$

Let \(S\) be the collection of vectors\(\left[\begin{array}{l}x \\\ y\end{array}\right]\) in \(\mathbb{R}^{2}\) that satisfy the given property. In each case, either prove that S forms a subspace of \(\mathbb{R}^{2}\) or give a counterexample to show that it does not. $$x \geq 0, y \geq 0$$

Let \(T_{A}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) be the matrix transformation corre sponding to \(A=\left[\begin{array}{rrr}4 & 0 & -1 \\ -2 & 1 & 3\end{array}\right] .\) Find \(T_{A}(\mathbf{u})\) and \(T_{A}(\mathbf{v}),\) where \(\mathbf{u}=\left[\begin{array}{r}1 \\ -1 \\\ 2\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{r}0 \\ 5 \\\ -1\end{array}\right]\).