91Ó°ÊÓ

Perform the operations, given \(c=-2\) and \(A=\left[\begin{array}{rrr}1 & 2 & 3 \\\ 0 & 1 & -1\end{array}\right], B=\left[\begin{array}{rr}1 & 3 \\ -1 & 2\end{array}\right], C=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]\). $$c(C B)$$.

Find, if possible, (a) \(A B\) and (b) \(B A\) $$A=\left[\begin{array}{rr} 2 & -2 \\ -1 & 4 \end{array}\right], \quad B=\left[\begin{array}{rr} 4 & 1 \\ 2 & -2 \end{array}\right]$$

Find, if possible, (a) \(A B\) and (b) \(B A\) $$A=\left[\begin{array}{rrr} 2 & -1 & 3 \\ 5 & 1 & -2 \\ 2 & 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rrr} 0 & 1 & 2 \\ -4 & 1 & 3 \\ -4 & -1 & -2 \end{array}\right]$$

Determine whether the stochastic matrix \(P\) is regular. Then find the steady state matrix \(X\) of the Markov chain with matrix of transition probabilities \(P\). $$P=\left[\begin{array}{ll} 0.5 & 0.1 \\ 0.5 & 0.9 \end{array}\right]$$

Find the inverse of the matrix (if it exists). $$\left[\begin{array}{rrr}1 & 1 & 2 \\\3 & 1 & 0 \\\\-2 & 0 & 3\end{array}\right]$$

Find a sequence of elementary matrices that can be used to write the matrix in row-echelon form. $$ \left[\begin{array}{rrr} -2 & 1 & 0 \\ 3 & -4 & 0 \\ 1 & -2 & 2 \\ -1 & 2 & -2 \end{array}\right] $$

Perform the operations, given \(c=-2\) and \(A=\left[\begin{array}{rrr}1 & 2 & 3 \\\ 0 & 1 & -1\end{array}\right], B=\left[\begin{array}{rr}1 & 3 \\ -1 & 2\end{array}\right], C=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]\). $$B(C A)$$.

Find, if possible, (a) \(A B\) and (b) \(B A\) $$A=\left[\begin{array}{rrr} 1 & -1 & 7 \\ 2 & -1 & 8 \\ 3 & 1 & -1 \end{array}\right], \quad B=\left[\begin{array}{rrr} 1 & 1 & 2 \\ 2 & 1 & 1 \\ 1 & -3 & 2 \end{array}\right]$$

Find a sequence of elementary matrices that can be used to write the matrix in row-echelon form. $$ \left[\begin{array}{rrrr} 1 & -6 & 0 & 2 \\ 0 & -3 & 3 & 9 \\ 2 & 5 & -1 & 1 \\ 4 & 8 & -5 & 1 \end{array}\right] $$

Perform the operations, given \(c=-2\) and \(A=\left[\begin{array}{rrr}1 & 2 & 3 \\\ 0 & 1 & -1\end{array}\right], B=\left[\begin{array}{rr}1 & 3 \\ -1 & 2\end{array}\right], C=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]\). $$C(B C)$$.