+0

# done

0
119
4
+78

done

Mar 18, 2020
edited by rubikx2910  Apr 15, 2020

### 4+0 Answers

#1
+25243
+2

Let the matrix $$\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$$.
Calculate $$\mathbf{A}^{10} \begin{pmatrix}1 \\ 0 \end{pmatrix},\ \mathbf{A}^{10} \begin{pmatrix}0 \\ 1 \end{pmatrix}$$.

$$\begin{array}{|lrcll|} \hline \mathbf{A}^{\color{red}1} & &=& \begin{pmatrix} 1 & 2\cdot \color{red}1 \\ 0 & 1 \end{pmatrix} \\\\ \mathbf{A}^{\color{red}2} & = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} &=& \begin{pmatrix} 1 & 2\cdot \color{red}2 \\ 0 & 1 \end{pmatrix} \\\\ \mathbf{A}^{\color{red}3} & = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2\cdot 2 \\ 0 & 1 \end{pmatrix} &=& \begin{pmatrix} 1 & 2\cdot \color{red}3 \\ 0 & 1 \end{pmatrix} \\\\ \mathbf{A}^{\color{red}4} & = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2\cdot 3 \\ 0 & 1 \end{pmatrix} &=& \begin{pmatrix} 1 & 2\cdot \color{red}4 \\ 0 & 1 \end{pmatrix} \\\\ \cdots \\ \mathbf{A}^{\color{red}10} & &=& \begin{pmatrix} 1 & 2\cdot \color{red}10 \\ 0 & 1 \end{pmatrix} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \mathbf{A}^{\color{red}10}\begin{pmatrix}1 \\ 0 \end{pmatrix} \\ &=& \begin{pmatrix} 1 & 20 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}1 \\ 0 \end{pmatrix} \\ &=& \begin{pmatrix}1 \\ 0 \end{pmatrix} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{A}^{\color{red}10} \begin{pmatrix}0 \\ 1 \end{pmatrix} \\ &=& \begin{pmatrix} 1 & 20 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}0 \\ 1 \end{pmatrix} \\ &=& \begin{pmatrix}20 \\ 1 \end{pmatrix} \\ \hline \end{array}$$

Mar 18, 2020
edited by heureka  Mar 18, 2020
#3
+78
0

done

rubikx2910  Mar 19, 2020
edited by rubikx2910  Apr 15, 2020
#2
+2
+1

In secondary school, the adjugate strategy (otherwise called the adjoint) and the Gauss-Jordan end technique are usually taught. Here, I will depict a somewhat intriguing technique that utilizes the Cayley-Hamilton hypothesis.

For an n×n

network, this technique requires finding the determinant of one n×n lattice, containing a vague x, and the result of (n−1)

indistinguishable networks.

Here is a bit by bit record of this technique. We expect that the n×n

the framework we require the reverse of is a get your assignment done online, whose passages are (aij).

Construct the matrix M

whose diagonal entries are (x−aii) and whose off-diagonal entries are −aij
.
Find the determinant of M
, and expand the resulting polynomial p(x)
.
If c0
, the coefficient of x0 in p(x), is 0, then halt. The matrix A
has no inverse.
Otherwise, let q(x)=−1c0(p(x)−c0x)
. The inverse of matrix A is q(A).

Mar 18, 2020
#4
+78
0

done

rubikx2910  Mar 19, 2020
edited by rubikx2910  Apr 15, 2020