+0  
 
0
119
4
avatar+78 

done

 Mar 18, 2020
edited by rubikx2910  Apr 15, 2020
 #1
avatar+25243 
+2

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

 

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

 

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

 

laugh

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

done

rubikx2910  Mar 19, 2020
edited by rubikx2910  Apr 15, 2020
 #2
avatar+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
avatar+78 
0

done

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

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