Question

alguém pode me ajudar ? e de matrizes​

Answer 1

Método somente para matriz 2x2

X=

a b

c  d

X⁻¹ =  1/det(X)    *   d   -b

                              -c     a

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A*B =

2*1+2*1= 4     2*2+2*2 =8

1*1+4*1 =5     1*2+4*2= 10

=

4      8

5      10

det(A)= 2*4-1*2 =  6

X⁻¹ = 1/6    *    4     -2

                       -1     2

X⁻¹ =  

2/3    -1/3

-1/6      1/3

A* B -7*A⁻¹  =

4           8      -     14/3       -7/3

5          10           -7/6        7/3

=

-2/3        31/3

37/6       23/3

Answer 2

Resposta e explicação passo a passo:

Primerio, fazemos AxB:

$A\times{B}\\\\\left[\begin{array}{ccc}2&2\\1&4\end{array}\right]\times\left[\begin{array}{ccc}1&2\\1&2\end{array}\right]=\left[\begin{array}{ccc}2\times1+2\times1&2\times2+2\times2\\1\times1+4\times1&1\times2+4\times2\end{array}\right]\\\\=\left[\begin{array}{ccc}2+2&4+4\\1+4&2+8\end{array}\right]=\left[\begin{array}{ccc}4&8\\5&10\end{array}\right]$

Segundo, encontramos a inversa de A:

$A^{-1}=\left[\begin{array}{ccc}2&2\\1&4\end{array}\right|\left\begin{array}{ccc}1&0\\0&1\end{array}\right]-->L_1=\frac{L_1}{2}\\\\\\\left[\begin{array}{ccc}1&1\\1&4\end{array}\right|\left\begin{array}{ccc}\frac{1}{2} &0\\0&1\end{array}\right]-->L_2=L_2-L_1\\\\\\\left[\begin{array}{ccc}1&1\\0&3\end{array}\right|\left\begin{array}{ccc}\frac{1}{2} &0\\-\frac{1}{2}&1\end{array}\right]-->L_2=\frac{L_2}{3}$

$\left[\begin{array}{ccc}1&1\\0&1\end{array}\right|\left\begin{array}{ccc}\frac{1}{2} &0\\-\frac{1}{2}\times\frac{1}{3}&\frac{1}{3}\end{array}\right]\\\\\\\left[\begin{array}{ccc}1&1\\0&1\end{array}\right|\left\begin{array}{ccc}\frac{1}{2} &0\\-\frac{1}{6}&\frac{1}{3}\end{array}\right]-->L_1=L_1-L_2\\\\\\\left[\begin{array}{ccc}1&0\\0&1\end{array}\right|\left\begin{array}{ccc}\frac{1}{2}-(\frac{1}{6})&-\frac{1}{3}\\-\frac{1}{6}&\frac{1}{3}\end{array}\right]$

$\left[\begin{array}{ccc}1&0\\0&1\end{array}\right|\left\begin{array}{ccc}\frac{2}{3}&-\frac{1}{3}\\-\frac{1}{6}&\frac{1}{3}\end{array}\right]\\\\\\A^{-1}=\left[\begin{array}{ccc}\frac{2}{3}&-\frac{1}{3}\\-\frac{1}{6}&\frac{1}{3}\end{array}\right]$

Terceiro, fazemos 7A^-1:

$7\times{A^{-1}}=7\times\left[\begin{array}{ccc}\frac{2}{3}&-\frac{1}{3}\\-\frac{1}{6}&\frac{1}{3}\end{array}\right]\\\\\\=\left[\begin{array}{ccc}\frac{14}{3}&-\frac{7}{3}\\-\frac{7}{6}&\frac{7}{3}\end{array}\right]$

Por último, fazemos (AxB)-(7A^-1)

$A\times{B}-7\times{A^{-2}}=\left[\begin{array}{ccc}4&8\\5&10\end{array}\right]-\left[\begin{array}{ccc}\frac{14}{3}&-\frac{7}{3}\\-\frac{7}{6}&\frac{7}{3}\end{array}\right]=\left[\begin{array}{ccc}4-\frac{14}{3}&8-(-\frac{7}{3}) \\5-(-\frac{7}{6})&10-\frac{7}{3}\end{array}\right]$

$=\left[\begin{array}{ccc}\frac{12-14}{3}&\frac{24+7}{3}) \\\frac{30+7}{6}&\frac{30-7}{3}\end{array}\right]=\left[\begin{array}{ccc}-\frac{2}{3}&\frac{31}{3})\\\frac{37}{6}&\frac{23}{3}\end{array}\right]$