Question

POR FAVOR! ALGUÉM ME AJUDA PLEASEEE!!!

Answer 1

Resposta:

$X=\left[\begin{array}{ccc}2&2\\9&-1\\\end{array}\right]$

Explicação passo-a-passo:

$A=\left[\begin{array}{ccc}1&1\\0&1\\\end{array}\right] \\\\B=\left[\begin{array}{ccc}2&0\\2&1\\\end{array}\right] \\\\C=\left[\begin{array}{ccc}3&4\\2&0\\\end{array}\right] \\\\X=\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] \\\\X^{t}=\left[\begin{array}{ccc}a&c\\b&d\\\end{array}\right]$

a, b, c, d ∈ R

$B+AX^{t}=2C\\\\AX^{t}=2C-B\\\\A^{-1}AX^{t}=A^{-1}(2C-B)\\\\X^{t}=A^{-1}(2C-B)$

Vamos Calcular a inversa de A:

$A^{-1}=\left[\begin{array}{ccc}1&-1\\0&1\\\end{array}\right] \\\\\pois A*A^{-1}=\left[\begin{array}{ccc}1&1\\1&1\\\end{array}\right]$

$X^{t}=\left[\begin{array}{ccc}1&-1\\0&1\\\end{array}\right] *(2\left[\begin{array}{ccc}3&4\\2&0\\\end{array}\right] -\left[\begin{array}{ccc}2&0\\2&1\\\end{array}\right] )\\\\X^{t}=\left[\begin{array}{ccc}1&-1\\0&1\\\end{array}\right]*(\left[\begin{array}{ccc}6-2&8-0\\4-2&-1\\\end{array}\right] )\\\\X^{t}=\left[\begin{array}{ccc}1&-1\\0&1\\\end{array}\right] *\left[\begin{array}{ccc}4&8\\2&-1\\\end{array}\right] \\\\X^{t}=\left[\begin{array}{ccc}2&9\\2&-1\\\end{array}\right]$

Assim: a=2, c=9, b=2, d= -1

$X=\left[\begin{array}{ccc}2&2\\9&-1\\\end{array}\right]$