Question

Sejam as matrizes A = (aij) 2x2, com aij = 2i - j², e B = (bij) 2 x 2, com bij = aij + 1, encontre a matriz X de modo que:

a) X - A + B = 0

b) X - At + Bt = 0

c) -A - X = -B

Answer 1

$A~=~\left[\begin{array}{ccc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right] \\\\\\A~=~\left[\begin{array}{ccc}2.(1)-1^2&2.(1)-2^2\\2.(2)-1^2&2.(2)-2^2\end{array}\right] \\\\\\A~=~\left[\begin{array}{ccc}1&-2\\3&0\end{array}\right]$

$B~=~\left[\begin{array}{ccc}b_{11}&b_{12}\\b_{21}&b_{22}\end{array}\right] \\\\\\B~=~\left[\begin{array}{ccc}a_{11}+1&a_{12}+1\\a_{21}+1&a_{22}+1\end{array}\right]\\\\\\B~=~\left[\begin{array}{ccc}1+1&-2+1\\3+1&0+1\end{array}\right]\\\\\\B~=~\left[\begin{array}{ccc}2&-1\\4&1\end{array}\right]$

a)

$X-A+B=0\\\\\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]-\left[\begin{array}{ccc}1&-2\\3&0\end{array}\right]+\left[\begin{array}{ccc}2&-1\\4&1\end{array}\right]~=~\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]\\\\\\\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~\left[\begin{array}{ccc}0+1-2&0-2+1\\0+3-4&0+0-1\end{array}\right]~=~\left[\begin{array}{ccc}-1&-1\\-1&-1\end{array}\right]$

b)

$X-A^t+B^t=0\\\\\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]-\left[\begin{array}{ccc}1&3\\-2&0\end{array}\right]+\left[\begin{array}{ccc}2&4\\-1&1\end{array}\right]~=~\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]\\\\\\\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~\left[\begin{array}{ccc}0+1-2&0+3-4\\0-2+1&0-0-1\end{array}\right]~=~\left[\begin{array}{ccc}-1&-1\\-1&-1\end{array}\right]$

c)

$-A-X=-B\\\\-\left[\begin{array}{ccc}1&-2\\3&0\end{array}\right]-\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~-\left[\begin{array}{ccc}2&-1\\4&1\end{array}\right]\\\\\\\left[\begin{array}{ccc}x_{11}&x_{12}\\x_{21}&x_{22}\end{array}\right]~=~\left[\begin{array}{ccc}-1+2&2-1\\-3+4&0+1\end{array}\right]~=~\left[\begin{array}{ccc}1&1\\1&1\end{array}\right]$