Question

Se A= $\left[\begin{array}{ccc}2&1\\-4&5\end{array}\right]$ , B= $\left[\begin{array}{ccc}1&5\\-3&2\end{array}\right]$ e C= $\left[\begin{array}{ccc}4&2\\5&4\end{array}\right]$, determine X=A²- B²+C.

Answer 1

Olá


$\displaystyle\mathsf{A= \left[\begin{array}{ccc}2&1\\-4&5\\\end{array}\right] \qquad\qquad B= \left[\begin{array}{ccc}1&5\\-3&2\\\end{array}\right] \qquad\qquad C= \left[\begin{array}{ccc}4&2\\5&4\\\end{array}\right] }\\\\\\\\\mathsf{A^2~=~A\cdot A}\\\mathsf{B^2=B\cdot B}\\\mathsf{C=C}\\\\\\\text{Encontrando A}^2$

$\displaystyle\mathsf{A^2= \left[\begin{array}{ccc}2&1\\-4&5\\\end{array}\right] \cdot \left[\begin{array}{ccc}2&1\\-4&5\\\end{array}\right] }\\\\\\\mathsf{A^2=\left[\begin{array}{ccc}2\cdot 2~+~1\cdot(-4)\quad&(2\cdot1~+~1\cdot 5)\\(-4\cdot 2~+~5\cdot(-4))\quad&(-4\cdot 1~+~5\cdot 5)\\\end{array}\right] }}}}}}$

$\mathsf{A^2=\left[\begin{array}{ccc}4-4\quad&2+5\\-8-20\quad&-4+25\\\end{array}\right]}\\\\\\\boxed{\mathsf{A^2=\left[\begin{array}{ccc}0&7\\-28&21\\\end{array}\right]}}\\\\\\\text{Encontrando B}^2$



$\mathsf{B^2=\left[\begin{array}{ccc}1&5\\-3&2\\\end{array}\right]\cdot \left[\begin{array}{ccc}1&5\\-3&2\\\end{array}\right]}\\\\\\\mathsf{B^2=\left[\begin{array}{ccc}(1\cdot 1~+~5\cdot(-3))\quad&(1\cdot5~+~5\cdot 2)\\(-3\cdot1~+~2\cdot(-3))\quad&(-3\cdot5~+~2\cdot2)\\\end{array}\right]}$

$\mathsf{B^2=\left[\begin{array}{ccc}(1-15)\quad&(5+10)\\(-3-6)\quad&(-15+4)\\\end{array}\right]}\\\\\\\boxed{\mathsf{B^2=\left[\begin{array}{ccc}-14&15\\-9&-11\\\end{array}\right]}}}}}}}}}}}}}}}}}$



Efetuando a expressão

X = A² - B² + C


$\mathsf{X=\left[\begin{array}{ccc}0&7\\-28&21\\\end{array}\right]~-~\left[\begin{array}{ccc}-14&15\\-9&-11\\\end{array}\right]~+~\left[\begin{array}{ccc}4&2\\5&4\\\end{array}\right]}\\\\\\\mathsf{X=\left[\begin{array}{ccc}(0-(-14)+4)\quad&(7-15+2)\\(-28-(-9)+5)\quad&(21-(-11)+4)\\\end{array}\right]}\\\\\\\\\boxed{\boxed{\mathsf{X=\left[\begin{array}{ccc}18&-6\\-14&36\\\end{array}\right]}}}$