Question

Se a = $\left[\begin{array}{cc}2&1\\-3&4\\\end{array}\right]$, b = $\left[\begin{array}{cc}21&7\\-3&1\\\end{array}\right]$, c = $\left[\begin{array}{cc}-1&-2\\5&3\\\end{array}\right]$, determine A = a²+b-c²

Answer 1

Olá

A = a² + b - c²

Lembrando que multiplicação de matrizes é dada por linhas vezes colunas.


$\displaystyle\mathsf{ a=\left[\begin{array}{ccc}2&1\\-3&4\\\end{array}\right] }\\\\\\\boxed{\mathsf{a^2=a\cdot a}}\\\\\\\mathsf{ a^2=\left[\begin{array}{ccc}2&1\\-3&4\\\end{array}\right] \cdot\left[\begin{array}{ccc}2&1\\-3&4\\\end{array}\right] }\\\\\\\mathsf{=\left[\begin{array}{ccc}(2\cdot2)+(1\cdot(-3))\qquad&(2\cdot1)+(1\cdot4)\\(-3\cdot2)+(4\cdot(-3))\qquad&(-3\cdot1)+(4\cdot4)\\\end{array}\right] }}$


$\displaystyle\mathsf{=\left[\begin{array}{ccc}(4-3)\qquad&(2+4)\\(-6-12)\qquad&(-3+16)\\\end{array}\right] }}\\\\\\\boxed{\mathsf{a^2=\left[\begin{array}{ccc}1&6\\-18&13\\\end{array}\right] }}}$



O mesmo processo deve ser feito para obter c²


$\displaystyle\mathsf{ c=\left[\begin{array}{ccc}-1&-2\\5&3\\\end{array}\right] }\\\\\\\boxed{\mathsf{c^2=c\cdot c}}\\\\\\\mathsf{ c^2=\left[\begin{array}{ccc}-1&-2\\5&3\\\end{array}\right] \cdot\left[\begin{array}{ccc}-1&-2\\5&3\\\end{array}\right]}\\\\\\\mathsf{=\left[\begin{array}{ccc}(-1\cdot(-1))+(-2\cdot5)\qquad&(-1\cdot(-2))+(-2\cdot3)\\(5\cdot(-1))+(3\cdot5)\qquad&(5\cdot(-2))+(3\cdot3)\\\end{array}\right] }}$


$\displaystyle\mathsf{=\left[\begin{array}{ccc}(1-10)\qquad&(2-6)\\(-5+15)\qquad&(-10+9)\\\end{array}\right] }}\\\\\\\boxed{\mathsf{c^2=\left[\begin{array}{ccc}-9&-4\\10&-1\\\end{array}\right] }}}$


Como b não muda, Agora basta fazer a operação final:

A = a² + b - c²


$\displaystyle\mathsf{A =\left[\begin{array}{ccc}1&6\\-18&13\\\end{array}\right] +\left[\begin{array}{ccc}21&7\\-3&1\\\end{array}\right]-\left[\begin{array}{ccc}-9&-4\\10&-1\\\end{array}\right]}\\\\\\\\\mathsf{A=\left[\begin{array}{ccc}(1+21-(-9))\qquad&(6+7-(-4))\\(-18+(-3)-10)\qquad&(13+1-(-1))\\\end{array}\right] }\\\\\\\\\boxed{\boxed{\mathsf{A=\left[\begin{array}{ccc}31&17\\-31&15\\\end{array}\right] }}}$