Question

URGENTE !!

MULTIPLICAÇÃO DE MATRIZES / IGUALDADE DE MATRIZES​

Answer 1

Resposta:

a)

A = $\left[\begin{array}{ c c }4 & 1 \\8 & 2 \end{ array }\right]$  B = $\left[\begin{array}{ c c }-3 & 5 \\12 & -20 \end{ array }\right]$

1a linha e 1a coluna

$\left[\begin{array}{ c c }4 & 1 \\8 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }-3 & 5 \\12 & -20 \end{ array }\right]$ = $\left[\begin{array}{ c c }4 *(-3) + 1*12 & \\ & \end{ array }\right]$

1a linha e 2a coluna

$\left[\begin{array}{ c c }4 & 1 \\8 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }-3 & 5 \\12 & -20 \end{ array }\right]$ = $\left[\begin{array}{ c c }4 *(-3) + 1*12 & 4*5+1*(-20) \\ & \end{ array }\right]$

2a linha e 1a coluna

$\left[\begin{array}{ c c }4 & 1 \\8 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }-3 & 5 \\12 & -20 \end{ array }\right]$ = $\left[\begin{array}{ c c }4 *(-3) + 1*12 & 4*5+1*(-20) \\ 8*(-3)+2*12 & \end{ array }\right]$

2a linha e 2a coluna

$\left[\begin{array}{ c c }4 & 1 \\8 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }-3 & 5 \\12 & -20 \end{ array }\right]$ = $\left[\begin{array}{ c c }4 *(-3) + 1*12 & 4*5+1*(-20) \\ 8*(-3)+2*12 & 8*5 + 2*(-20) \end{ array }\right]$

Resultado de AB

$\left[\begin{array}{ c c }4 & 1 \\8 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }-3 & 5 \\12 & -20 \end{ array }\right]$ = $\left[\begin{array}{ c c }0 & 0 \\0 & 0 \end{ array }\right]$

b)

A = $\left[\begin{array}{ c c }5 & 1 \\10 & 2 \end{ array }\right]$ B = $\left[\begin{array}{ c c }3 & -2 \\-15 & 10 \end{ array }\right]$

1a linha e 1a coluna

$\left[\begin{array}{ c c }5 & 1 \\10 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }3 & -2 \\-15 & 10 \end{ array }\right]$ = $\left[\begin{array}{ c c }5 *3 + 1*(-15) & \\ & \end{ array }\right]$

1a linha e 2a coluna

$\left[\begin{array}{ c c }5 & 1 \\10 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }3 & -2 \\-15 & 10 \end{ array }\right]$ = $\left[\begin{array}{ c c }5 *3 + 1*(-15) & 5*(-2)+1*10 \\ & \end{ array }\right]$

2a linha e 1a coluna

$\left[\begin{array}{ c c }5 & 1 \\10 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }3 & -2 \\-15 & 10 \end{ array }\right]$ = $\left[\begin{array}{ c c }5 *3 + 1*(-15) & 5*(-2)+1*10 \\ 10*3+2*(-15) & \end{ array }\right]$

2a linha e 2a coluna

$\left[\begin{array}{ c c }5 & 1 \\10 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }3 & -2 \\-15 & 10 \end{ array }\right]$ = $\left[\begin{array}{ c c }5 *3 + 1*(-15) & 5*(-2)+1*10 \\ 10*3+2*(-15) & 10*(-2)+2*10 \end{ array }\right]$

Resultado de AB

$\left[\begin{array}{ c c }5 & 1 \\10 & 2 \end{ array }\right]$ $\left[\begin{array}{ c c }3 & -2 \\-15 & 10 \end{ array }\right]$ = $\left[\begin{array}{ c c }0 & 0 \\0 & 0 \end{ array }\right]$