Question

Sendo A = ( 2 0 1)
( 4 1 3) e
B = ( 0 -1 2)
( 5 0 6) ;

determine:

a) 5 . A

b) -2 . B

c) 1/2 A

d) 2 A + B

e) 5 A - 3B

Answer 1

Explicação passo a passo:

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Temos as seguintes matrizes:

$\boxed{A=\left[\begin{array}{ccc}2&0&1\\4&1&3\\\end{array}\right]\ \ \ e \ \ B=\left[\begin{array}{ccc}0&-1&2\\5&0&6\\\end{array}\right] }$

a)

$5A=5.\left[\begin{array}{ccc}2&0&1\\4&1&3\\\end{array}\right]\\\\\\=\left[\begin{array}{ccc}5.2&5.0&5.1\\5.4&5.1&5.3\\\end{array}\right]\\\\\\=\boxed{\boxed{\left[\begin{array}{ccc}10&0&5\\20&5&15\\\end{array}\right]}}$

b)

$-2B=(-2).\left[\begin{array}{ccc}0&-1&2\\5&0&6\\\end{array}\right]\\\\\\=\left[\begin{array}{ccc}(-2).0&(-2).(-1)&(-2).2\\(-2).5&(-2).0&(-2).6\\\end{array}\right]\\ \\\\=\boxed{\boxed{\left[\begin{array}{ccc}0&2&-4\\-10&0&-12\\\end{array}\right]}}$

c)

$\dfrac{1}{2} A=\dfrac{1}{2}.\left[\begin{array}{ccc}2&0&1\\4&1&3\\\end{array}\right]\\\\\\=\left[\begin{array}{ccc}\frac{1}{2}.2&\frac{1}{2}.0&\frac{1}{2}.1\\\frac{1}{2}.4&\frac{1}{2}.1&\frac{1}{2}.3\\\end{array}\right]\\\\\\=\boxed{\boxed{\left[\begin{array}{ccc}1&0&\frac{1}{2} \\2&\frac{1}{2} &\frac{3}{2} \\\end{array}\right]}}$

d)

$2A+B=2.\left[\begin{array}{ccc}2&0&1\\4&1&3\\\end{array}\right]+\left[\begin{array}{ccc}0&-1&2\\5&0&6\\\end{array}\right] \\\\\\=\left[\begin{array}{ccc}4&0&2\\8&2&6\\\end{array}\right]+\left[\begin{array}{ccc}0&-1&2\\5&0&6\\\end{array}\right]\\\\\\=\left[\begin{array}{ccc}4+0&0-1&2+2\\8+5&2+0&6+6\\\end{array}\right]\\\\\\=\boxed{\boxed{\left[\begin{array}{ccc}4&-1&4\\13&2&12\\\end{array}\right]}}$

e)

$5A-3B=5.\left[\begin{array}{ccc}2&0&1\\4&1&3\\\end{array}\right]-3.\left[\begin{array}{ccc}0&-1&2\\5&0&6\\\end{array}\right] \\\\\\=\left[\begin{array}{ccc}10&0&5\\20&5&15\\\end{array}\right]-\left[\begin{array}{ccc}0&-3&6\\15&0&18\\\end{array}\right]\\\\\\=\left[\begin{array}{ccc}10-0&0-(-3)&5-6\\20-15&5-0&15-18\\\end{array}\right]\\\\\\=\boxed{\boxed{\left[\begin{array}{ccc}10&3&-1\\5&5&-3\\\end{array}\right]}}$