Question

Calcule a soma das matrizes A=(aij)3×3 sendo aij=4- i2 e vij=-4 + j3

Answer 1

A = (aij)3x3

A = $\left[\begin{array}{ccc}a11&a12&a13\\a21&a22&a23\\a31&a32&a33\end{array}\right]$

aij = 4 - i²
a11 = 4 - (1)²
a11 = 4 - 1
a11 = 3

a12 = 4 - (1)²
a12 = 4 - 1
a12 = 3

a13 = 4 - (1)²
a13 = 4 - 1
a13 = 3

a21 = 4 - (2)²
a21 = 4 - 4
a21 = 0

a22 = 4 - (2)²
a22 = 4 - 4
a22 = 0

a23 = 4 - (2)²
a23 = 4 - 4
a23 = 0

a31 = 4 - (3)²
a31 = 4 - 9
a31 = - 5

a32 = 4 - (3)²
a32 = 4 - 9
a32 = - 5

a33 = 4 - (3)²
a33 = 4 - 9
a33 = - 5

A = $\left[\begin{array}{ccc}3&3&3\\0&0&0\\-5&-5&-5\end{array}\right]$

V = (vij)3x3

V = $\left[\begin{array}{ccc}v11&v12&v13\\v21&v22&v23\\v31&v32&v33\end{array}\right]$

vij = - 4 + j³

v11 = - 4 + (1)³
v11 = - 4 + 1
v11 = - 3

v12 = - 4 + (2)³
v12 = - 4 + 8
v12 = 4

v13 = - 4 + (3)³
v13 = - 4 + 27
v13 = 23

v21 = - 4 + (1)³
v21 = - 4 + 1
v21 = - 3

v22 = - 4 + (2)³
v22 = - 4 + 8
v22 = 4

v23 = - 4 + (3)³
v23 = - 4 + 27
v23 = 23

v31 = - 4 + (1)³
v31 = - 4 + 1
v31 = - 3

v32 = - 4 + (2)³
v32 = - 4 + 8
v32 = 4

v33 = - 4 + (3)³
v33 = - 4 + 27
v33 = 23

V = $\left[\begin{array}{ccc}-3&4&23\\-3&4&23\\-3&4&23\end{array}\right]$

A + V = $\left[\begin{array}{ccc}3&3&3\\0&0&0\\-5&-5&-5\end{array}\right] + \left[\begin{array}{ccc}-3&4&23\\-3&4&23\\-3&4&23\end{array}\right] = \left[\begin{array}{ccc}0&7&26\\-3&4&23\\-8&-1&18\end{array}\right]$