Question

Ayudaaa... Con los escalares α = 4 β = –1 γ = 0 λ = –2 y las matrices

Answer 1

Al calcular la matriz X se comprueba la igualdad:

X = X₁  =X₂ =X₃ =X₄ = $\left[\begin{array}{cc}11&-11\\20&27\end{array}\right]$

Explicación paso a paso:

Datos;

escalares α = 4 β = –1 γ = 0 λ = –2

matrices

$A =\left[\begin{array}{cc}2&-3\\5&8\end{array}\right]$ $B =\left[\begin{array}{cc}-3&1\\2&5\end{array}\right]$$C =\left[\begin{array}{cc}2&-3\\4&-5\end{array}\right]$$D =\left[\begin{array}{cc}0&-1\\-1&0\end{array}\right]$

Relación asociativa:

a) X₁  = αA + [βB+(γC+λD)]

b) X₂ = [(αA + βB) + γC] + λD

c) X₃ = (αA + βB) + (γC + λD)

d) X₄ = (αA + λD) + (βB + γC)

X = X₁  =X₂ =X₃ =X₄ = $\left[\begin{array}{cc}11&-11\\20&27\end{array}\right]$

Producto de un escalar por una matriz;

$4A =\left[\begin{array}{cc}2(4)&-3(4)\\5(4)&8(4)\end{array}\right]=\left[\begin{array}{cc}8&-12\\20&32\end{array}\right]$

$-B =\left[\begin{array}{cc}3&-1\\-2&-5\end{array}\right]$

$0.C =\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$

$-2D =\left[\begin{array}{cc}0&2\\2&0\end{array}\right]$

$X=\left[\begin{array}{ccc}8&-12\\20&32\end{array}\right] +\left[\begin{array}{cc}3&-1\\-2&-5\end{array}\right] +\left[\begin{array}{cc}0&0\\0&0\end{array}\right]+\left[\begin{array}{cc}0&2\\2&0\end{array}\right]$

$X=\left[\begin{array}{ccc}11&-11\\20&27\end{array}\right]$

$X_1=\left[\begin{array}{ccc}8&-12\\20&32\end{array}\right] +\left[\left[\begin{array}{cc}3&-1\\-2&-5\end{array}\right] +\left(\left[\begin{array}{cc}0&0\\0&0\end{array}\right]+\left[\begin{array}{cc}0&2\\2&0\end{array}\right]\right)\right]$

$X_1=\left[\begin{array}{ccc}8&-12\\20&32\end{array}\right] +\left[\left[\begin{array}{cc}3&-1\\-2&-5\end{array}\right] +\left[\begin{array}{cc}0&2\\2&0\end{array}\right]\right]$

$X_1=\left[\begin{array}{ccc}8&-12\\20&32\end{array}\right] +\left[\begin{array}{cc}3&1\\0&-5\end{array}\right]$

$X_1=\left[\begin{array}{ccc}11&-11\\20&26\end{array}\right]$

$X_2=\left[\left(\left[\begin{array}{ccc}8&-12\\20&32\end{array}\right] +\left[\left[\begin{array}{cc}3&-1\\-2&-5\end{array}\right] \right)+\left[\begin{array}{cc}0&0\\0&0\end{array}\right]\right]+\left[\begin{array}{cc}0&2\\2&0\end{array}\right]$

$X_2=\left[\left[\begin{array}{ccc}11&-13\\18&27\end{array}\right] +\left[\begin{array}{cc}0&0\\0&0\end{array}\right]\right]+\left[\begin{array}{cc}0&2\\2&0\end{array}\right]$

$X_2=\left[\begin{array}{ccc}11&-13\\18&27\end{array}\right] +\left[\begin{array}{cc}0&2\\2&0\end{array}\right]$

$X_2=\left[\begin{array}{ccc}11&-11\\20&27\end{array}\right]$

$X_3=\left(\left[\begin{array}{ccc}8&-12\\20&32\end{array}\right] +\left[\begin{array}{cc}3&-1\\-2&-5\end{array}\right] \right)+\left(\left[\begin{array}{cc}0&0\\0&0\end{array}\right]+\left[\begin{array}{cc}0&2\\2&0\end{array}\right]\right)$

$X_3=\left[\begin{array}{ccc}11&-13\\18&27\end{array}\right] +\left[\begin{array}{cc}0&2\\2&0\end{array}\right]$

$X_3=\left[\begin{array}{ccc}11&-11\\20&27\end{array}\right]$

$X_4=\left(\left[\begin{array}{ccc}8&-12\\20&32\end{array}\right] +\left[\begin{array}{cc}0&2\\2&0\end{array}\right]\right)+\left(\left[\begin{array}{cc}3&-1\\-2&-5\end{array}\right] +\left[\begin{array}{cc}0&0\\0&0\end{array}\right]\right)$

$X_4=\left[\begin{array}{ccc}8&-10\\22&32\end{array}\right] +\left[\begin{array}{cc}3&-1\\-2&-5\end{array}\right]$

$X_4=\left[\begin{array}{ccc}11&-11\\20&27\end{array}\right]$

Answer 2

a continuacion se definen mediante las realizacion de las asiciaciones q se indican