Dadas las siguientes matrices :
A = <img src="https://tex.z-dn.net/?f=%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%261%5C%5C0%261%260%5C%5C1%260%260%5Cend%7Barray%7D%5Cright%5D%0A" />
B = <img src="https://tex.z-dn.net/?f=%0A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%260%260%5C%5C0%263%260%5C%5C0%260%263%5Cend%7Barray%7D%5Cright%5D%0A" />
Procedemos a resolver la ecuación matricial : 6X = B - 3AX, despejando :
B = 6X - 3AX = (6I + 3A)X
X = (6I + 3A)⁻¹.
B
Realizamos primero la operación de suma de matrices,
6I + 3A = C = <img src="https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%260%5C%5C0%266%260%5C%5C0%260%266%5Cend%7Barray%7D%5Cright%5D%20%20%2B%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%263%5C%5C0%263%260%5C%5C3%260%260%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%263%5C%5C0%269%260%5C%5C3%260%266%5Cend%7Barray%7D%5Cright%5D%20" />
Invertimos la matriz resultante, para esto necesitamos usar el método de Gauss - Jordan :
C⁻¹ = <img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%263%5C%5C0%269%260%5C%5C3%260%266%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20" />
C⁻¹ = <img src="https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B2%7D%7B9%7D%20%260%26%20%5Cfrac%7B-1%7D%7B9%7D%20%5C%5C0%26%20%5Cfrac%7B1%7D%7B9%7D%20%260%5C%5C%20%5Cfrac%7B-1%7D%7B9%7D%20%260%26%20%5Cfrac%7B2%7D%7B9%7D%20%5Cend%7Barray%7D%5Cright%5D%20" />
Finalmente :
X = C⁻¹.
B = <img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B2%7D%7B9%7D%20%260%26%20%5Cfrac%7B-1%7D%7B9%7D%20%5C%5C0%26%20%5Cfrac%7B1%7D%7B9%7D%20%260%5C%5C%20%5Cfrac%7B-1%7D%7B9%7D%20%260%26%20%5Cfrac%7B2%7D%7B9%7D%20%5Cend%7Barray%7D%5Cright%5D" />.
<img src="https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%260%260%5C%5C0%263%260%5C%5C0%260%263%5Cend%7Barray%7D%5Cright%5D%20" />
X = <img src="https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B2%7D%7B3%7D%20%260%26%20%5Cfrac%7B-1%7D%7B3%7D%20%5C%5C0%26%20%5Cfrac%7B1%7D%7B3%7D%20%260%5C%5C%20%5Cfrac%7B-1%7D%7B3%7D%20%260%26%20%5Cfrac%7B2%7D%7B3%7D%20%5Cend%7Barray%7D%5Cright%5D%20" />
Esta es la respuesta al ejercicio 3 parte b de la prueba de selectividad Madrid Convocatoria Jun 2014 - 2015 de Matematica II.