2ºBACH CCSS ÁLGEBRA MATRICIAL (PROBLEMAS)

Álgebra matricial

Problema (#1)

Calcule los valores de x e y que verifican:

$$3\begin{pmatrix} x & 2\\ 0 & y\\ \end{pmatrix}- \begin{pmatrix} y & 2x\\ 1 & -6\\ \end{pmatrix}= \begin{pmatrix} 4 & -2y\\ -1 & -3x\\ \end{pmatrix}$$

Solución:

$$\begin{pmatrix} 3x & 6\\ 0 & 3y\\ \end{pmatrix}-\begin{pmatrix} y & 2x\\ 1 & -6\\ \end{pmatrix}=\begin{pmatrix} 4 & -2y\\ -1 & -3x\\ \end{pmatrix} \Rightarrow$$

$$\Rightarrow\begin{pmatrix} 3x-y & 6-2x\\ 0-1 & 3y+6\\ \end{pmatrix}=\begin{pmatrix} 4 & -2y\\ -1 & -3x\\ \end{pmatrix} \Rightarrow$$

$$\Rightarrow\begin{pmatrix} 3x-y & 6-2x\\ -1 & 3y+6\\ \end{pmatrix}=\begin{pmatrix} 4 & -2y\\ -1 & -3x\\ \end{pmatrix} \Rightarrow$$

$$\Rightarrow\left.\begin{array}{l} 3x-y=4\\ 6-2x=-2y\\ -1=-1\\ 3y+6=-3x\\ \end{array}\right \}$$

Resolviendo por el método de Gauss:

$$\left.\begin{array}{l} 3x-y=4\\ -2x+2y=-6\\ 3x+3y=-6\\ \end{array}\right \} \underset{\begin{matrix} F_{2}'=\frac{F_{2}}{2}\\ F_{3}'=\frac{F_{3}}{3} \end{matrix}}{\Rightarrow} \left.\begin{array}{l} 3x-y=4\\ -x+y=-3\\ x+y=-2\\ \end{array}\right \} \underset{\begin{matrix} F_{1} \leftrightarrow F_{3}\\ \end{matrix}}{\Rightarrow}$$

$$\Rightarrow \left.\begin{array}{l} x+y=-2\\ -x+y=-3\\ 3x-y=4\\ \end{array}\right \} \underset{\begin{matrix} F_{2}'=F_{2}+F_{1}\\ F_{3}'=F_{3}-3F_{1}\\ \end{matrix}}{\Rightarrow} \left.\begin{array}{l} x+y=-2\\ 2y=-5\\ -4y=10\\ \end{array}\right \} \underset{\begin{matrix} F_{3}'=F_{3}+2F_{2}\\ \end{matrix}}{\Rightarrow}$$

$$\Rightarrow \left.\begin{array}{l} x+y=-2\\ 2y=-5\\ 0=0\\ \end{array}\right \} \Rightarrow \left \{\begin{array}{l} x = \frac{1}{2}\\ y =-\frac{5}{2} \\ \end{array}\right.$$