 # How would you solve the following system of linear equations: 2x_1 + (3+a)x_2 + 2x_3 = 2+aZ Margaux Cole 2022-02-22 Answered
How would you solve the following system of linear equations:
$2{x}_{1}+\left(3+a\right){x}_{2}+2{x}_{3}=2+a$
${x}_{1}+a{x}_{2}+2{x}_{3}=a$
$a{x}_{1}+2{x}_{2}+2a{x}_{3}=0$
assuming that $a\ne ±\sqrt{2}$? I feel confident in solving linear equation systems with just constants as the coefficients but the variable coefficient a is what gives me problems.
Here is the solution I find: solution as augmented matrix but how would the assumption about a make the reduced echelon form any different compared to an assumption about e.g.$a\ne ±\sqrt{2}$?
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With
${x}_{1}=a-a{x}_{2}-2{x}_{3}$
can we eliminate ${x}_{1}$:
${x}_{2}\left(3-a\right)-2{x}_{3}=2-a$
${x}_{2}\left(2-{a}^{2}\right)+4a{x}_{3}=-{a}^{2}$
And then using
${x}_{3}=\frac{{x}_{2}\left(3-a\right)-\left(2-a\right)}{2}$
to eliminate ${x}_{3}$. Then you will get
${x}_{2}\left(2-{a}^{2}\right)+2a\left(3-a\right){x}_{2}-2a\left(2-a\right)=-{a}^{2}$