How do you solve $3 x - y = 3$ $3x-y=3$ and $- 2 x + y = 2$ $-2x+y=2$ using matrices?

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Answer 1 · 2016-08-04T09:46:39

The Soln. is $x = 5, y = 12$ $x=5, y=12$ .

We can write the given system of eqns., using the Matrix Form, as :

$[(3,-1),(-2,1)][(x), (y)]=[(3),(2)].............(1)$ .

Let $A=[(3,-1),(-2,1)] , X=[(x), (y)], and, B=[(3),(2)]$ .

Then, $(1)$ becomes, $AX=B$ .

Now, we know from Algebra that the soln. of $(1)$ exists iff $A^-1$ exists.

Also, $A^-1$ exists iff $detA!=0$ , where,

$detA=det|(3,-1),(-2,1)|=3-2=1!=0$ . Hence, $A^-1$ exists, and, so does the unique soln. of the eqns.

Now, $A^-1=1/detA*adjA=1/1[(1,1),(2,3)]=[(1,1),(2,3)]$

Therefore, the soln. is $X=A^-1B=[(1,1),(2,3)][(3),(2)]$

$:. [(x), (y)]=[(1,1),(2,3)][(3),(2)]=[(5),(12)]$

So, the Soln. is $x=5, y=12$ .

How do you solve 3x-y=33x−y=33x-y=3 and -2x+y=2−2x+y=2-2x+y=2 using matrices?