How do you find the inverse of #((3, 7), (4, 17))#?

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Answer 1 · 2017-02-22T06:44:12

Inverse: #1/23((17,-7),(-4,3)) = ((17/23,-7/23),(-4/23,3/23))#

Since we have a square matrix, you can use Cramer's rule for a #2xx2# matrix where #A =((a, b),(c, d))# :

#A^-1 = 1/(ad-bc) ((d, -b),(-c, a))#

After we substitute: #A^-1 = 1/(3*17-7*4) ((17, -7),(-4, 3))#

So #A^-1 =1/23((17,-7),(-4,3)) = ((17/23,-7/23),(-4/23,3/23))#

Answer 2 · 2017-02-22T06:56:32

The inverse is #((17/23,-7/23),(-4/23,3/23))#

Let #A=((3,7),(4,17))#

To determine if matrix #A# is invertible, we calculate the determinant

#detA=|(3,7),(4,17)|=-3*17-(7)(4)=51-28=23#

As, #detA!=0#, the matrix is invertible

We start, by calculating the matrix of co-factors

#C=((17,-4),(-7,3))#

Then, we calculate the transpose of matrix #C#

#C^T=((17,-7),(-4,3))#

And the inverse is

#A^-1=1/detA*C^T#

#=-1/23*((17,-7),(-4,3))#

#=((17/23,-7/23),(-4/23,3/23))#

Verification,

#A*A^-1=((3,7),(4,17))*((17/23,-7/23),(-4/23,3/23))#

#=((1,0),(0,1))#

#=I#