Prove that (cA)^-1=(1/c)A^-1If A is an invertible matrix and c is a nonzero scalar, then cA is an invertible matrix and the above equation is true. Please show step by step how you would prove it.
You should remember that if you need to multiply a matrix with a scalar, you need to multiply each element of the matrix with the scalar, such that:
`A = v*((a,b),(c,d)) => A = ((v*a,v*b),(v*c,v*d))`
Notice that v represents the scalar(number) and A represents the matrix.
You may prove the requested identity evaluating the inverse of the new matrix `v*A` .
You need to remember that you may evaluate the inverse of a matrix only if its determinant is not equal to zero.
Supposing that the determminant of 2x2 matrix A is not zero, yields:
`det A = va*vd - vb*vc => detA = v^2ad - v^2bc`
Factoring out `v^2` yields:
`detA = v^2(ad-bc)`
Since `det A!= 0 =>` there exists `A^(-1)` such that:
`A^(-1) = (1/(det A))((A^t_(11),A^t_(21)),(A^t_(12),A^t_(22)))`
You need to evaluate `A^t_(11)` such that:
`A^t_(11) = (-1)^(1+1)*d ` (you need to remove the terms in column 1 and row 1 and it remains the element d)
`A^t_(11) = d`
`A^t_(12) = (-1)^(1+2)*b ` (you need to remove the terms in column 1 and row 2 and it remains the element b)
`A^t_(12) = -b`
`A^t_(21) = (-1)^(2+1)*c => A^t_(21) = -c`
`A^t_(22) = (-1)^(2+2)*c => A^t_(22) = a`
`A^(-1) = (1/det A)((d,-c),(-b,a))` `=> A^(-1) = 1/(v^2(ad-bc))*((d,-c),(-b,a))`
Notice that you may write `1/det A = (1/v)*(1/(v(ad-bc)))` where `1/v` represents the inverse of the scalar v, but the next factor does not represents `1/ det A` , hence, since `A^(-1) = 1/(det A)*((d,-c),(-b,a)),` the given expression `(v*A)^(-1) = v^(-1)*A^(-1)` does not hold.