# 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.

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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.