Answer:
The matrix [tex]cA[/tex] is invertible and its inverse is [tex]\frac{1}{c}\cdot A^{-1}[/tex].
Step-by-step explanation:
Since the definition of the inverse matrix states that the inverse of matrix [tex]A[/tex] is a matrix B such that:
[tex]A\cdot B=B\cdot A=I[/tex]
we have to assume the form of such matrix. In our case we have the matrix [tex]cA, c\neq 0[/tex] and so, the constant [tex]c[/tex] must be somehow eliminated from the equation. The most logical way to do so is to include [tex]\frac{1}{c}[/tex] in the inverse. If we choose matrix B to be [tex]B=\frac{1}{c}\cdot A^{-1}[/tex], we will have this:
[tex]cA\cdot \frac{1}{c}\cdot A^{-1}=c\cdot \frac{1}{c}\cdot A\cdot A^{-1}=1\cdot I=I[/tex] and
[tex]\frac{1}{c}\cdot A^{-1}\cdot cA=\frac{1}{c}\cdot c\cdot A^{-1}\cdot A=1\cdot I=I[/tex].
We can form the matrix B like this because we know from the text of the problem that the inverse matrix of A exists and that c is a nonzero number.
Here is another way to solve this using the formula of the inverse matrix
Since we know that the matrix [tex]A[/tex] is invertible, it follows that its determinant is different from zero. Using the formula for the inverse matrix:
[tex]A^{-1}=\frac{1}{\det (A)}\cdot \text{Adj} (A)[/tex]
we will assume the form of an inverse matrix of [tex]cA[/tex]. We need to obtain the formula for the inverse of [tex]cA[/tex], so we first need to find [tex]\det (cA)\ \text{and}\ \text{Adj} (cA)[/tex]. Since the matrix [tex]cA[/tex] is obtained from matrix [tex]A[/tex] by multiplying every term with [tex]c[/tex], while calculating determinant we have a constant [tex]c[/tex] that can be extracted from every column (or row) in front. Therefore, we have that
[tex]\det (cA)=c^n\cdot \det (A)[/tex].
On the other hand, [tex]\text{Adj} (cA)[/tex] consists of minors of the matrix [tex]cA[/tex]. Therefore, when we extract the constant in front of such ([tex]n-1 \times n-1[/tex]) determinants, we have [tex]c^{n-1}[/tex] in each column (row). Including all this into the formula we have that:
[tex](cA)^{-1}=\frac{1}{c^n\cdot \det (A)}\cdot c^{n-1} \text{Adj } (A)=\frac{1}{c\cdot \det (A)} \cdot \text{Adj} A=\frac{1}{c}\cdot A^{-1}[/tex].
