Question

Using the definition of inverse (Definition 1, on Page 43) and nothing more, show that if A is an invertible matrix and c is a nonzero scalar, then cA is an invertible matrix. Hint: First, guess intelligently how the inverse of cA looks like, then apply the definition to show that you are right.

Answer 1

Answer:

The matrix $cA$ is invertible and its inverse is $\frac{1}{c}\cdot A^{-1}$.

Step-by-step explanation:

Since the definition of the inverse matrix states that the inverse of matrix $A$ is a matrix B such that:

$A\cdot B=B\cdot A=I$

we have to assume the form of such matrix. In our case we have the matrix $cA, c\neq 0$ and so, the constant $c$ must be somehow eliminated from the equation. The most logical way to do so is to include $\frac{1}{c}$ in the inverse. If we choose matrix B to be $B=\frac{1}{c}\cdot A^{-1}$, we will have this:

$cA\cdot \frac{1}{c}\cdot A^{-1}=c\cdot \frac{1}{c}\cdot A\cdot A^{-1}=1\cdot I=I$ and

$\frac{1}{c}\cdot A^{-1}\cdot cA=\frac{1}{c}\cdot c\cdot A^{-1}\cdot A=1\cdot I=I$.

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 $A$ is invertible, it follows that its determinant is different from zero. Using the formula for the inverse matrix:

$A^{-1}=\frac{1}{\det (A)}\cdot \text{Adj} (A)$

we will assume the form of an inverse matrix of $cA$. We need to obtain the formula for the inverse of $cA$, so we first need to find $\det (cA)\ \text{and}\ \text{Adj} (cA)$. Since the matrix $cA$ is obtained from matrix $A$ by multiplying every term with $c$, while calculating determinant we have a constant $c$ that can be extracted from every column (or row) in front. Therefore, we have that

$\det (cA)=c^n\cdot \det (A)$.

On the other hand, $\text{Adj} (cA)$ consists of minors of the matrix $cA$. Therefore, when we extract the constant in front of such ($n-1 \times n-1$) determinants, we have $c^{n-1}$ in each column (row). Including all this into the formula we have that:

$(cA)^{-1}=\frac{1}{c^n\cdot \det (A)}\cdot c^{n-1} \text{Adj } (A)=\frac{1}{c\cdot \det (A)} \cdot \text{Adj} A=\frac{1}{c}\cdot A^{-1}$.