# How do you find the inverse of A=((2, 4, 1),(-1, 1, -1), (1, 4, 0))?

Jul 11, 2016

The inverted matrix is: $\left(\begin{matrix}- 4 & - 4 & 5 \\ 1 & 1 & - 1 \\ 5 & 4 & - 6\end{matrix}\right)$

#### Explanation:

There are many ways in invert matrices, but for this problem I used the cofactor transpose method.

If we imagine that
$A = \left(\begin{matrix}\vec{A} \\ \vec{B} \\ \vec{C}\end{matrix}\right)$

So that:
$\vec{A} = \left(2 , 4 , 1\right)$
$\vec{B} = \left(- 1 , 1 , - 1\right)$
$\vec{C} = \left(1 , 4 , 0\right)$

Then we can define reciprocal vectors:
${\vec{A}}_{R} = \vec{B} \times \vec{C}$
${\vec{B}}_{R} = \vec{C} \times \vec{A}$
${\vec{C}}_{R} = \vec{A} \times \vec{B}$

Each is easily calculated using the determinant rule for cross products:
${\vec{A}}_{R} = | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 1 , 1 , - 1\right) , \left(1 , 4 , 0\right) | = \left(4 , - 1 , - 5\right)$
${\vec{B}}_{R} = | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(- 1 , 4 , 0\right) , \left(2 , 4 , 1\right) | = \left(4 , - 1 , - 4\right)$
${\vec{C}}_{R} = | \left(\hat{i} , \hat{j} , \hat{k}\right) , \left(2 , 4 , 1\right) , \left(- 1 , 1 , - 1\right) | = \left(- 5 , 1 , 6\right)$

We can use these to construct the cofactor transpose of $M$, $\overline{M}$, as such:
$\overline{M} = \left(\left({\vec{A}}_{R}^{T} , {\vec{B}}_{R}^{T} , {\vec{C}}_{R}^{T}\right)\right) = \left(\begin{matrix}4 & 4 & - 5 \\ - 1 & - 1 & 1 \\ - 5 & - 4 & 6\end{matrix}\right)$

The reciprocal vectors and the cofactor transpose matrix have two interesting properties:
$\vec{A} \cdot {\vec{A}}_{R} = \vec{B} \cdot {\vec{B}}_{R} = \vec{C} \cdot {\vec{C}}_{R} = \det \left(M\right)$
and
${M}^{-} 1 = \frac{\overline{M}}{\det} M$

So we can determine that:
$\det \left(M\right) = \vec{C} \cdot {\vec{C}}_{R} = \left(1 , 4 , 0\right) \cdot \left(- 5 , 1 , 6\right) = - 1$

This means that:
${M}^{-} 1 = - \frac{\overline{M}}{1} = - \left(\begin{matrix}4 & 4 & - 5 \\ - 1 & - 1 & 1 \\ - 5 & - 4 & 6\end{matrix}\right) = \left(\begin{matrix}- 4 & - 4 & 5 \\ 1 & 1 & - 1 \\ 5 & 4 & - 6\end{matrix}\right)$