It looks like you're using Internet Explorer 11 or older. This website works best with modern browsers such as the latest versions of Chrome, Firefox, Safari, and Edge. If you continue with this browser, you may see unexpected results.

# Linear Algebra

## Inverse by Cofactors

 If $$A\in M_{n\times n}(\mathbb{R})$$ is invertible, then $$(A^{-1})_{ij}=\frac{1}{|A|}C_{ji}$$. $$C_{ji}$$ is known as the adjugate of $$A$$.

Example: Determine the adjugate and inverse of $$A=\left[\begin{array}{rrr} 2&4&-1\\0&3&1\\6&-2&5\end{array}\right]$$

Solution: The nine cofactors are

 $$C_{11}=(1)\left|\begin{array}{rr} 3&1\\-2&5\end{array}\right|=17$$ $$C_{12}=(-1)\left|\begin{array}{rr} 0&1\\6&5\end{array}\right|=6$$ $$C_{13}=(1)\left|\begin{array}{rr} 0&3\\6&-2\end{array}\right|=-18$$ $$C_{21}=(-1)\left|\begin{array}{rr} 4&-1\\-2&5\end{array}\right|=-18$$ $$C_{22}=(1)\left|\begin{array}{rr} 2&-1\\6&5\end{array}\right|=16$$ $$C_{23}=(-1)\left|\begin{array}{rr} 2&4\\6&-2\end{array}\right|=28$$ $$C_{31}=(1)\left|\begin{array}{rr} 4&-1\\4&1\end{array}\right|=7$$ $$C_{32}=(-1)\left|\begin{array}{rr} 2&-1\\0&1\end{array}\right|=-2$$ $$C_{33}=(1)\left|\begin{array}{rr} 2&4\\0&3\end{array}\right|=6$$

$adj(A)=\left[\begin{array}{rrr} C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33}\end{array}\right]=\left[\begin{array}{rrr} 17&-18&7\\6&16&-2\\-18&28&6\end{array}\right]$

We can verify by cofactors that the determinant $$|A|=76$$. Therefore, we can calculate the inverse

$A^{-1}=\frac{1}{|A|}adj(A)=\frac{1}{76}\left[\begin{array}{rrr} 17&-18&7\\6&16&-2\\-18&28&6\end{array}\right]$

## Cramer's Rule

 We can apply Cramer's rule to solve a system of n linear equations in n variables, $$A\bar{x}=\bar{b}$$. If $$|A|\neq 0$$ so that $$A$$ is invertible. Let $$N_i$$ be the matrix obtained from $$A$$ by replacing the i-th column of $$A$$ by $$\bar{b}$$. Then the i-th component of $$\bar{x}$$ in the solution of $$A\bar{x}=\bar{b}$$ is $x_i=\frac{|N_i|}{|A|}$

The matrix $$N_i$$ is defined by replacing $$\bar{b}$$ by the column i.

For example, if

$A=\left[\begin{array}{rrr} 3&2&-1\\0&1&3\\-2&4&0\end{array}\right] \qquad \bar{x}=\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} \qquad \bar{b}=\begin{bmatrix} 4\\2\\1\end{bmatrix}$

Then in $$N_1$$, $$\bar{b}$$ will replace column 1

$N_1=\left[\begin{array}{rrr} \mathbf{4}&2&-1\\\mathbf{2}&1&3\\\mathbf{1}&4&0\end{array}\right]$

In $$N_2$$, $$\bar{b}$$ will replace column 2

$N_2=\left[\begin{array}{rrr} 3&\mathbf{4}&-1\\0&\mathbf{2}&3\\-2&\mathbf{1}&0\end{array}\right]$

In $$N_3$$, $$\bar{b}$$ will replace column 3

$N_3=\left[\begin{array}{rrr} 3&2&\mathbf{4}\\0&1&\mathbf{2}\\-2&4&\mathbf{1}\end{array}\right]$

Example: Solve the following system of equations using Cramer's Rule

\begin{align} 7x+y-4z&=3\\-6x-4y+z&=0\\4x-y-2z&=6\end{align}

Solution

$A=\left[\begin{array}{rrr} 7&1&-4\\-6&-4&1\\4&-1&-2\end{array}\right] \qquad \bar{b}=\begin{bmatrix}3\\0\\6\end{bmatrix}$

To solve for variable x, we need to find the matrix $$N_x$$ by replacing the x or 1st column with $$\bar{b}$$

$N_x=\left[\begin{array}{rrr} \mathbf{3}&1&-4\\\mathbf{0}&-4&1\\\mathbf{6}&-1&-2\end{array}\right]$

We repeat to find $$N_y$$ and $$N_z$$

$N_y=\left[\begin{array}{rrr} 7&\mathbf{3}&-4\\-6&\mathbf{0}&1\\4&\mathbf{6}&-2\end{array}\right]$

$N_z=\left[\begin{array}{rrr} 7&1&\mathbf{3}\\-6&-4&\mathbf{0}\\4&-1&\mathbf{6}\end{array}\right]$

Now we apply Cramer's Rule

$x=\frac{|N_x|}{|A|} = \frac{-63}{-33}=\frac{21}{11} \qquad y=\frac{|N_y|}{|A|} = \frac{78}{-33}=-\frac{26}{11} \qquad z=\frac{|N_z|}{|A|} = \frac{-66}{-33}=2$

The solutions check out when plugging them back into each equation.