Linear Algebra

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 $A$.

The adjugate, adj(A), is the matrix of cofactors,

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

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

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

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

For example, if 

$$A=\left[\begin{array}{rrr}3& 2& -1\\ 0& 1& 3\\ -2& 4& 0\end{array}\right]\overline{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]\overline{b}=\left[\begin{array}{c}4\\ 2\\ 1\end{array}\right]$$

Then in $N_1$, $\overline{b}$ 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 $N_2$, $\overline{b}$ 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 $N_3$, $\overline{b}$ will replace column 3

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

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

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

Solution

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

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

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

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

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

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

Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.