Problems from Notation, complementary minors and adjoint matrix

Development:

If we write the $2\times 2$ matrix, $\left(\begin{array}{cc}1& -1\\ 2& 1\end{array}\right)$ the determinant is written $\left|\begin{array}{cc}1& -1\\ 2& 1\end{array}\right|$

And the same for the $3\times 3$ case $\left(\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 1& 1\end{array}\right)\to \left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|$

Solution:

The determinant is written $\left|\begin{array}{cc}1& -1\\ 2& 1\end{array}\right|$ and $\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|$

Hide solution and development

Development:

We have to find the complementary minors of the elements of the main diagonal, that is, of the elements $a_{11},a_{22},a_{33},a_{44}$.

$$a_{11}=\left(\begin{array}{cccc}\xcancel{1}& \cancel{0}& \cancel{1}& \cancel{0}\\ \bcancel{1}& 1& 0& 2\\ \bcancel{1}& 1& 1& 0\\ \bcancel{1}& 2& 3& 1\end{array}\right)=\begin{array}{c}\left|\begin{array}{ccc}1& 0& 2\\ 1& 1& 0\\ 2& 3& 1\end{array}\right|\\ \begin{array}{ccc}1& 0& 2\\ 1& 1& 0\end{array}\end{array}=$$ $$=1\cdot 1\cdot 1+1\cdot 3\cdot 2+\cancel{2\cdot 0\cdot 0}-2\cdot 1\cdot 2-\cancel{0\cdot 3\cdot 1}-\cancel{1\cdot 0\cdot 1}=1+6-4=3$$

Using the formula of Sarrus to calculate the determinant $3\times 3$.

The rest of minors must be calculated in the same way

$$a_{22}=\left(\begin{array}{cccc}1& \cancel{0}& 1& 0\\ \bcancel{1}& \xcancel{1}& \bcancel{0}& \bcancel{2}\\ 1& \cancel{1}& 1& 0\\ 1& \cancel{2}& 3& 1\end{array}\right)=\left|\begin{array}{ccc}1& 1& 0\\ 1& 1& 0\\ 1& 3& 1\end{array}\right|=0$$ (since there are repeated rows)

$$a_{33}=\left(\begin{array}{cccc}1& 0& \cancel{1}& 0\\ 1& 1& \cancel{0}& 2\\ \bcancel{1}& \bcancel{1}& \xcancel{1}& \bcancel{0}\\ 1& 2& \cancel{3}& 1\end{array}\right)=\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 2\\ 1& 2& 1\end{array}\right|=$$ $$=1\cdot 1\cdot 1+1\cdot 2\cdot 0+1\cdot 0\cdot 2-0\cdot 1\cdot 1-2\cdot 2\cdot 1-1\cdot 0\cdot 1=-3$$

$$a_{44}=\left(\begin{array}{cccc}1& 0& 1& \cancel{0}\\ 1& 1& 0& \cancel{2}\\ 1& 1& 1& \cancel{0}\\ \bcancel{1}& \bcancel{2}& \bcancel{3}& \xcancel{1}\end{array}\right)=\left|\begin{array}{ccc}1& 0& 1\\ 1& 1& 0\\ 1& 1& 1\end{array}\right|=$$ $$=1\cdot 1\cdot 1+1\cdot 1\cdot 1+1\cdot 0\cdot 0-1\cdot 1\cdot 1-0\cdot 1\cdot 1-1\cdot 0\cdot 1=1$$

Solution:

$a_{11}=3,a_{22}=0,a_{33}=-3,a_{44}=1$

Hide solution and development

Development:

We have to find $adj(A)$. The first step is to calculate all the complementary minors.

$$a_{11}=\left|\begin{array}{ccc}\xcancel{1}& \cancel{0}& \cancel{2}\\ \bcancel{0}& 3& 1\\ \bcancel{3}& 1& 0\end{array}\right|=\left|\begin{array}{cc}3& 1\\ 1& 0\end{array}\right|=-1$$

$$a_{12}=\left|\begin{array}{cc}0& 1\\ 3& 0\end{array}\right|=-3$$

$$a_{13}=\left|\begin{array}{cc}0& 3\\ 3& 1\end{array}\right|=-9$$

$$a_{21}=\left|\begin{array}{cc}0& 2\\ 1& 0\end{array}\right|=-2$$

$$a_{22}=\left|\begin{array}{cc}1& 2\\ 3& 0\end{array}\right|=-6$$

$$a_{23}=\left|\begin{array}{cc}1& 0\\ 3& 1\end{array}\right|=1$$

$$a_{31}=\left|\begin{array}{cc}0& 2\\ 3& 1\end{array}\right|=-6$$

$$a_{32}=\left|\begin{array}{cc}1& 2\\ 0& 1\end{array}\right|=1$$

$$a_{33}=\left|\begin{array}{cc}1& 0\\ 0& 3\end{array}\right|=3$$

To calculate the adjugate matrix we must bear in mind the signs!

$$\left(\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right)$$ And finally, $$adj(A)=\left(\begin{array}{ccc}-1& 3& -9\\ 2& -6& -1\\ -6& -1& 3\end{array}\right)$$

Solution:

$adj(A)=\left(\begin{array}{ccc}-1& 3& -9\\ 2& -6& -1\\ -6& -1& 3\end{array}\right)$

Hide solution and development