Symmetric, nonsymmetric, orthogonal, and inverse matrices.
A square matrix $A$ is called symmetric if $A^T=A.$ A square matrix $B$ is called skew-symmetric if $B^T=-B.$
A square matrix $A$ is called singular (degenerate) if its determinant equals zero, and nonsingular (non-degenerate) otherwise. If $A$ is a nonsingular matrix, then there exists and moreover a unique matrix $A^{-1}$ such that $AA^{-1}=A^{-1}A=E,$ where $E$ is the identity matrix (i.e., a matrix with ones on the main diagonal and zeros elsewhere). The matrix $A^{-1}$ is called the inverse of matrix $A.$
Basic methods for computing the inverse matrix:
The adjoint matrix method. The adjoint matrix $A^*$ is defined as the transpose of the matrix composed of the cofactors of the corresponding elements of matrix $A.$ Thus,
$A^*=\begin{pmatrix}A_{11}&A_{21}&...&A_{n1}\\A_{12}&A_{22}&...&A_{n2}\\\vdots&\vdots&\ddots&\vdots\\A_{1n}&A_{2n}&...&A_{nn}\end{pmatrix}.$
The equality holds:
$A^*A=AA^*=\det A\cdot E.$
Hence, if $A$ is a nonsingular matrix, then
$A^{-1}=\frac{1}{\det A}A^*.$
Examples:
Using the adjoint matrix method, find the inverses for the following matrices:
1. $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}.$
Solution.
$\det A=\begin{vmatrix}1&2\\3&4\end{vmatrix}=1\cdot 4-2\cdot 3=4-6=-2\neq 0.$
Since the determinant is not equal to zero, this matrix is nonsingular and the inverse matrix exists.
Let's find the cofactors of the corresponding elements of matrix $A:$
$A_{11}=(-1)^{1+1}\cdot4=4;$
$A_{12}=(-1)^{1+2}\cdot3=-3;$
$A_{21}=(-1)^{2+1}\cdot2=-2;$
$A_{22}=(-1)^{2+2}\cdot1=1.$
From here, we find the adjoint matrix:
$A^*=\begin{pmatrix}4&-2\\-3&1\end{pmatrix}.$
$A^{-1}=\frac{1}{\det A}A^*=\frac{1}{-2}\begin{pmatrix}4&-2\\-3&1\end{pmatrix}=\begin{pmatrix}-2&1\\ 3/2&-1/2\end{pmatrix}. $
Answer: $\begin{pmatrix}-2&1\\3/2&-1/2\end{pmatrix}.$
2. $\begin{pmatrix}2&5&7\\6&3&4\\5&-2&-3\end{pmatrix}.$
Solution.
$\det A=\begin{vmatrix}2&5&7\\6&3&4\\5&-2&-3\end{vmatrix}=2\cdot 3\cdot (-3)+6\cdot(-2)\cdot 7+5\cdot 4\cdot 5-$ $=5\cdot3\cdot7-2\cdot (-2)\cdot4-5\cdot 6\cdot (-3)=-18-84+100-105+16+90=$ $=-1\neq 0.$
Since the determinant is not equal to zero, this matrix is nonsingular and the inverse matrix exists.
Let's find the cofactors of the corresponding elements of matrix $A:$
$A_{11}=(-1)^{1+1}\cdot\begin{vmatrix}3&4\\-2&-3\end{vmatrix}=3\cdot(-3)-4\cdot(-2)=-9+8=-1;$
$A_{12}=(-1)^{1+2}\cdot\begin{vmatrix}6&4\\5&-3\end{vmatrix}=-(6\cdot(-3)-4\cdot5)=-(-18-20)=38;$
$A_{13}=(-1)^{1+3}\cdot\begin{vmatrix}6&3\\5&-2\end{vmatrix}=6\cdot(-2)-5\cdot3=-12-15=-27;$
$A_{21}=(-1)^{2+1}\cdot\begin{vmatrix}5&7\\-2&-3\end{vmatrix}=-(5\cdot(-3)-7\cdot(-2))=-(-15+14)=1;$
$A_{22}=(-1)^{2+2}\cdot\begin{vmatrix}2&7\\5&-3\end{vmatrix}=2\cdot(-3)-7\cdot5=-6-35=-41;$
$A_{23}=(-1)^{2+3}\cdot\begin{vmatrix}2&5\\5&-2\end{vmatrix}=-(2\cdot(-2)-5\cdot5)=-(-4-25)=29;$
$A_{31}=(-1)^{3+1}\cdot\begin{vmatrix}5&7\\3&4\end{vmatrix}=5\cdot4-7\cdot3=20-21=-1;$
$A_{32}=(-1)^{3+2}\cdot\begin{vmatrix}2&7\\6&4\end{vmatrix}=-(2\cdot4-7\cdot6)=-(8-42)=34;$
$A_{33}=(-1)^{3+3}\cdot\begin{vmatrix}2&5\\6&3\end{vmatrix}=2\cdot3-5\cdot6=6-30=-24;$
From here, we find the adjoint matrix:
$A^*=\begin{pmatrix}-1&1&-1\\38&-41&34\\-27&29&-24\end{pmatrix}.$
$A^{-1}=\frac{1}{\det A}A^*=\frac{1}{-1}\begin{pmatrix}-1&1&-1\\38&-41&34\\-27&29&-24\end{pmatrix}=\begin{pmatrix}1&-1&1\\-38&41&-34\\27&-29&24\end{pmatrix}. $
Answer: $\begin{pmatrix}1&-1&1\\-38&41&-34\\27&-29&24\end{pmatrix}.$
2. Solve the matrix equation.
$\begin{pmatrix}1&2\\3&4\end{pmatrix}X=$ $\begin{pmatrix}3&5\\5&9\end{pmatrix}.$
Solution.
Let's multiply both sides of the equation on the left by $\begin{pmatrix}1&2\\3&4\end{pmatrix}^{-1}:$
$\begin{pmatrix}1&2\\3&4\end{pmatrix}^{-1}$ $\begin{pmatrix}1&2\\3&4\end{pmatrix}X=$ $\begin{pmatrix}1&2\\3&4\end{pmatrix}^{-1}\begin{pmatrix}3&5\\5&9\end{pmatrix}\Leftrightarrow$
$EX=\begin{pmatrix}1&2\\3&4\end{pmatrix}^{-1}\begin{pmatrix}3&5\\5&9\end{pmatrix}\Leftrightarrow$
$X=\begin{pmatrix}1&2\\3&4\end{pmatrix}^{-1}\begin{pmatrix}3&5\\5&9\end{pmatrix}.$
In problem 1, we found $\begin{pmatrix}1&2\\3&4\end{pmatrix}^{-1}:$
$A^{-1}=\begin{pmatrix}-2&1\\ 3/2&-1/2\end{pmatrix}. $
Thus, $X=\begin{pmatrix}-2&1\\3/2&-1/2\end{pmatrix}\begin{pmatrix}3&5\\5&9\end{pmatrix}=\begin{pmatrix}-2\cdot3+1\cdot5&-2\cdot5+1\cdot9\\3/2\cdot3-1/2\cdot5&3/2\cdot5-1/2\cdot9\end{pmatrix}=$ $=\begin{pmatrix}-1&-1\\2&3\end{pmatrix}.$
Answer: $\begin{pmatrix}-1&-1\\ 2&3\end{pmatrix}. $
The method of elementary transformations. Elementary transformations of a matrix include the following:
1. Permutation of rows (columns);
2. Multiplication of a row (column) by a number other than zero;
3. Addition of the elements of one row (column) to the corresponding elements of another row (column), previously multiplied by some number.
For a given matrix $A$ of order $n$, let's construct the rectangular matrix $\Gamma_A=(A|E)$ of size $n\times 2n$, appending the identity matrix $E$ to the right of $A$. Then, using elementary transformations on the rows, we transform the matrix $\Gamma_A$ into the form $(E|B)$, which is always possible if $A$ is nonsingular. Then $B=A^{-1}$.
Example.
3.115. Using the method of elementary transformations, find the inverse of the following matrix:
$\begin{pmatrix}1&2&2\\2&1&-2\\2&-2&1\end{pmatrix}.$
Solution.
We form the matrix $\Gamma_A$:
$\Gamma_A=\left(\begin{array}{cccc} 1&2&2\\2&1&-2\\2&-2&1\end{array}\left|\begin{array}{cccc}1&0&0\\0&1&0\\0&0&1\end{array}\right.\right).$
Denoting the rows of the matrix $\Gamma_A$ by $\gamma_1, \gamma_2, \gamma_3$, we perform the following transformations: $\gamma_1'=\gamma_1$, $\gamma_2'=\gamma_2-2\gamma_1$, $\gamma_3'=\gamma_3-2\gamma_1$
$\gamma_1''=\gamma_1',$ $\gamma_2''=\gamma_2'-2\gamma_3',$ $\gamma_3''=\gamma_3'-2\gamma_2'$
$\gamma_1'''=\gamma_1''-\frac{2}{9}\gamma_2''-\frac{2}{9}\gamma_3'',$ $\gamma_2'''=\frac{1}{9}\gamma_2'',$ $\gamma_3'''=\frac{1}{9}\gamma_3''$
We obtain $\left(\begin{array}{cccc} 1&2&2\\2&1&-2\\2&-2&1\end{array}\left|\begin{array}{cccc}1&0&0\\0&1&0\\0&0&1\end{array}\right.\right)\sim $ $\left(\begin{array}{cccc} 1&2&2\\0&-3&-6\\0&-6&-3\end{array}\left|\begin{array}{cccc}1&0&0\\-2&1&0\\-2&0&1\end{array}\right.\right)\sim$
$\left(\begin{array}{cccc} 1&2&2\\0&9&0\\0&0&9\end{array}\left|\begin{array}{cccc}1&0&0\\2&1&-2\\2&-2&1\end{array}\right.\right)\sim $ $\left(\begin{array}{cccc} 1&0&0\\0&1&0\\0&0&1\end{array}\left|\begin{array}{cccc}\frac{1}{9}&\frac{2}{9}&\frac{2}{9}\\\frac{2}{9}&\frac{1}{9}&-\frac{2}{9}\\\frac{2}{9}&-\frac{2}{9}&\frac{1}{9}\end{array}\right.\right)$
Consequently, $A^{-1}=\begin{pmatrix}\frac{1}{9}&\frac{2}{9}&\frac{2}{9}\\\frac{2}{9}&\frac{1}{9}&-\frac{2}{9}\\\frac{2}{9}&-\frac{2}{9}&\frac{1}{9}\end{pmatrix}.$
Answer: $A^{-1}=\begin{pmatrix}\frac{1}{9}&\frac{2}{9}&\frac{2}{9}\\\frac{2}{9}&\frac{1}{9}&-\frac{2}{9}\\\frac{2}{9}&-\frac{2}{9}&\frac{1}{9}\end{pmatrix}.$
An orthogonal matrix is defined as a matrix for which $A^{-1}=A^T$.
Homework:
1. Prove that any matrix $A$ can be represented, and uniquely so, in the form $A=B+C$, where $B$ is a symmetric matrix, and $C$ is a skew-symmetric matrix.
Using the method of the adjugate matrix, find the inverses for the following matrices:
2. $A=\begin{pmatrix}\cos\alpha &-\sin\alpha\\\sin\alpha&\cos\alpha\end{pmatrix}.$
Answer: $A^{-1}=\begin{pmatrix}\cos\alpha &\sin\alpha\\-\sin\alpha&\cos\alpha\end{pmatrix}.$
3. $A=\begin{pmatrix}3&-4&5\\2&-3&1\\3&-5&-1\end{pmatrix}.$
Answer: $A^{-1}=\begin{pmatrix}-8&29&-11\\-5&18&-7\\1&-3&1\end{pmatrix}.$
4. $A=\begin{pmatrix}1&1&1&1\\1&1&-1&-1\\1&-1&0&0\\0&0&1&-1\end{pmatrix}.$
Answer: $A^{-1}=\begin{pmatrix}1/4&1/4&1/2&0\\1/4&1/4&-1/2&0\\1/4&-1/4&0&1/2\\1/4&-1/4&0&-1/2\end{pmatrix}.$
Using elementary transformations, find the inverse for the following matrix:
5. $A=\begin{pmatrix}2&7&3\\3&9&4\\1&5&3\end{pmatrix}.$
Answer: $A^{-1}=\begin{pmatrix}-7/3&2&-1/3\\5/3&-1&-1/3\\-2&1&1\end{pmatrix}.$
Solve the matrix equation:
6. $X\cdot\begin{pmatrix}5&3&1\\1&-3&-2\\-5&2&1\end{pmatrix}=$ $\begin{pmatrix}-8&3&0\\-5&9&0\\-2&15&0\end{pmatrix}.$
Answer: $A=\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}.$
