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 $A{A}^{-1}=A^{-1}A=E,$ 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 $A^{\ast }$ is defined as the transpose of the matrix composed of the cofactors of the corresponding elements of matrix $A.$ Thus,

$A^{\ast }=\left(\begin{array}{cccc}{A}_{11}& A_{21}& ...& A_{n1}\\ A_{12}& A_{22}& ...& A_{n2}\\ ⋮& ⋮& \ddots & ⋮\\ A_{1n}& A_{2n}& ...& A_{nn}\end{array}\right).$

The equality holds:

$A^{\ast }A=A{A}^{\ast }=detA\cdot E.$

Hence, if $A$ is a nonsingular matrix, then

$A^{-1}=\frac{1}{detA}{A}^{\ast }.$

Examples:

Using the adjoint matrix method, find the inverses for the following matrices:

1. $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right).$

Solution.

$detA=\left|\begin{array}{cc}1& 2\\ 3& 4\end{array}\right|=1\cdot 4-2\cdot 3=4-6=-2\ne 0.$

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 $A:$

$A_{11}=(-1{)}^{1+1}\cdot 4=4;$

$A_{12}=(-1{)}^{1+2}\cdot 3=-3;$

$A_{21}=(-1{)}^{2+1}\cdot 2=-2;$

$A_{22}=(-1{)}^{2+2}\cdot 1=1.$

From here, we find the adjoint matrix:

$A^{\ast }=\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right).$

$A^{-1}=\frac{1}{detA}{A}^{\ast }=\frac{1}{-2}\left(\begin{array}{cc}4& -2\\ -3& 1\end{array}\right)=\left(\begin{array}{cc}-2& 1\\ 3/2& -1/2\end{array}\right).$

Answer: $\left(\begin{array}{cc}-2& 1\\ 3/2& -1/2\end{array}\right).$

2. $\left(\begin{array}{ccc}2& 5& 7\\ 6& 3& 4\\ 5& -2& -3\end{array}\right).$

Solution.

$detA=\left|\begin{array}{ccc}2& 5& 7\\ 6& 3& 4\\ 5& -2& -3\end{array}\right|=2\cdot 3\cdot (-3)+6\cdot (-2)\cdot 7+5\cdot 4\cdot 5-$ $=5\cdot 3\cdot 7-2\cdot (-2)\cdot 4-5\cdot 6\cdot (-3)=-18-84+100-105+16+90=$ $=-1\ne 0.$

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 $A:$

$A_{11}=(-1{)}^{1+1}\cdot \left|\begin{array}{cc}3& 4\\ -2& -3\end{array}\right|=3\cdot (-3)-4\cdot (-2)=-9+8=-1;$

$A_{12}=(-1{)}^{1+2}\cdot \left|\begin{array}{cc}6& 4\\ 5& -3\end{array}\right|=-(6\cdot (-3)-4\cdot 5)=-(-18-20)=38;$

$A_{13}=(-1{)}^{1+3}\cdot \left|\begin{array}{cc}6& 3\\ 5& -2\end{array}\right|=6\cdot (-2)-5\cdot 3=-12-15=-27;$

$A_{21}=(-1{)}^{2+1}\cdot \left|\begin{array}{cc}5& 7\\ -2& -3\end{array}\right|=-(5\cdot (-3)-7\cdot (-2))=-(-15+14)=1;$

$A_{22}=(-1{)}^{2+2}\cdot \left|\begin{array}{cc}2& 7\\ 5& -3\end{array}\right|=2\cdot (-3)-7\cdot 5=-6-35=-41;$

$A_{23}=(-1{)}^{2+3}\cdot \left|\begin{array}{cc}2& 5\\ 5& -2\end{array}\right|=-(2\cdot (-2)-5\cdot 5)=-(-4-25)=29;$

$A_{31}=(-1{)}^{3+1}\cdot \left|\begin{array}{cc}5& 7\\ 3& 4\end{array}\right|=5\cdot 4-7\cdot 3=20-21=-1;$

$A_{32}=(-1{)}^{3+2}\cdot \left|\begin{array}{cc}2& 7\\ 6& 4\end{array}\right|=-(2\cdot 4-7\cdot 6)=-(8-42)=34;$

$A_{33}=(-1{)}^{3+3}\cdot \left|\begin{array}{cc}2& 5\\ 6& 3\end{array}\right|=2\cdot 3-5\cdot 6=6-30=-24;$

From here, we find the adjoint matrix:

$A^{\ast }=\left(\begin{array}{ccc}-1& 1& -1\\ 38& -41& 34\\ -27& 29& -24\end{array}\right).$

$A^{-1}=\frac{1}{detA}{A}^{\ast }=\frac{1}{-1}\left(\begin{array}{ccc}-1& 1& -1\\ 38& -41& 34\\ -27& 29& -24\end{array}\right)=\left(\begin{array}{ccc}1& -1& 1\\ -38& 41& -34\\ 27& -29& 24\end{array}\right).$

Answer: $\left(\begin{array}{ccc}1& -1& 1\\ -38& 41& -34\\ 27& -29& 24\end{array}\right).$

2. Solve the matrix equation.

$\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)X=$ $\left(\begin{array}{cc}3& 5\\ 5& 9\end{array}\right).$

Solution.

Let's multiply both sides of the equation on the left by ${\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{-1}:$

${\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{-1}$ $\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)X=$ ${\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{-1}\left(\begin{array}{cc}3& 5\\ 5& 9\end{array}\right)\iff$

$EX={\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{-1}\left(\begin{array}{cc}3& 5\\ 5& 9\end{array}\right)\iff$

$X={\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{-1}\left(\begin{array}{cc}3& 5\\ 5& 9\end{array}\right).$

In problem 1, we found ${\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)}^{-1}:$

$A^{-1}=\left(\begin{array}{cc}-2& 1\\ 3/2& -1/2\end{array}\right).$

Thus, $X=\left(\begin{array}{cc}-2& 1\\ 3/2& -1/2\end{array}\right)\left(\begin{array}{cc}3& 5\\ 5& 9\end{array}\right)=\left(\begin{array}{cc}-2\cdot 3+1\cdot 5& -2\cdot 5+1\cdot 9\\ 3/2\cdot 3-1/2\cdot 5& 3/2\cdot 5-1/2\cdot 9\end{array}\right)=$ $=\left(\begin{array}{cc}-1& -1\\ 2& 3\end{array}\right).$

Answer: $\left(\begin{array}{cc}-1& -1\\ 2& 3\end{array}\right).$

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 ${\mathrm{\Gamma }}_A=(A|E)$ 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 ${\mathrm{\Gamma }}_A$ 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:

$\left(\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ 2& -2& 1\end{array}\right).$

Solution.

We form the matrix ${\mathrm{\Gamma }}_A$:

${\mathrm{\Gamma }}_A=\left(\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ 2& -2& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right).$

Denoting the rows of the matrix ${\mathrm{\Gamma }}_A$ by ${\gamma }_1,{\gamma }_2,{\gamma }_3$, we perform the following transformations: ${\gamma }_1^{\prime }={\gamma }_1$, ${\gamma }_2^{\prime }={\gamma }_2-2{\gamma }_1$, ${\gamma }_3^{\prime }={\gamma }_3-2{\gamma }_1$

${\gamma }_1^{″}={\gamma }_1^{\prime },$ ${\gamma }_2^{″}={\gamma }_2^{\prime }-2{\gamma }_3^{\prime },$ ${\gamma }_3^{″}={\gamma }_3^{\prime }-2{\gamma }_2^{\prime }$

${\gamma }_1^{‴}={\gamma }_1^{″}-\frac{2}{9}{\gamma }_2^{″}-\frac{2}{9}{\gamma }_3^{″},$ ${\gamma }_2^{‴}=\frac{1}{9}{\gamma }_2^{″},$ ${\gamma }_3^{‴}=\frac{1}{9}{\gamma }_3^{″}$

We obtain $\left(\begin{array}{ccc}1& 2& 2\\ 2& 1& -2\\ 2& -2& 1\end{array}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\sim$ $\left(\begin{array}{ccc}1& 2& 2\\ 0& -3& -6\\ 0& -6& -3\end{array}|\begin{array}{ccc}1& 0& 0\\ -2& 1& 0\\ -2& 0& 1\end{array}\right)\sim$

$\left(\begin{array}{ccc}1& 2& 2\\ 0& 9& 0\\ 0& 0& 9\end{array}|\begin{array}{ccc}1& 0& 0\\ 2& 1& -2\\ 2& -2& 1\end{array}\right)\sim$ $\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\begin{array}{ccc}\frac{1}{9}& \frac{2}{9}& \frac{2}{9}\\ \frac{2}{9}& \frac{1}{9}& -\frac{2}{9}\\ \frac{2}{9}& -\frac{2}{9}& \frac{1}{9}\end{array}\right)$

Consequently, $A^{-1}=\left(\begin{array}{ccc}\frac{1}{9}& \frac{2}{9}& \frac{2}{9}\\ \frac{2}{9}& \frac{1}{9}& -\frac{2}{9}\\ \frac{2}{9}& -\frac{2}{9}& \frac{1}{9}\end{array}\right).$

Answer: $A^{-1}=\left(\begin{array}{ccc}\frac{1}{9}& \frac{2}{9}& \frac{2}{9}\\ \frac{2}{9}& \frac{1}{9}& -\frac{2}{9}\\ \frac{2}{9}& -\frac{2}{9}& \frac{1}{9}\end{array}\right).$

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. $A=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right).$

Answer: $A^{-1}=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right).$

3. $A=\left(\begin{array}{ccc}3& -4& 5\\ 2& -3& 1\\ 3& -5& -1\end{array}\right).$

Answer: $A^{-1}=\left(\begin{array}{ccc}-8& 29& -11\\ -5& 18& -7\\ 1& -3& 1\end{array}\right).$

4. $A=\left(\begin{array}{cccc}1& 1& 1& 1\\ 1& 1& -1& -1\\ 1& -1& 0& 0\\ 0& 0& 1& -1\end{array}\right).$

Answer: $A^{-1}=\left(\begin{array}{cccc}1/4& 1/4& 1/2& 0\\ 1/4& 1/4& -1/2& 0\\ 1/4& -1/4& 0& 1/2\\ 1/4& -1/4& 0& -1/2\end{array}\right).$

Using elementary transformations, find the inverse for the following matrix:

5. $A=\left(\begin{array}{ccc}2& 7& 3\\ 3& 9& 4\\ 1& 5& 3\end{array}\right).$

Answer: $A^{-1}=\left(\begin{array}{ccc}-7/3& 2& -1/3\\ 5/3& -1& -1/3\\ -2& 1& 1\end{array}\right).$

Solve the matrix equation:

6. $X\cdot \left(\begin{array}{ccc}5& 3& 1\\ 1& -3& -2\\ -5& 2& 1\end{array}\right)=$ $\left(\begin{array}{ccc}-8& 3& 0\\ -5& 9& 0\\ -2& 15& 0\end{array}\right).$

Answer: $A=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right).$

Tags: inverse matrices, linear algebra, matrix, nonsymmetric, orthogonal matrices matrices, symmetric matrices