Theoretical part

Matrix is a set of objects arranged in form of the table of size \(n\times m\).

• Matrix summation

\[ A+B = \left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\\a_{41}&a_{42}\end{array}\right)+\left(\begin{array}{cc}b_{11}&b_{12}\\b_{21}&b_{22}\\b_{31}&b_{32}\\b_{41}&b_{42}\end{array}\right)=\left(\begin{array}{cc}a_{11}+b_{11}&a_{12}+b_{12}\\a_{21}+b_{21}&a_{22}+b_{22}\\a_{31}+b_{31}&a_{32}+b_{32}\\a_{41}+b_{41}&a_{42}+b_{42}\end{array}\right). \]

• Matrix multiplication by the constant \[ \lambda A = \lambda\cdot\left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\\a_{41}&a_{42}\end{array}\right) = \left(\begin{array}{cc}\lambda a_{11}&\lambda a_{12}\\\lambda a_{21}&\lambda a_{22}\\\lambda a_{31}&\lambda a_{32}\\\lambda a_{41}&\lambda a_{42}\end{array}\right) \]

Number of columns in matrix A should equal to the number of rows in matrix B!

Matrix product is not commutative which means that \(AB\ne BA\) in general.

• Matrix multiplication \[ \begin{multline*} AB = \left(\begin{array}{cc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{array}\right)\left(\begin{array}{cc}b_{11}&b_{12}&b_{13}&b_{14}\\b_{21}&b_{22}&b_{23}&b_{24}\\b_{31}&b_{32}&b_{33}&b_{34}\end{array}\right) =\\ \left(\begin{array}{cc}a_{11}b_{11}+a_{12}b_{21}+a_{13}b_{31}&a_{11}b_{12}+a_{12}b_{22}+a_{13}b_{32}&a_{11}b_{13}+a_{12}b_{23}+a_{13}b_{33}\\ a_{21}b_{11}+a_{22}b_{21}+a_{23}b_{31}&a_{21}b_{12}+a_{22}b_{22}+a_{23}b_{32}&a_{21}b_{13}+a_{22}b_{23}+a_{23}b_{33}\\ a_{31}b_{11}+a_{32}b_{21}+a_{33}b_{31}&a_{31}b_{12}+a_{32}b_{22}+a_{33}b_{32}&a_{31}b_{13}+a_{32}b_{23}+a_{33}b_{33}\end{array}\right) \end{multline*} \]

• Matrix transposition

Rows become columns and columns become rows. \[ A^T = \left(\begin{array}{cc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{array}\right)^T=\left(\begin{array}{cc}a_{11}&a_{21}\\a_{12}&a_{22}\\a_{13}&a_{23}\end{array}\right) \]

• Determinant of the matrix

Main formulas: \[ \left|\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right| = a_{11}a_{22}-a_{12}a_{21}. \] \[ \begin{multline*} \left|\begin{array}{cc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right| = a_{11}\left|\begin{array}{cc}a_{22}&a_{23}\\a_{32}&a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}a_{21}&a_{23}\\a_{31}&a_{33}\end{array}\right|+a_{31}\left|\begin{array}{cc}a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right|=\\ a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}+a_{31}a_{21}a_{32}-a_{31}a_{22}a_{31}. \end{multline*} \] Alternatively, it may be convenient to consider a bit different notation such as follows: \[ \begin{multline*} \left|\begin{array}{cc}a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{array}\right| = a_{1}\left|\begin{array}{cc}b_{2}&c_{2}\\b_{3}&c_{3}\end{array}\right|-b_{1}\left|\begin{array}{cc}a_{2}&a_{3}\\c_{2}&c_{3}\end{array}\right|+c_{1}\left|\begin{array}{cc}a_{2}&b_{2}\\a_{3}&b_{3}\end{array}\right|=\\ a_{1}b_{2}c_{3}-a_{1}b_{3}c_{2}+a_{3}b_{1}c_{2}-a_{2}b_{1}c_{3}+a_{2}b_{3}c_{1}-a_{3}b_{2}c_{1}, \end{multline*} \] where the sign in the last line depend on the odd/even number of permutations to the sequence \(\{1,2,3\}\).

We may also write that: \[ \det A = \sum_{k=1}^n (-1)^{i+k}a_{ik}M_{ik}, \] respectively to the matrix row \(a_{i*}\) and where \(M_{ik}\) is a minor for the element \(a_{ik}\)11 Please see https://en.wikipedia.org/wiki/Minor_(linear_algebra); or: \[ \det A = \sum_{k=1}^n (-1)^{k+j}a_{kj}M_{kj}, \] respectively to the matrix column \(a_{*j}\).

• Identity matrix

Identity matrix is a matrix with ones on the diagonal, and zeros for nondiagonal elements.

In the literature it is sometimes denoted as \(I\) or \(E\):

\[ E = \left(\begin{array}{cc}1&0&0\\0&1&0\\0&0&1\end{array}\right) \] Note that: \(AE=EA=A\).

• Inverse matrix

Can we find such matrix \(A^{-1}\) such that: \(A^{-1}A=I\)?

Formula: \[ A^{-1} = \frac1{\det A}\mathrm{adj}(A), \] where \(\mathrm{adj}(A)\) is an anjoin (or adjugate) matrix of \(A\). The adjoin matrix is a transposed matrix of co-factors (\(=\) minors with the factors \((-1)^{i+j}\)) of the elements of matrix \(A\). This means that for the matrix \(2\times2\): \[ A = \left(\begin{array}{rr}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right), \] we write \[ \mathrm{adj} (A) = \left(\begin{array}{rr}a_{22}&-a_{21}\\-a_{12}&a_{22}\end{array}\right)^T=\left(\begin{array}{rr}a_{22}&-a_{12}\\-a_{21}&a_{22}\end{array}\right). \] Whereas for the matrix \(3\times3\): \[ A = \left(\begin{array}{cc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{array}\right), \] we write: \[ \mathrm{adj} (A) = \left(\begin{array}{rrr}\left|\begin{array}{cc}a_{22}&a_{23}\\a_{32}&a_{33}\end{array}\right|&-\left|\begin{array}{cc}a_{21}&a_{23}\\a_{31}&a_{33}\end{array}\right|&\left|\begin{array}{cc}a_{21}&a_{22}\\a_{31}&a_{32}\end{array}\right|\\ -\left|\begin{array}{cc}a_{12}&a_{13}\\a_{32}&a_{33}\end{array}\right|&\left|\begin{array}{cc}a_{11}&a_{13}\\a_{31}&a_{33}\end{array}\right|&-\left|\begin{array}{cc}a_{11}&a_{12}\\a_{31}&a_{32}\end{array}\right|\\ \left|\begin{array}{cc}a_{12}&a_{13}\\a_{22}&a_{23}\end{array}\right|&-\left|\begin{array}{cc}a_{11}&a_{13}\\a_{21}&a_{23}\end{array}\right|&\left|\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right|\end{array}\right)^T. \] Do not forget to transpose the matrix of co-factors at the end!

• Eigenvalue

If we have: \[ Ay = \lambda y, \] the scalar \(\lambda\) is called an eigenvalue, the corresponding vector \(y\) is an eigenvector22 Eigenvectors are not covered in this session.

We write: \[ (A-\lambda E)y = 0 \], which has a non-zero solution \(y\) only if: \[ \det(A-\lambda E) = 0. \]

Practical part

• Problem 1

Calculate the sum of two vectors: \[ \left(\begin{array}{c} 4 \\ 0 \\ 1 \end{array} \right) + 2\left(\begin{array}{c} {-}1 \\ 5 \\ 1 \end{array} \right) = \left(\begin{array}{cc} 4+2\times(-1) \\ 0+2\times5 \\ 1+2\times1 \end{array} \right) = \left(\begin{array}{cc} 2 \\ 10 \\ 3 \end{array} \right). \]

• Problem 2

Calculate the product of two one-row and one-column matrices: \[ \left(\begin{array}{c} 1 & -2 & 4 \end{array} \right) \times\left(\begin{array}{c} 4 \\ 3 \\ 1 \end{array} \right)=1\times4+(-2)\times3+4\times1=2. \]

• Problem 3

Calculate the product of two matrices: \[ \left(\begin{array}{cc} 3 & 1 \\ 2 & 1 \\ 4 & 0 \end{array} \right) \times\left(\begin{array}{c} 1 & {-}1 & {-}2 \\ 2 & 1 & 3 \end{array} \right)=\left(\begin{array}{ccc} 3\times1+1\times2 & 3\times(-1)+1\times1 & 3\times({-}2)+1\times3 \\ 2\times1+1\times2 & 2\times(-1)+1\times1 & 2\times(-2)+1\times3 \\ 4\times1+0\times2 & 4\times(-1)+0\times1 & 4\times(-2)+0\times3 \end{array}\right) = \left(\begin{array}{ccc} 5 & -2 & -3 \\ 4 & -1 & -1 \\ 4 & -4 & -8 \end{array}\right) \]

• Problem 4

Calculate the transposed matrix: \[ \left(\begin{array}{cccc} 0 & \cdots & 0 & \lambda_1 \\ \vdots & & \lambda_2 & 0 \\ 0 & & & \vdots \\\lambda_n & 0 &\cdots & 0 \end{array} \right)^T=\left(\begin{array}{cccc} 0 & \cdots & 0 & \lambda_n \\ \vdots & & \lambda_{n-1} & 0 \\ 0 & & & \vdots \\\lambda_1 & 0 &\cdots & 0 \end{array} \right) \]

• Problem 5

Calculte the determinant: \[ \begin{multline*} \left|\begin{array}{rrr} 2 & 1 & -1 \\ 3 & 1 & -2 \\ 1 & 0 & 1 \end{array} \right| = (-1)^{3+1}\cdot1\cdot\left|\begin{array}{rr} 1 & -1 \\ 1 & -2 \end{array} \right| + (-1)^{3+2}\cdot0\cdot\left|\begin{array}{rr} 2 & -1 \\ 3 & -2 \end{array}\right| + (-1)^{3+3}\cdot1\cdot\left|\begin{array}{rr} 2 & 1 \\ 3 & 1 \end{array}\right|={}\\ (1\times(-2)-1\times(-1))+(2\times1-1\times3) = -1-1=-2 \end{multline*} \]

# We define the matrix A
A = matrix(c(2,1,-1,3,1,-2,1,0,1), 3, 3, byrow = TRUE)
A

##      [,1] [,2] [,3]
## [1,]    2    1   -1
## [2,]    3    1   -2
## [3,]    1    0    1

det(A)

## [1] -2

# We define the matrix A
A = matrix(c(2,-1,0,0,2,-1,-1,-1,1), 3, 3, byrow = TRUE)
A

##      [,1] [,2] [,3]
## [1,]    2   -1    0
## [2,]    0    2   -1
## [3,]   -1   -1    1

det(A)

## [1] 1

solve(A)

##      [,1] [,2] [,3]
## [1,]    1    1    1
## [2,]    1    2    2
## [3,]    2    3    4

# We define the matrix A
A = matrix(c(2,5,1,3), 2, 2, byrow = TRUE)
A

##      [,1] [,2]
## [1,]    2    5
## [2,]    1    3

Ainv = solve(A)
Ainv

##      [,1] [,2]
## [1,]    3   -5
## [2,]   -1    2

# Then matrix in the right hand side
B = matrix(c(2,1,1,1), 2, 2, byrow = TRUE)
B

##      [,1] [,2]
## [1,]    2    1
## [2,]    1    1

Ainv%*%B

##      [,1] [,2]
## [1,]    1   -2
## [2,]    0    1

A = matrix(c(25,60,60,144),2,2,byrow=TRUE)
A

##      [,1] [,2]
## [1,]   25   60
## [2,]   60  144

eigen(A)$values

## [1] 1.690000e+02 3.552714e-15

A = matrix(c(1,2,1,1,2,4,-1,-2,-3),3,3,byrow=TRUE)
A

##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    1    2    4
## [3,]   -1   -2   -3

eigen(A)$values

## [1]  1.377528e-05+0.000000e+00i -6.887639e-06+1.192973e-05i
## [3] -6.887639e-06-1.192973e-05i

A = matrix(c(5,-2,4,-4,-2,8,2,-2,0,0,4,5,0,0,5,4),4,4,byrow=TRUE)
A

##      [,1] [,2] [,3] [,4]
## [1,]    5   -2    4   -4
## [2,]   -2    8    2   -2
## [3,]    0    0    4    5
## [4,]    0    0    5    4

eigen(A)$values

## [1]  9  9  4 -1