# Continuum mechanics/Matrices

Much of finite elements revolves around forming matrices and solving systems of linear equations using matrices. This learning resource gives you a brief review of matrices.

## Matrices

Suppose that you have a linear system of equations

\begin{align} a_{11} x_1 + a_{12} x_2 + a_{13} x_3 + a_{14} x_4 &= b_1 \\ a_{21} x_1 + a_{22} x_2 + a_{23} x_3 + a_{24} x_4 &= b_2 \\ a_{31} x_1 + a_{32} x_2 + a_{33} x_3 + a_{34} x_4 &= b_3 \\ a_{41} x_1 + a_{42} x_2 + a_{43} x_3 + a_{44} x_4 &= b_4 \end{align} ~.

Matrices provide a simple way of expressing these equations. Thus, we can instead write

$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix} ~.$

An even more compact notation is

$\left[\mathsf{A}\right] \left[\mathsf{x}\right] = \left[\mathsf{b}\right] ~~~~\text{or}~~~~ \mathbf{A} \mathbf{x} = \mathbf{b} ~.$

Here $\mathbf{A}$ is a $4\times 4$ matrix while $\mathbf{x}$ and $\mathbf{b}$ are $4\times 1$ matrices. In general, an $m \times n$ matrix $\mathbf{A}$ is a set of numbers arranged in $m$ rows and $n$ columns.

$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & a_{m3} & \dots & a_{mn} \end{bmatrix}~.$

### Practice Exercises

Practice: Expressing Linear Equations As Matrices

## Types of Matrices

Common types of matrices that we encounter in finite elements are:

• a row vector that has one row and $n$ columns.
$\mathbf{v} = \begin{bmatrix} v_1 & v_2 & v_3 & \dots & v_n \end{bmatrix}$
• a column vector that has $n$ rows and one column.
$\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \vdots \\ v_n \end{bmatrix}$
• a square matrix that has an equal number of rows and columns.
• a diagonal matrix which is a square matrix with only the

diagonal elements ($a_{ii}$) nonzero.

$\mathbf{A} = \begin{bmatrix} a_{11} & 0 & 0 & \dots & 0 \\ 0 & a_{22} & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & a_{nn} \end{bmatrix}~.$
• the identity matrix ($\mathbf{I}$) which is a diagonal matrix and

with each of its nonzero elements ($a_{ii}$) equal to 1.

$\mathbf{A} = \begin{bmatrix} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \end{bmatrix}~.$
• a symmetric matrix which is a square matrix with elements

such that $a_{ij} = a_{ji}$.

$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{12} & a_{22} & a_{23} & \dots & a_{2n} \\ a_{13} & a_{23} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & a_{3n} & \dots & a_{nn} \end{bmatrix}~.$
• a skew-symmetric matrix which is a square matrix with elements

such that $a_{ij} = -a_{ji}$.

$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ -a_{12} & a_{22} & a_{23} & \dots & a_{2n} \\ -a_{13} & -a_{23} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -a_{1n} & -a_{2n} & -a_{3n} & \dots & a_{nn} \end{bmatrix}~.$

Note that the diagonal elements of a skew-symmetric matrix have to be zero: $a_{ii} = -a_{ii} \Rightarrow a_{ii} = 0$.

Let $\mathbf{A}$ and $\mathbf{B}$ be two $m \times n$ matrices with components $a_{ij}$ and $b_{ij}$, respectively. Then

$\mathbf{C} = \mathbf{A} + \mathbf{B} \implies c_{ij} = a_{ij} + b_{ij}$

## Multiplication by a scalar

Let $\mathbf{A}$ be a $m \times n$ matrix with components $a_{ij}$ and let $\lambda$ be a scalar quantity. Then,

$\mathbf{C} = \lambda\mathbf{A} \implies c_{ij} = \lambda a_{ij}$

## Multiplication of matrices

Let $\mathbf{A}$ be a $m \times n$ matrix with components $a_{ij}$. Let $\mathbf{B}$ be a $p \times q$ matrix with components $b_{ij}$.

The product $\mathbf{C} = \mathbf{A} \mathbf{B}$ is defined only if $n = p$. The matrix $\mathbf{C}$ is a $m \times q$ matrix with components $c_{ij}$. Thus,

$\mathbf{C} = \mathbf{A} \mathbf{B} \implies c_{ij} = \sum^n_{k=1} a_{ik} b_{kj}$

Similarly, the product $\mathbf{D} = \mathbf{B} \mathbf{A}$ is defined only if $q = m$. The matrix $\mathbf{D}$ is a $p \times n$ matrix with components $d_{ij}$. We have

$\mathbf{D} = \mathbf{B} \mathbf{A} \implies d_{ij} = \sum^m_{k=1} b_{ik} a_{kj}$

Clearly, $\mathbf{C} \ne \mathbf{D}$ in general, i.e., the matrix product is not commutative.

However, matrix multiplication is distributive. That means

$\mathbf{A} (\mathbf{B} + \mathbf{C}) = \mathbf{A} \mathbf{B} + \mathbf{A} \mathbf{C} ~.$

The product is also associative. That means

$\mathbf{A} (\mathbf{B} \mathbf{C}) = (\mathbf{A} \mathbf{B}) \mathbf{C} ~.$

## Transpose of a matrix

Let $\mathbf{A}$ be a $m \times n$ matrix with components $a_{ij}$. Then the transpose of the matrix is defined as the $n \times m$ matrix $\mathbf{B} = \mathbf{A}^T$ with components $b_{ij} = a_{ji}$. That is,

$\mathbf{B} = \mathbf{A}^T = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & a_{m3} & \dots & a_{mn} \end{bmatrix}^T = \begin{bmatrix} a_{11} & a_{21} & a_{31} & \dots & a_{m1} \\ a_{12} & a_{22} & a_{32} & \dots & a_{m2} \\ a_{13} & a_{23} & a_{33} & \dots & a_{m3} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & a_{3n} & \dots & a_{mn} \end{bmatrix}$

An important identity involving the transpose of matrices is

${ (\mathbf{A} \mathbf{B})^T = \mathbf{B}^T \mathbf{A}^T }~.$

## Determinant of a matrix

The determinant of a matrix is defined only for square matrices.

For a $2 \times 2$ matrix $\mathbf{A}$, we have

$\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \implies \det(\mathbf{A}) = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix} = a_{11} a_{22} - a_{12} a_{21} ~.$

For a $n \times n$ matrix, the determinant is calculated by expanding into minors as

\begin{align} &\det(\mathbf{A}) = \begin{vmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \dots & a_{nn} \end{vmatrix} \\ &= a_{11} \begin{vmatrix} a_{22} & a_{23} & \dots & a_{2n} \\ a_{32} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n2} & a_{n3} & \dots & a_{nn} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} & \dots & a_{2n} \\ a_{31} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n3} & \dots & a_{nn} \end{vmatrix} + \dots \pm a_{1n} \begin{vmatrix} a_{21} & a_{22} & \dots & a_{2(n-1)} \\ a_{31} & a_{32} & \dots & a_{3(n-1)} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{n(n-1)} \end{vmatrix} \end{align}

In short, the determinant of a matrix $\mathbf{A}$ has the value

${ \det(\mathbf{A}) = \sum^n_{i=1} (-1)^{i+j} a_{ij} M_{ij} }$

where $M_{ij}$ is the determinant of the submatrix of $\mathbf{A}$ formed by eliminating row $i$ and column $j$ from $\mathbf{A}$.

Some useful identities involving the determinant are given below.

• If $\mathbf{A}$ is a $n \times n$ matrix, then
$\det(\mathbf{A}) = \det(\mathbf{A}^T)~.$
• If $\lambda$ is a constant and $\mathbf{A}$ is a $n \times n$ matrix, then
$\det(\lambda\mathbf{A}) = \lambda^n\det(\mathbf{A}) \implies \det(-\mathbf{A}) = (-1)^n\det(\mathbf{A}) ~.$
• If $\mathbf{A}$ and $\mathbf{B}$ are two $n \times n$ matrices, then
$\det(\mathbf{A}\mathbf{B}) = \det(\mathbf{A})\det(\mathbf{B})~.$

If you think you understand determinants, take the quiz.

## Inverse of a matrix

Let $\mathbf{A}$ be a $n \times n$ matrix. The inverse of $\mathbf{A}$ is denoted by $\mathbf{A}^{-1}$ and is defined such that

${ \mathbf{A} \mathbf{A}^{-1} = \mathbf{I} }$

where $\mathbf{I}$ is the $n \times n$ identity matrix.

The inverse exists only if $\det(\mathbf{A}) \ne 0$. A singular matrix does not have an inverse.

An important identity involving the inverse is

${ (\mathbf{A}\mathbf{B})^{-1} = \mathbf{B}^{-1} \mathbf{A}^{-1}, }$

since this leads to: ${ (\mathbf{A} \mathbf{B})^{-1} (\mathbf{A} \mathbf{B}) = (\mathbf{B}^{-1} \mathbf{A}^{-1}) (\mathbf{A} \mathbf{B} ) = \mathbf{B}^{-1} \mathbf{A}^{-1} \mathbf{A} \mathbf{B} = \mathbf{B}^{-1} (\mathbf{A}^{-1} \mathbf{A}) \mathbf{B} = \mathbf{B}^{-1} \mathbf{I} \mathbf{B} = \mathbf{B}^{-1} \mathbf{B} = \mathbf{I}. }$

Some other identities involving the inverse of a matrix are given below.

• The determinant of a matrix is equal to the multiplicative inverse of the

determinant of its inverse.

$\det(\mathbf{A}) = \cfrac{1}{\det(\mathbf{A}^{-1})}~.$
• The determinant of a similarity transformation of a matrix

is equal to the original matrix.

$\det(\mathbf{B} \mathbf{A} \mathbf{B}^{-1}) = \det(\mathbf{A}) ~.$

We usually use numerical methods such as Gaussian elimination to compute the inverse of a matrix.

## Eigenvalues and eigenvectors

A thorough explanation of this material can be found at Eigenvalue, eigenvector and eigenspace. However, for further study, let us consider the following examples:

• Let :$\mathbf{A} = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix} , \mathbf{v} = \begin{bmatrix} 6 \\ -5 \end{bmatrix} , \mathbf{t} = \begin{bmatrix} 7 \\ 4 \end{bmatrix}~.$

Which vector is an eigenvector for $\mathbf{A}$ ?

We have $\mathbf{A}\mathbf{v} = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} 6 \\ -5 \end{bmatrix} = \begin{bmatrix} -24 \\ 20 \end{bmatrix} = -4\begin{bmatrix} 6 \\ -5 \end{bmatrix}$ , and $\mathbf{A}\mathbf{t} = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} 7 \\ 4 \end{bmatrix} = \begin{bmatrix} 31 \\ 43 \end{bmatrix}~.$

Thus, $\mathbf{v}$ is an eigenvector.

• Is $\mathbf{u} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}$ an eigenvector for $\mathbf{A} = \begin{bmatrix} -3 & -3 \\ 1 & 8 \end{bmatrix}$ ?

We have that since $\mathbf{A}\mathbf{u} = \begin{bmatrix} -3 & -3 \\ 1 & 8 \end{bmatrix}\begin{bmatrix} 1 \\ 4 \end{bmatrix} = \begin{bmatrix} -15 \\ 33 \end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} 1 \\ 4 \end{bmatrix}$ is not an eigenvector for $\mathbf{A} = \begin{bmatrix} -3 & -3 \\ 1 & 8 \end{bmatrix}~.$