title: MMME1026 // Systems of Equations and Matrices

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Systems of Equations (Simultaneous Equations)

Gaussian Elimination

Gaussian eliminiation can be used when the number of unknown variables you have is equal to the number of equations you are given.

I’m pretty sure it’s the name for the method you use to solve simultaneous equations in school.

For example if you have 1 equation and 1 unknown:

2x=6x=3\begin{align*} 2x &= 6 \\ x &= 3 \end{align*}

Number of Solutions

Let’s generalise the example above to

ax=bax = b

There are 3 possible cases:

a0x=baa=0,b0no solution for xa=0,b=0infinite solutions for x\begin{align*} a \ne 0 &\rightarrow x = \frac b a \\ a = 0, b \ne 0 &\rightarrow \text{no solution for $x$} \\ a = 0, b = 0 &\rightarrow \text{infinite solutions for $x$} \end{align*}

2x2 Systems

A 2x2 system is one with 2 equations and 2 unknown variables.

Example 1

3x1+4x2=2(1)x1+2x2=0(2)\begin{align*} 3x_1 + 4x_2 &= 2 &\text{(1)} \\ x_1 + 2x_2 &= 0 &\text{(2)} \\ \end{align*}

3×(2)=3x1+6x2=0(3)(3)(1)=0x1+2x2=2x2=1\begin{align*} 3\times\text{(2)} = 3x_1 + 6x_2 &= 0 &\text{(3)} \\ \text{(3)} - \text{(1)} = 0x_1 + 2x_2 &= -2 \\ x_2 &= -1 \end{align*}

We’ve essentially created a 1x1 system for x2x_2 and now that’s solved we can back substitute it into equation (1) (or equation (2), it doesn’t matter) to work out the value of x1x_1:

3x1+4x2=23x11=23x1=6x1=2\begin{align*} 3x_1 + 4x_2 &= 2 \\ 3x_1 - 1 &= 2 \\ 3x_1 &= 6 \\ x_1 &= 2 \end{align*}

You can check the values for x1x_1 and x2x_2 are correct by substituting them into equation (2).

3x3 Systems

A 3x3 system is one with 3 equations and 3 unknown variables.

Example 1

2x1+3x2x3=5(1)4x1+4x23x3=5(2)2x13x2+x3=5(3)\begin{align*} 2x_1 + 3x_2 - x_3 &= 5 &\text{(1)} \\ 4x_1 + 4x_2 - 3x_3 &= 5 &\text{(2)} \\ 2x_1 - 3x_2 + x_3 &= 5 &\text{(3)} \\ \end{align*}

The first step is to eliminate x1x_1 from (2) and (3) using (1):

(2)2×(1)=2x2x3=1(4)(3)(1)=6x2+3x3=6(5)\begin{align*} \text{(2)}-2\times\text{(1)} = -2x_2 -x_3 &= -1 &\text{(4)} \\ \text{(3)}-\text{(1)} = -6x_2 + 3x_3 &= -6 &\text{(5)} \\ \end{align*}

This has created a 2x2 system of x2x_2 and x3x_3 which can be solved as any other 2x2 system. I’m too lazy to type up the working, but it is solved like any other 2x2 system.

x2=2x3=5\begin{align*} x_2 &= -2 x_3 &= 5 \end{align*}

These values can be back-substituted into any of the first 3 equations to find out x1x_1:

2x1+3x2x3=2x1+63=5x1=1\begin{align*} -2x_1 + 3x_2 - x_3 = 2x_1 + 6 - 3 = 5 \rightarrow x_1 = 1 \end{align*}

Example 2

x1+x22x3=1R12x1x2x3=1R2x1+4x2+7x3=2R3\begin{align*} x_1 + x_2 - 2x_3 &= 1 &R_1 \\ 2x_1 - x_2 - x_3 &= 1 &R_2 \\ x_1 + 4x_2 + 7x_3 &= 2 &R_3 \\ \end{align*}

  1. Eliminate x1x_1 from R2R_2, R3R_3:

    x1+x22x3=1R1=R13x25x3=1R2=R22R13x2+5x3=1R3=R3R1\begin{align*} x_1 + x_2 - 2x_3 &= 1 &R_1' = R_1\\ - 3x_2 - 5x_3 &= -1 &R_2' = R_2 - 2R_1 \\ 3x_2 + 5x_3 &= 1 &R_3' = R_3 - R_1 \\ \end{align*}

    We’ve created another 2x2 system of R2R_2' and R3R_3'

  2. Eliminate x2x_2 from R3R_3''

    x1+x22x3=1R1=R1=R13x25x3=1R2=R2=R22R10x3=0R3=R3+R2\begin{align*} x_1 + x_2 - 2x_3 &= 1 &R_1'' = R_1' = R_1\\ - 3x_2 - 5x_3 &= -1 &R_2'' = R_2' = R_2 - 2R_1 \\ 0x_3 &= 0 &R_3'' = R_3 '+ R_2' \\ \end{align*}

    We can see that x3x_3 can be any number, so there are infinite solutions. Let:

    x3=tx_3 = t

    where tt can be any number

  3. Substitute x3x_3 into R2R_2'':

    R2=3x25t=1x2=135t3R_2'' = -3x_2 - 5t = -1 \rightarrow x_2 = \frac 1 3 - \frac{5t} 3

  4. Substitute x2x_2 and x3x_3 into R1R_1'':

    R1=x1+135t3+2t=1x1=23t3R_1'' = x_1 + \frac 1 3 - \frac{5t} 3 + 2t = 1 \rightarrow x_1 = \frac 2 3 - \frac t 3

Systems of Equations and Matrices

Many problems in engineering have a very large number of unknowns and equations to solve simultaneously. We can use matrices to solve these efficiently.

Take the following simultaneous equations::

3x1+4x2=2(1)x1+2x2=0(2)\begin{align*} 3x_1 + 4x_2 &= 2 &\text{(1)} \\ x_1 + 2x_2 &= 0 &\text{(2)} \end{align*}

They can be represented by the following matrices:

A=(3412)𝐱=(x1x2)𝐛=(20)\begin{align*} A &= \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix} \\ \pmb x &= \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \\ \pmb b &= \begin{pmatrix} 2 \\ 0 \end{pmatrix} \\ \end{align*}

You can then express the system as:

A𝐱=𝐛A\pmb x = \pmb b

A 3x3 System as a Matrix

2x1+3x2x3=54x1+4x23x3=32x13x2+x3=1\begin{align*} 2x_1 + 3x_2 - x_3 &= 5 \\ 4x_1 + 4x_2 - 3x_3 &= 3 \\ 2x_1 - 3x_2 + x_3 &= -1 \end{align*}

Could be expressed in the form A𝐱=𝐛A\pmb x = \pmb b where:

A=(231443231)𝐱=(x1x2x3)𝐛=(531)\begin{align*} A &= \begin{pmatrix} 2 & 3 & -1 \\ 4 & 4 & -3 \\ 2 & -3 & -1 \end{pmatrix} \\ \pmb x &= \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \\ \pmb b &= \begin{pmatrix} 5 \\ 3 \\ -1 \end{pmatrix} \\ \end{align*}

An m×nm\times n System as a Matrix

a11x1+a12x2++a1nxn=b1a21x1+a22x2++a2nxn=b2am1x1+am2x2++amnxn=bm\begin{align*} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n &= b_1 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n &= b_2 \\ \cdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n &= b_m \\ \end{align*}

Could be expressed in the form A𝐱=𝐛A\pmb x = \pmb b where:

A=(a11a12a1na21a22a2nam1am2amn),𝐱=(x1x2xn),𝐛=(b1b2bm)\begin{align*} A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}, \pmb x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, \pmb b = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{pmatrix} \end{align*}

Matrices

Order of a Matrix

The order of a matrix is its size e.g. 3×23\times2 or m×nm\times n

Column Vectors

Matrix Algebra

Equality

Two matrices are the same if:

Addition and Subtraction

Only possible if their order is the same. A+B=(a11+b11a12+b12a1n+b1na21+b21a22+b22a2n+b2nam1+bm1am2+bm2amn+bmn)AB=(a11b11a12b12a1nb1na21b21a22b22a2nb2nam1bm1am2bm2amnbmn),\begin{align*} A + B&= \begin{pmatrix} a_{11} + b_{11} & a_{12} + b_{12} & \cdots & a_{1n} + b_{1n} \\ a_{21} + b_{21} & a_{22} + b_{22} & \cdots & a_{2n} + b_{2n} \\ \vdots & & & \vdots \\ a_{m1} + b_{m1} & a_{m2} + b_{m2} & \cdots & a_{mn} + b_{mn} \end{pmatrix} \\ A - B&= \begin{pmatrix} a_{11} - b_{11} & a_{12} - b_{12} & \cdots & a_{1n} - b_{1n} \\ a_{21} - b_{21} & a_{22} - b_{22} & \cdots & a_{2n} - b_{2n} \\ \vdots & & & \vdots \\ a_{m1} - b_{m1} & a_{m2} - b_{m2} & \cdots & a_{mn} - b_{mn} \end{pmatrix}, \end{align*}

Zero Matrix

This is a matrix whose elements are all zeros. For any matrix AA,

A+0=AA + 0 =A

We can only add matrices of the same order, therefore 0 must be of the same order as AA.

Multiplication

Let

Am×nBp×q \begin{matrix} A & m\times n \\ B & p\times q \end{matrix}

To be able to multiply AA by BB, n=pn = p.

If npn \ne p, then ABAB does not exist.

AB=Cm×np×qm×q \begin{matrix} A & B & = & C \\ m\times n & p \times q & & m\times q \end{matrix}

When C=ABC = AB exists,

Cij=rairbrjC_{ij} = \sum_r\! a_{ir}b_{rj}

That is, CijC_{ij} is the ‘product’ of the iith row of AA and jjth column of BB.

Multiplication of a Matrix by a Scalar

If λ\lambda is a scalar, we define

λa=(λa11λa12λa1nλa21λa22λa2nλam1λam2λamn), \lambda a = \begin{pmatrix} \lambda a_{11} & \lambda a_{12} & \cdots & \lambda a_{1n} \\ \lambda a_{21} & \lambda a_{22} & \cdots & \lambda a_{2n} \\ \vdots & & & \vdots \\ \lambda a_{m1} & \lambda a_{m2} & \cdots & \lambda a_{mn} \end{pmatrix},

Example 1

(1121)(0132)=(3134) \begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix} = \begin{pmatrix} -3 & -1 \\ 3 & 4 \end{pmatrix} (0132)(1121)=(2171) \begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 7 & -1 \end{pmatrix}

Example 2

A=(416321),B=(111210) A = \begin{pmatrix} 4 & 1 & 6 \\ 3 & 2 & 1 \end{pmatrix},\, B = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 0 \end{pmatrix} AB=(11667),BA=(7371058416) AB = \begin{pmatrix} 11 & 6 \\ 6 & 7 \end{pmatrix},\, BA = \begin{pmatrix} 7 & 3 & 7 \\ 10 & 5 & 8 \\ 4 & 1 & 6 \end{pmatrix}

Other Properties of Matrix Algebra

Special Matrices

Square Matrix

Where m=nm = n

Example 1

A 3×33\times3 matrix.

(123456789)\begin{pmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{pmatrix}

Example 2

A 2×22\times2 matrix.

(1245)\begin{pmatrix}1 & 2 \\ 4 & 5 \end{pmatrix}

Identity Matrix

The identity matrix is a square matrix whose eleements are all 0, except the leading diagonal which is 1s. The leading diagonal is the top left to bottom right corner.

It is usually denoted by II or InI_n.

The identity matrix has the properties that

AI=IA=AAI = IA = A

for any square matrix AA of the same order as I, and

Ix=xIx = x

for any vector xx.

Example 1

The 3×33\times3 identity matrix.

(100010001)\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}

Example 2

The 2×22\times2 identity matrix.

(1001)\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}

Transposed Matrix

The transpose of matrix AA of order m×nm\times n is matrix ATA^T which has the order n×mn\times m. It is found by reflecting it along the leading diagonal, or interchanging the rows and columns of AA.

by Lucas Vieira

Let matrix D=EFD = EF, then DT=(EF)T=ETFTD^T = (EF)^T = E^TF^T

Example 1

A=(321456),AT=(342516) A = \begin{pmatrix}3 & 2 & 1 \\ 4 & 5 & 6 \end{pmatrix},\, A^T = \begin{pmatrix}3 & 4 \\ 2 & 5 \\ 1 & 6\end{pmatrix}

Example 2

B=(14),BT=(14) B = \begin{pmatrix}1 \\ 4\end{pmatrix},\, B^T = \begin{pmatrix}1 & 4\end{pmatrix}

Example 3

C=(123051237),CT=(102254317) C = \begin{pmatrix}1 & 2 & 3 \\ 0 & 5 & 1 \\ 2 & 3 & 7\end{pmatrix},\, C^T = \begin{pmatrix}1 & 0 & 2 \\ 2 & 5 & 4 \\ 3 & 1 & 7\end{pmatrix}

Orthogonal Matrices

A matrix, AA, such that

A1=ATA^{-1} = A^T

is said to be orthogonal.

Another way to say this is

AAT=ATA=IAA^T = A^TA = I

Symmetric Matrices

A square matrix which is symmetric about its leading diagonal:

A=ATA = A^T

You can also express this as the matrix AA, where

aij=ajia_{ij} = a_{ji}

is satisfied to all elements.

Example 1

(1013034124163762)\begin{pmatrix} 1 & 0 & -1 & 3 \\ 0 & 3 & 4 & -1 \\ -2 & 4 & -1 & 6 \\ 3 & -7 & 6 & 2 \end{pmatrix}

Anti-Symmetric

A square matrix is anti-symmetric if

A=ATA = -A^T

This can also be expressed as

aij=ajia_{ij} = -a_{ji}

This means that all elements on the leading diagonal must be 0.

Example 1

(015101510)\begin{pmatrix} 0 & -1 & 5 \\ 1 & 0 & 1 \\ -5 & -1 & 0 \end{pmatrix}

The Determinant

Determinant of a 2x2 System

The determinant of a 2x22x2 system is

D=a11a22a12a21D = a_{11}a_{22} - a_{12}a_{21}

It is denoted by

a11a12a21a22 or det(a11a12a21a22) \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} \text{ or } \det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}

Determinant of a 3x3 System

Let

A=(a11a12a13a21a22a23a31a32a33) A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

detA=a11×det(a22a23a32a33)a12×det(a21a23a31a33)+a13×det(a21a22a31a32)\begin{align*} \det A = &a_{11} \times \det \begin{pmatrix}a_{22} & a_{23} \\ a_{32} & a_{33} \end{pmatrix} \\ &-a_{12} \times \det \begin{pmatrix}a_{21} & a_{23} \\ a_{31} & a_{33} \end{pmatrix} \\ &+a_{13} \times \det \begin{pmatrix}a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} \end{align*}

The 2x22x2 matrices above are created by removing any elements on the same row or column as its corresponding coefficient:

Chessboard Determinant

detA\det A may be obtained by expanding out any row or column. To figure out which coefficients should be subtracted and which ones added use the chessboard pattern of signs:

(+++++)\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}

Properties of Determinants

Inverse of a Matrix

If AA is a square matrix, then its inverse matrix is A1A^{-1} and is defined by the property that:

A1A=AA1=IA^{-1}A = AA^{-1} = I

Inverse of a 2x2 Matrix

If AA is the 2x22x2 matrix

A=(a11a12a21a22) A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}

and its determinant, DD, satisfies D0D \ne 0, AA has the inverse A1A^{-1} given by

A1=1D(a22a12a21a11) A^{-1} = \frac 1 D \begin{pmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{pmatrix}

If D=0D = 0, then matrix AA has no inverse.

Example 1

Find the inverse of matrix A=(1523)A = \begin{pmatrix} -1 & 5 \\ 2 & 3 \end{pmatrix}.

  1. Calculate the determinant

    detA=1×35×2=13\det A = -1 \times 3 - 5 \times 2 = -13

    Since detA0\det A \ne 0, the inverse exists.

  2. Calculate A1A^{-1}

    A1=113(3521) A^{-1} = \frac 1 {-13} \begin{pmatrix} 3 & -5 \\ -2 & -1\end{pmatrix}

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