Matrices add in the natural way:

Example:

**Multiplying Matrices **

Multiplying is a little more complex. Remember that you multiply rows by columns.

Example:

**Determinants of Matrices **

To find the determinant of a matrix multiply diagonal corners together and subtract.

Example:

**Inverses of Matrices **

Each entry in the matrix can be divided byin the natural way. An example is shown below

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x+2y=5

x-y=2

has the unique solution x=3, y=1.

We can write the system in matrix form as

Then

The condition nexcessary for a unique solution to exist is then thatexists, OR, sincethat

We can generalize this.

If a system of linear equations can be associated with a square coeefficient matrix with non zero determinant, then the system has a unique solution. More specifically, if we can write a system of linear equations in the formwhereis a square matrix andis the vector of variables to be solved for, then

Example: Find if the system of equations

has a unique solution, and if so find it.

The third equation is reduncant, since it is twice the second equation. The system is equivalent to the system

The coeefficient matrix iswith determinantso the system has a unique solution, given by

]]>A system of equations is degenerate if more than one set of solutions equations and non degenerate if only one set of solutions exists.

A system of equations is inconsistent if no solutions exists.

A system of equations is consistent if solutions exist – either a unique set of solutions or more than one.

The definitions can be illustrated with matrices.

We can represent the simultaneous equations

in matrix form as

This system of equations is consistent and non – degenerate. It is consistent and non – degenerate because the coefficient matrixis invertible, so a solution exists. We can write the solution as- this is unique so the system is non degenerate.

The system of equations

is inconsistent.

It is inconsisten because if we double the first equation, we obtain

4x+6y=14

but from the second equation,

For a system of equations to be inconsistent it must be the case that the coefficient matrix has zero determinant, and no inverse. The coefficient matrix for these simultaneous equations iswith determinant 2*6-3*4=0.

It is possible however for the coefficient matrix to have determinant zero, but for the system to have solutions – many solutions in fact. The system of equations

has many solutions:andandIn fact any number will return a valuesuch thatis a solution of the equations.

We can generalise to higher orders. A system of linear equations has:

A unique solution if the determinant of the coefficient matrix is not equal to zero.

No solutions if the determinant of the coefficient matrix is zero but no equation can be expressed in terms of the others. For the last example above the second equation can be expressed in terms of the first by doubling it, and the first equation can be expressed in terms of the second by halving it.

Many solutions if the determinant of the coefficient matrix is zero and one equation can be expressed in terms of the others.

]]>Style\Material | Steel | Wood | Glass | Paint | Concrete |

Ranch | 5 | 20 | 16 | 7 | 17 |

Cape \: Cod | 7 | 18 | 9 | 12 | 21 |

Colonial | 6 | 25 | 8 | 5 | 13 |

Suppose we are to build 5 Ranch, 7 Cape Cod and 12 Colonial style houses respectively. We can represent this by the vector {jatex options:inline}\mathbf{v} = \begin{pmatrix}5\\7\\12\end{pmatrix}{/jatex}.

{jatex options:inline}\mathbf{v}^TC =(5, 7, 12)\left( \begin{array}{ccccc} 5 & 20 & 16 & 7 & 17 \\ 7 & 18 & 9 & 12 & 21 \\ 6 & 2 & 8 & 5 & 13 \end{array} \right) = (146, 526, 260, 158, 388){/jatex}.

{jatex options:inline}(146, 526, 260, 158, 388) \begin{pmatrix}15\\8\\5\\1\\10\end{pmatrix}=11,736{/jatex}.]]>

There are three types of elementary row operations. 1. Interchanging two rows.

2. Rep;ace any row by a non zero multiple of itself.

3. Replace any row by itself plus or minus a non zero multiple of another row.

The rows represent equations, so each of these operations is an operation on the equations. We aim to manipulating the matrix into upper triangular form. The solution can be obtained by back substitution.

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In ordinary algebra,andrepresent numbers.

In matrix algebra, ifandrepresent matrices, it is not in general the case thatFor example, ifandthen

but

This means that we cannot collectasorand we must be careful to preserve the order of multiplication.

Example. Ifandare matrices, expand and simplify

Simplification is possible ifsinceand

]]>1. Form the adjoint matrix.

2. Permute the signs according to

3. Take the transpose.

4. Find the determinantof the original matrix and divide the matrix by it, which means multiplying the matrix by a factor

Example: Find the inverse of

We find the adjoint matrix by, for each element, crossing out the elements in the same row and column, and finding the determinant of the submatrix left behind. For instannce, take the element 3 in the top left hand corner. Cross out the top row and the first column. The submatrix left behind iswith determinant 0*1-2*2=-4. This goes in place of the 3. The adjoint matrix formed in this way is:

Now permute the signs, which results in those signs labelled with a “–“ in 2. above changing signs.

Take the transpose to obtain

Finally multiply the matrix by the reciprocal of the determinant of the original matrix:

The inverse is

]]>Example: Solve the matrix equation

Write the equation asand factorisethenor

Example: Solve the matrix equation

Write the equation asand factorisethenor

To see why this method does not find all solutions, consider the first equation

Ifthen

]]>so the above equation becomes

ando

We can rearrange this to makethe subject.

The process is a lot like ordinary algebra, with the exception that when using matrices we must be careul to multiply byon the left throughout or on the right throughour, since matrix multiplication is not commutative. This does not matter if onlyand the identity matrix appear in the equation, but it is still a good habit to acquire.

Example: Ifthen

so that multiplying throughout bygivesand

]]>Suppose then that an {jatex options:inline}m \times m{/jatex} matrix satisfies {jatex options:inline}A^2-3A+2I=0{/jatex}. This factorises as {jatex options:inline}(A-I)(A-2I)=0{/jatex} then by the above property {jatex options:inline}det(A-I)det(A-2I)=det(0)=0{/jatex}.

Hence {jatex options:inline}det(A-I)=0{/jatex} or {jatex options:inline}det(A-2I)=0{/jatex} so {jatex options:inline}det(A)=1{/jatex} or {jatex options:inline}det(A)=2^m{/jatex}.]]>