International A LevelFurther Maths: Further Pure 1EdexcelRevision NotesMatricesMatrix AlgebraInverse Matrices

Inverse Matrices (Edexcel International A Level Further Maths)

Revision Note

Mark Curtis

Expertise

Maths

Determinant of a 2x2 Matrix

What is the determinant?

  • The determinant is a numerical value (positive or negative) calculated from elements of a square matrix

    • It is used to find the inverse of a matrix

  • For a 2 × 2 matrix, bold A, the determinant, space det space left parenthesis bold A right parenthesis or open vertical bar bold A close vertical bar, is given by

bold A equals open parentheses table row a b row c d end table close parentheses space space space rightwards double arrow space space det space bold A equals open vertical bar bold A close vertical bar equals a d minus b c

What properties of determinants do I need to know?

  • If space det space left parenthesis bold A right parenthesis equals 0 then bold A is called a singular matrix

  • If space det space left parenthesis bold A right parenthesis not equal to 0 then bold A is called a non-singular matrix

  • The determinant of the identity matrix is 1

    • det space left parenthesis bold I right parenthesis equals 1

  • The determinant of the zero matrix is 0

    • det space left parenthesis bold 0 right parenthesis equals 0

  • You do not need to know the following rules, but they can be good checks in an exam:

    • det space left parenthesis bold AB right parenthesis equals det space left parenthesis bold A right parenthesis cross times det space left parenthesis bold B right parenthesis

    • det open parentheses bold A to the power of negative 1 end exponent close parentheses equals fraction numerator 1 over denominator det open parentheses bold A close parentheses end fraction

Worked Example

Consider the matrix bold A equals open parentheses table row 3 cell negative 6 end cell row p 7 end table close parentheses, where p is a constant.

Given that det space bold A equals negative 3, find the value of p.

Use that if bold A equals open parentheses table row a b row c d end table close parentheses space then det space bold A equals open vertical bar bold A close vertical bar equals a d minus b c
Work out the determinant algebraically

table row cell det open parentheses bold A close parentheses end cell equals cell 3 cross times 7 minus open parentheses negative 6 close parentheses cross times p end cell row blank equals cell 21 plus 6 p end cell end table

Set this expression equal to -3 and solve for p

table row cell 21 plus 6 p end cell equals cell negative 3 end cell row cell 6 p end cell equals cell negative 24 end cell end table

p equals negative 4

Inverse of a 2x2 Matrix

What is the inverse of a matrix?

  • The inverse of a square matrix bold Ais the matrix bold A to the power of negative 1 end exponent which, when multiplied together (in either order), gives the identity matrix

    • bold AA to the power of negative 1 end exponent equals bold A to the power of negative 1 end exponent bold A equals bold I

    • bold A to the power of negative 1 end exponent  has the same dimensions (order) as bold A

How do I find the inverse of a 2x2 matrix?

  • To find the inverse of a 2 × 2 matrix:

    • Switch the two entries on leading diagonal (top-left to bottom right)

    • Change the signs of the other two entries

    • Divide by the determinant

bold A equals open parentheses table row a b row c d end table close parentheses space space rightwards double arrow space space bold A to the power of negative 1 end exponent equals fraction numerator 1 over denominator det space bold A end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses

  • Note that you cannot divide by zero

    • If det space bold A equals 0, then bold A is not invertible ( bold A to the power of negative 1 end exponent does not exist)

      • bold A is singular

    • If det space bold A not equal to 0, then bold A is invertible

How do I find the inverse of a product of matrices?

  • The inverse of a product of matrices is the product of the inverses of the matrices in reverse order

    • open parentheses bold AB close parentheses to the power of negative 1 end exponent equals bold B to the power of negative 1 end exponent bold A to the power of negative 1 end exponent

Exam Tip

There are two ways to check whether your inverse matrix is correct:

  • use a calculator (they can find inverses),

  • or calculate bold AA to the power of negative 1 end exponent to see if you get the identity bold I.

Worked Example

Let bold P equals open parentheses table row 1 cell negative 2 end cell row 4 k end table close parentheses where k not equal to negative 8.

(a) Find bold P to the power of negative 1 end exponent, giving your answer in terms of k.

Use that bold A equals open parentheses table row a b row c d end table close parentheses space space rightwards double arrow space space bold A to the power of negative 1 end exponent equals fraction numerator 1 over denominator det space bold A end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses where det space bold A equals a d minus b c
First find det open parentheses bold P close parentheses

table row cell det open parentheses bold P close parentheses end cell equals cell 1 cross times k minus 4 cross times open parentheses negative 2 close parentheses end cell row blank equals cell k plus 8 end cell end table

Now divide bold P by det open parentheses bold P close parentheses, swap a and d, and change signs of b and c

bold P to the power of negative 1 end exponent equals fraction numerator 1 over denominator k plus 8 end fraction open parentheses table row k 2 row cell negative 4 end cell 1 end table close parentheses

You can also write this as bold P to the power of negative 1 end exponent equals open parentheses table row cell fraction numerator k over denominator k plus 8 end fraction end cell cell fraction numerator 2 over denominator k plus 8 end fraction end cell row cell negative fraction numerator 4 over denominator k plus 8 end fraction end cell cell fraction numerator 1 over denominator k plus 8 end fraction end cell end table close parentheses

(b) Verify that bold P to the power of negative 1 end exponent bold P equals bold I, where bold I is the identity matrix.

Verify means check that it is true
Substitute bold P to the power of negative 1 end exponent and bold P into the left-hand side and multiply out the matrices

table row cell bold P to the power of negative 1 end exponent bold P end cell equals cell fraction numerator 1 over denominator k plus 8 end fraction open parentheses table row k 2 row cell negative 4 end cell 1 end table close parentheses open parentheses table row 1 cell negative 2 end cell row 4 k end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator k plus 8 end fraction open parentheses table row cell k cross times 1 plus 2 cross times 4 end cell cell k cross times open parentheses negative 2 close parentheses plus 2 cross times k end cell row cell negative 4 cross times 1 plus 1 cross times 4 end cell cell negative 4 cross times open parentheses negative 2 close parentheses plus 1 cross times k end cell end table close parentheses end cell row blank equals cell fraction numerator 1 over denominator k plus 8 end fraction open parentheses table row cell k plus 8 end cell 0 row 0 cell 8 plus k end cell end table close parentheses end cell end table

The 8 plus k is the same as k plus 8
Since k not equal to negative 8 in the question, cancel the open parentheses k plus 8 close parentheses's

bold P to the power of negative 1 end exponent bold P equals open parentheses table row 1 0 row 0 1 end table close parentheses

The right-hand side is the 2 × 2 identity matrix

bold P to the power of negative 1 end exponent bold P equals bold I

Even though bold PP to the power of negative 1 end exponent equals bold I is also true, this question only asks for the order shown

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    Mark Curtis

    Author: Mark Curtis

    Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.