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Solving Matrix Equations (Edexcel IGCSE Maths B): Revision Note

Exam code: 4MB1

Roger B

Written by: Roger B

Reviewed by: Jamie Wood

Updated on

Solving matrix equations with inverses

  • Inverses can be used to rearrange and solve equations with matrices

  • This relies on the essential properties of inverses and of 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 AI equals bold IA equals bold A

  • For example, to solve the matrix equation bold AB equals bold C for bold B

    • First multiply both sides of the equation 'from the left' by bold A to the power of negative 1 end exponent

      • bold A to the power of negative 1 end exponent bold AB equals bold A to the power of negative 1 end exponent bold C

    • But bold A to the power of negative 1 end exponent bold A equals bold I, so

      • bold IB equals bold A to the power of negative 1 end exponent bold C

    • And bold IB equals bold B, so

      • bold B equals bold A to the power of negative 1 end exponent bold C

    • Then multiply together the two matrices on the right-hand side to find bold B explicitly

  • Similarly, you can solve the matrix equation bold BA equals bold C for bold B

    • Though here you will need to multiply 'from the right' by bold A to the power of negative 1 end exponent

table row cell bold BAA to the power of negative 1 end exponent end cell equals cell bold CA to the power of negative 1 end exponent end cell row bold BI equals cell bold CA to the power of negative 1 end exponent end cell row bold B equals cell bold CA to the power of negative 1 end exponent end cell end table

  • This is similar to solving a regular equation by 'doing the same thing to both sides'

    • Except that order of multiplication matters with matrices

      • Multiplying 'from the left' is not the same as multiplying 'from the right'

    • For example, if you tried to solve bold BA equals bold C for bold B by multiplying 'from the left' by bold A to the power of negative 1 end exponent

      • you would get bold A to the power of negative 1 end exponent bold BA equals bold A to the power of negative 1 end exponent bold C

      • which does not simplify any further

Examiner Tips and Tricks

Remember that a matrix and its inverse only 'collapse' to the identity matrix bold I when they are next to each other in a multiplication.

  • So bold A to the power of negative 1 end exponent bold AB equals bold IB equals bold B

  • and bold BAA to the power of negative 1 end exponent equals bold BI equals bold B

  • but bold A to the power of negative 1 end exponent bold BA does not simplify any further

Worked Example

bold P equals open parentheses table row 4 cell negative 2 end cell row cell negative 2 end cell 2 end table close parentheses space space space space space space space bold Q equals open parentheses table row k 1 row 3 0 end table close parentheses space space space space space space space bold R equals open parentheses table row 10 4 row cell negative 2 end cell cell negative 2 end cell end table close parentheses

where k is a constant.

(a) Find bold P to the power of negative 1 end exponent.

Answer:

Use the inverse formula for a 2 cross times 2 matrix

  • The inverse of space open parentheses table row a b row c d end table close parentheses space is space fraction numerator 1 over denominator a d minus b c end fraction open parentheses table row d cell negative b end cell row cell negative c end cell a end table close parentheses

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

table row cell bold P to the power of bold minus bold 1 end exponent end cell equals cell open parentheses table row cell 1 half end cell cell 1 half end cell row cell 1 half end cell 1 end table close parentheses end cell end table

(b) Given that bold PQ equals bold R find the value of k.

Answer:

Use matrix algebra to solve the equation for bold Q

  • Multiply both sides of the equation 'from the left' by bold P to the power of negative 1 end exponent

bold P to the power of bold minus bold 1 end exponent bold PQ equals bold P to the power of negative 1 end exponent bold R

  • By the definition of the inverse, bold P to the power of negative 1 end exponent bold P equals bold I

bold IQ equals bold P to the power of negative 1 end exponent bold R

  • By the properties of the identity matrix, bold IQ equals bold Q

bold Q equals bold P to the power of negative 1 end exponent bold R

That gives you a matrix 'formula' for bold Q

  • Substitute in matrices bold P to the power of negative 1 end exponent and bold R

bold Q equals open parentheses table row cell 1 half end cell cell 1 half end cell row cell 1 half end cell 1 end table close parentheses open parentheses table row 10 4 row cell negative 2 end cell cell negative 2 end cell end table close parentheses

  • Carry out the multiplication

table row bold Q equals cell open parentheses table row cell 1 half cross times 10 plus 1 half cross times open parentheses negative 2 close parentheses end cell cell 1 half cross times 4 plus 1 half cross times open parentheses negative 2 close parentheses end cell row cell 1 half cross times 10 plus 1 cross times open parentheses negative 2 close parentheses end cell cell 1 half cross times 4 plus 1 cross times negative 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 5 minus 1 end cell cell 2 minus 1 end cell row cell 5 minus 2 end cell cell 2 minus 2 end cell end table close parentheses end cell row blank equals cell open parentheses table row 4 1 row 3 0 end table close parentheses end cell end table

k equals 4

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths Content Creator

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Jamie Wood

    Reviewer: Jamie Wood

    Expertise: Maths Content Creator

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.