IGCSEMathsEdexcelMaths BRevision NotesMatricesProperties of MatricesAdding Matrices & Multiplying by a Scalar

Adding Matrices & Multiplying by a Scalar (Edexcel IGCSE Maths B): Revision Note

Exam code: 4MB1

Roger B

Written by: Roger B

Reviewed by: Jamie Wood

Updated on

Basic operations with matrices

How do I add or subtract two matrices?

  • To add or subtract two matrices, they need to be of the same order (i.e. the same number of rows and columns)

  • If they are of the same order, then

    • to add two matrices, just add the corresponding elements

      • open parentheses table row 2 cell negative 3 end cell row 5 1 end table close parentheses plus open parentheses table row cell negative 1 end cell 7 row cell negative 4 end cell 6 end table close parentheses equals open parentheses table row cell 2 plus open parentheses negative 1 close parentheses end cell cell negative 3 plus 7 end cell row cell 5 plus open parentheses negative 4 close parentheses end cell cell 1 plus 6 end cell end table close parentheses equals open parentheses table row 1 4 row 1 7 end table close parentheses

    • to subtract two matrices just subtract the corresponding elements

      • open parentheses table row 2 cell negative 3 end cell row 5 1 end table close parentheses minus open parentheses table row cell negative 1 end cell 7 row cell negative 4 end cell 6 end table close parentheses equals open parentheses table row cell 2 minus open parentheses negative 1 close parentheses end cell cell negative 3 minus 7 end cell row cell 5 minus open parentheses negative 4 close parentheses end cell cell 1 minus 6 end cell end table close parentheses equals open parentheses table row 3 cell negative 10 end cell row 9 cell negative 5 end cell end table close parentheses

How do I multiply a matrix by a scalar?

  • To multiply any matrix by a scalar (a number), multiply each element by that scalar 

    • If bold A equals open parentheses table row 5 2 row 0 4 end table close parentheses then 2 bold A equals 2 open parentheses table row 5 2 row 0 4 end table close parentheses equals open parentheses table row cell 2 cross times 5 end cell cell 2 cross times 2 end cell row cell 2 cross times 0 end cell cell 2 cross times 4 end cell end table close parentheses equals open parentheses table row 10 4 row 0 8 end table close parentheses

  • Lower case letters often refer to scalar multiples

    • k bold A is the matrix bold Amultiplied by the scalar k

Worked Example

bold A equals open parentheses table row cell negative 2 end cell 2 row cell negative 2 end cell cell negative 5 end cell end table close parentheses space space space space space space space space space space space space space bold B equals open parentheses table row cell negative 4 end cell 1 row 3 cell negative 4 end cell end table close parentheses

(a) Calculate bold A plus bold B

Answer:

Add the corresponding elements of the two matrices

table row cell bold A plus bold B end cell equals cell open parentheses table row cell negative 2 end cell 2 row cell negative 2 end cell cell negative 5 end cell end table close parentheses plus open parentheses table row cell negative 4 end cell 1 row 3 cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 2 plus open parentheses negative 4 close parentheses end cell cell 2 plus 1 end cell row cell negative 2 plus 3 end cell cell negative 5 plus open parentheses negative 4 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 6 end cell 3 row 1 cell negative 9 end cell end table close parentheses end cell end table

table row cell bold A plus bold B end cell equals cell open parentheses table row cell negative 6 end cell 3 row 1 cell negative 9 end cell end table close parentheses end cell end table
 

Given that 2 bold A minus n bold B equals open parentheses table row 8 1 row cell negative 13 end cell 2 end table close parentheses

(b) find the value of n

Answer:

Multiply all elements of bold A by 2 to find 2 bold A

table row cell 2 bold A end cell equals cell 2 open parentheses table row cell negative 2 end cell 2 row cell negative 2 end cell cell negative 5 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 2 cross times open parentheses negative 2 close parentheses end cell cell 2 cross times 2 end cell row cell 2 cross times open parentheses negative 2 close parentheses end cell cell 2 cross times open parentheses negative 5 close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 end cell 4 row cell negative 4 end cell cell negative 10 end cell end table close parentheses end cell end table

Multiply all elements of bold B by n to find n bold B

table row cell n bold B end cell equals cell n open parentheses table row cell negative 4 end cell 1 row 3 cell negative 4 end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 n end cell n row cell 3 n end cell cell negative 4 n end cell end table close parentheses end cell end table

Subtract the corresponding elements to find 2 bold A minus n bold B

table row cell 2 A minus n B end cell equals cell open parentheses table row cell negative 4 end cell 4 row cell negative 4 end cell cell negative 10 end cell end table close parentheses minus open parentheses table row cell negative 4 n end cell n row cell 3 n end cell cell negative 4 n end cell end table close parentheses end cell row blank equals cell open parentheses table row cell negative 4 minus open parentheses negative 4 n close parentheses end cell cell 4 minus n end cell row cell negative 4 minus 3 n end cell cell negative 10 minus open parentheses negative 4 n close parentheses end cell end table close parentheses end cell row blank equals cell open parentheses table row cell 4 n minus 4 end cell cell 4 minus n end cell row cell negative 3 n minus 4 end cell cell 4 n minus 10 end cell end table close parentheses end cell end table

Substitute that into the equation from the question

table row cell open parentheses table row cell 4 n minus 4 end cell cell 4 minus n end cell row cell negative 3 n minus 4 end cell cell 4 n minus 10 end cell end table close parentheses end cell equals cell open parentheses table row 8 1 row cell negative 13 end cell 2 end table close parentheses end cell end table

For those two matrices to be equal, the correct value of n must make ALL the corresponding elements equal in the two matrices

  • So you can choose any one pair to solve to find n

table row cell 4 n minus 4 end cell equals 8 row cell 4 n end cell equals 12 row n equals 3 end table

It is worth checking the other three pairs of elements with that value of n, to make sure you haven't made a mistake

4 minus 3 equals 1 space space ✔ minus 3 open parentheses 3 close parentheses minus 4 equals negative 13 space space ✔ 4 open parentheses 3 close parentheses minus 10 equals 2 space space ✔

n equals 3

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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.