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1.14 MatricesIB Maths: Applications and Interpretation HL: Revision notes

Section 1

Matrices, order and algebra

A matrix is a rectangular array of numbers. It has mm rows and nn columns, so its order is m×nm\times n (rows first). Each number is an element; aija_{ij} is the element in row ii, column jj.

  • Equality: two matrices are equal if they have the same order and every pair of corresponding elements is equal.
  • Addition and subtraction: only for matrices of the same order, element by element.
  • Scalar multiplication: multiply every element by the scalar, e.g. 3(1−204)=(3−6012)3\begin{pmatrix}1&-2\\ 0&4\end{pmatrix}=\begin{pmatrix}3&-6\\ 0&12\end{pmatrix}. Your GDC can do all of this, and is the quickest way to handle large matrices.
Key termsmatrixelementrowcolumnorder
Common mistake

Adding matrices of different orders. A 2×32\times3 matrix and a 3×23\times2 matrix cannot be added.

Section 2

Matrix multiplication and its properties

ABAB is defined only when the number of columns of AA equals the number of rows of BB. If AA is m×nm\times n and BB is n×pn\times p then ABAB is m×pm\times p. Each element is a row of AA times a column of BB. Example: (2134)(10−12)=(12−18)\begin{pmatrix}2&1\\ 3&4\end{pmatrix}\begin{pmatrix}1&0\\ -1&2\end{pmatrix}=\begin{pmatrix}1&2\\ -1&8\end{pmatrix}, but reversing the order gives (2147)\begin{pmatrix}2&1\\ 4&7\end{pmatrix}.

  • Associative: A(BC)=(AB)CA(BC)=(AB)C.
  • Distributive: A(B+C)=AB+ACA(B+C)=AB+AC.
  • Not commutative: in general AB≠BAAB\neq BA. The identity matrix II (ones on the main diagonal, zeros elsewhere) satisfies AI=IA=AAI=IA=A. The zero matrix 00 satisfies A+0=AA+0=A and A0=0A0=0.
Key termsidentity matrixzero matrixnon-commutative
Common mistake

Multiplying element by element. Matrix multiplication is row by column.

Exam tip

Check the orders first: (m×n)(n×p)(m\times n)(n\times p) gives m×pm\times p.

Section 3

Determinants and inverses

For A=(abcd)A=\begin{pmatrix}a&b\\ c&d\end{pmatrix} the determinant is det⁡A=ad−bc\det A=ad-bc. If det⁡A≠0\det A\neq0 the inverse exists: A−1=1ad−bc(d−b−ca),AA−1=A−1A=I.A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix},\qquad AA^{-1}=A^{-1}A=I. Swap the diagonal elements, change the signs of the other two, then divide by the determinant. If det⁡A=0\det A=0 the matrix is singular and has no inverse. Example: A=(2134)A=\begin{pmatrix}2&1\\ 3&4\end{pmatrix} has det⁡A=5\det A=5 and A−1=15(4−1−32)A^{-1}=\frac15\begin{pmatrix}4&-1\\ -3&2\end{pmatrix}. For 3×33\times3 and larger matrices, find the determinant and inverse with your GDC.

Key termsdeterminantinversesingular
Common mistake

Forgetting to divide by the determinant, or changing the wrong signs.

Section 4

Solving systems of equations

A system of linear equations can be written Ax=bA\mathbf{x}=\mathbf{b}, where AA holds the coefficients, x\mathbf{x} the unknowns and b\mathbf{b} the right-hand sides. In examinations AA is invertible, so multiply on the left by A−1A^{-1}: x=A−1b.\mathbf{x}=A^{-1}\mathbf{b}. Example: 3s+2l=13.93s+2l=13.9 and 2s+5l=212s+5l=21 becomes (3225)(sl)=(13.921)\begin{pmatrix}3&2\\ 2&5\end{pmatrix}\begin{pmatrix}s\\ l\end{pmatrix}=\begin{pmatrix}13.9\\ 21\end{pmatrix}. Here det⁡=11\det=11 and the GDC gives s=2.5s=2.5, l=3.2l=3.2. The same method works for 3×33\times3 systems with technology.

Key termssystem of equationscoefficient matrix
Common mistake

Writing x=bA−1\mathbf{x}=\mathbf{b}A^{-1}. Matrix multiplication is not commutative, so A−1A^{-1} must go on the left of b\mathbf{b}.

Section 5

Modelling with matrices

Matrices store and combine data in real problems.

  • Costs and profits: a matrix of quantities times a vector of prices gives the total cost for each customer.
  • Coding and decoding: write letters as numbers, group them as column vectors and multiply by a coding matrix KK. To decode, multiply the coded vectors by K−1K^{-1}; if det⁡K=1\det K=1 the inverse has whole-number entries.
  • Links: transition matrices in Markov chains (AHL 4.19) and phase portraits (AHL 5.17) are built on the same ideas. Always state what each row, column and answer represents, with units, and check that the orders allow the product.
Key termscoding matrix
Exam tip

When decoding, multiply by K−1K^{-1} on the left of the coded matrix, just as the coding was done by KK on the left.

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Exam questions on 1.14 Matrices

  1. A=(2134)A=\begin{pmatrix}2&1\\ 3&4\end{pmatrix} and B=(10−12)B=\begin{pmatrix}1&0\\ -1&2\end{pmatrix}.
    Find BABA and hence state, with a reason, whether AB=BAAB=BA.2 marks
  2. A cafe sells small cups of coffee at ss EUR each and large cups at ll EUR each. On Monday, 3 small and 2 large cups cost 13.90 EUR in total. On Tuesday, 2 small and 5 large cups cost 21.00 EUR in total.
    Use x=A−1b\mathbf{x}=A^{-1}\mathbf{b} with your GDC, where AA is the coefficient matrix, to find the price of a small cup and of a large cup.2 marks
  3. Letters are replaced by numbers (A=1A=1, B=2B=2, ..., Z=26Z=26) and a message is coded in pairs of letters. Each pair of numbers forms a column vector, which is multiplied on the left by the matrix K=(2312)K=\begin{pmatrix}2&3\\ 1&2\end{pmatrix} to give the coded pair.
    Show that det⁡K=1\det K=1 and find K−1K^{-1}.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).