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Determinants and inverses of 2x2 matricesEdexcel International A Level Further Maths: Revision notes

Section 1

The determinant

For A=(abcd)\mathbf A=\begin{pmatrix}a&b\\ c&d\end{pmatrix} the determinant is det⁡A=∣A∣=ad−bc.\det\mathbf A=|\mathbf A|=ad-bc. Multiply the leading diagonal, then subtract the product of the other diagonal. Example: A=(4322)\mathbf A=\begin{pmatrix}4&3\\ 2&2\end{pmatrix} has det⁡A=4(2)−3(2)=2\det\mathbf A=4(2)-3(2)=2. Determinants often contain an unknown: for (k43k+1)\begin{pmatrix}k&4\\ 3&k+1\end{pmatrix} the determinant is k(k+1)−12=k2+k−12k(k+1)-12=k^2+k-12.

Key termsdeterminant
Common mistake

Writing bc−adbc-ad, or adding the products. The order is leading diagonal first, then subtract.

Section 2

Singular and non-singular matrices

A matrix with det⁡A=0\det\mathbf A=0 is singular and has no inverse. A matrix with det⁡A≠0\det\mathbf A\neq0 is non-singular and has an inverse. To find when a matrix with an unknown is singular, set the determinant equal to zero and solve. Example: (k43k+1)\begin{pmatrix}k&4\\ 3&k+1\end{pmatrix} is singular when k2+k−12=(k+4)(k−3)=0k^2+k-12=(k+4)(k-3)=0, so k=3k=3 or k=−4k=-4. For every other value it has an inverse.

Key termssingularnon-singular
Exam tip

A zero determinant means the rows are multiples of each other. (2346)\begin{pmatrix}2&3\\ 4&6\end{pmatrix} is singular since 12−12=012-12=0.

Section 3

The inverse of a 2×22\times2 matrix

If det⁡A≠0\det\mathbf A\neq0, the inverse is A−1=1ad−bc(d−b−ca).\mathbf A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\ -c&a\end{pmatrix}. Swap the entries on the leading diagonal, change the signs of the other two, and divide by the determinant. It satisfies AA−1=A−1A=I\mathbf{AA}^{-1}=\mathbf A^{-1}\mathbf A=\mathbf I. Example: for A=(4322)\mathbf A=\begin{pmatrix}4&3\\ 2&2\end{pmatrix}, A−1=12(2−3−24)\mathbf A^{-1}=\frac12\begin{pmatrix}2&-3\\ -2&4\end{pmatrix}. Check by multiplying: AA−1\mathbf{AA}^{-1} should give the identity.

Key termsinverse
Common mistake

Swapping the off-diagonal entries or forgetting to change their signs. Only the leading diagonal swaps; the other two entries keep their places but change sign.

Section 4

Using the inverse

To solve a matrix equation, multiply both sides by the inverse on the correct side:

  • AX=B⇒X=A−1B\mathbf{AX}=\mathbf B\Rightarrow\mathbf X=\mathbf A^{-1}\mathbf B (pre-multiply).
  • XA=B⇒X=BA−1\mathbf{XA}=\mathbf B\Rightarrow\mathbf X=\mathbf B\mathbf A^{-1} (post-multiply). Matrix multiplication is not commutative, so the side matters. Example: if AX=(2011)\mathbf{AX}=\begin{pmatrix}2&0\\ 1&1\end{pmatrix} with A=(4322)\mathbf A=\begin{pmatrix}4&3\\ 2&2\end{pmatrix} then X=12(2−3−24)(2011)=(12−3202)\mathbf X=\frac12\begin{pmatrix}2&-3\\ -2&4\end{pmatrix}\begin{pmatrix}2&0\\ 1&1\end{pmatrix}=\begin{pmatrix}\frac12&-\frac32\\ 0&2\end{pmatrix}.
Common mistake

Writing X=BA−1\mathbf X=\mathbf B\mathbf A^{-1} for AX=B\mathbf{AX}=\mathbf B. The inverse goes on the same side as A\mathbf A was.

Section 5

The inverse of a product

For non-singular matrices A\mathbf A and B\mathbf B, (AB)−1=B−1A−1.(\mathbf{AB})^{-1}=\mathbf B^{-1}\mathbf A^{-1}. The order reverses, like taking off socks and shoes. To see why: (AB)(B−1A−1)=A(BB−1)A−1=AA−1=I(\mathbf{AB})(\mathbf B^{-1}\mathbf A^{-1})=\mathbf A(\mathbf{BB}^{-1})\mathbf A^{-1}=\mathbf{AA}^{-1}=\mathbf I. Example: with C=(2153)\mathbf C=\begin{pmatrix}2&1\\ 5&3\end{pmatrix} and D=(1213)\mathbf D=\begin{pmatrix}1&2\\ 1&3\end{pmatrix}, CD=(37819)\mathbf{CD}=\begin{pmatrix}3&7\\ 8&19\end{pmatrix} and (CD)−1=(19−7−83)(\mathbf{CD})^{-1}=\begin{pmatrix}19&-7\\ -8&3\end{pmatrix}, which equals D−1C−1=(3−2−11)(3−1−52)\mathbf D^{-1}\mathbf C^{-1}=\begin{pmatrix}3&-2\\ -1&1\end{pmatrix}\begin{pmatrix}3&-1\\ -5&2\end{pmatrix}.

Key termsreversal law
Common mistake

Writing (AB)−1=A−1B−1(\mathbf{AB})^{-1}=\mathbf A^{-1}\mathbf B^{-1}. The order reverses.

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Exam questions on Determinants and inverses of 2x2 matrices

  1. A=(4322)\mathbf{A}=\begin{pmatrix}4&3\\ 2&2\end{pmatrix}.
    Given that AX=(2011)\mathbf{AX}=\begin{pmatrix}2&0\\ 1&1\end{pmatrix}, find the matrix X\mathbf{X}.2 marks
  2. B=(k43k+1)\mathbf{B}=\begin{pmatrix}k&4\\ 3&k+1\end{pmatrix}, where kk is a constant.
    Given that k=2k=2, find B−1\mathbf{B}^{-1}.2 marks
  3. C=(2153)\mathbf{C}=\begin{pmatrix}2&1\\ 5&3\end{pmatrix} and D=(1213)\mathbf{D}=\begin{pmatrix}1&2\\ 1&3\end{pmatrix}.
    Find CD\mathbf{CD} and hence find (CD)−1(\mathbf{CD})^{-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).