Matrices Part2

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Algebra of Matrices

Take note that algebra of matrices are DIFFERENT from normal algebra. Thus a lot of steps we normally would do such as bringing over, factorizing, expanding, etc. are either NOT allowed or done a little differently.

A) Commutative Law (Addition only)

\mathbf{A}+\mathbf{B} =\mathbf{B}+\mathbf{A}

  • Multiplication of matrices DO NOT satisfy Commutative Law
  • That is, \mathbf{AB} \neq\mathbf{BA} (in general)
  • Important : It is actually more important to remember \mathbf{AB}\neq\mathbf{BA} than \mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A} ( In other words, it's more important to remember what you CAN'T DO compared to normal algebra.)

B) Associative Law

\left( \mathbf{A}+\mathbf{B}\right)+\mathbf{C} =\mathbf{A}+\left( \mathbf{B}+\mathbf{C}\right)

\mathbf{\left(AB\right)C} =\mathbf{A\left(BC\right)}

  • Again, it is important to note what we CAN'T do. While multiplication DOES satisfy Associative Law, it still DOESN'T satisfy Commutative Law. In other words, we can change the order of the brackets, but NOT the actual arrangement of the matrices itself.
  • For example, state whether the following are equal or not equal
    • \mathbf{\left(AB\right)C}, \mathbf{C\left(AB\right)} NOT EQUAL
    • \mathbf{ABC}, \mathbf{ACB}, \mathbf{BAC}, \mathbf{BCA}, \mathbf{CAB}, \mathbf{CBA} ALL NOT EQUAL
    • \mathbf{ABA}, \mathbf{A}^{2}\mathbf{B} NOT EQUAL. Note that \mathbf{A}^{2}\mathbf{B} =\mathbf{AAB}\neq\mathbf{ABA}
    • \left(\mathbf{AB}\right)^{2}, \mathbf{A}^{2}\mathbf{B}^{2} NOT EQUAL. Note that \left(\mathbf{AB}\right)^{2}=\mathbf{ABAB} while \mathbf{A}^{2}\mathbf{B}^{2} =\mathbf{AABB}

C) Distributive Law

\mathbf{A}\left(\mathbf{B+C}\right) =\mathbf{AB}+\mathbf{AC}

\left(\mathbf{B+C}\right)\mathbf{A} =\mathbf{BA}+\mathbf{CA}

  • Note that \mathbf{A}\left(\mathbf{B+C}\right) \neq\left(\mathbf{B+C}\right)\mathbf{A}
  • \left(\mathbf{A+B}\right)^{2} =\mathbf{A}^{2}+2\mathbf{AB}+\mathbf{B}^{2} CORRECT? WRONG.
    • \left(\mathbf{A+B}\right)^{2} =\left(\mathbf{A+B}\right)\left(\mathbf{A+B}\right) =\mathbf{A}^{2}+\mathbf{AB}+\mathbf{BA}+\mathbf{B}^{2} which we can't simplify further.

D) Identity

\mathbf{AI} =\mathbf{A}

\mathbf{IA} =\mathbf{A}

  • \therefore \mathbf{AI}=\mathbf{A}=\mathbf{IA} This is one special case where commutative law is satisfied for multiplication

E) Inverse

If \mathbf{A}^{-1} exists (\mathbf{A} is not singular )

\mathbf{AA^{-1}} =\mathbf{I}

\mathbf{A^{-1}A} =\mathbf{I}

  • \therefore \mathbf{AA^{-1}}=\mathbf{I}=\mathbf{A^{-1}A}

F) Inverse of Product

\mathbf{\left(AB\right)^{-1}} =\mathbf{B^{-1}}\mathbf{A^{-1}}

  • Note that it is NOT \mathbf{A^{-1}}\mathbf{B^{-1}}
  • Important Note: This law IS stated explicitly in the syllabus and you are required to know it.
  • Prove
    • First, take note of properties of inverse \mathbf{AA^{-1}}=\mathbf{I}, \mathbf{A^{-1}A}=\mathbf{I}
      • Thus, if, say we are required to prove \mathbf{B} is inverse of \mathbf{A}, we would need to prove \mathbf{AB}=\mathbf{I} and \mathbf{BA}=\mathbf{I}
      • So in this case, since we are proving that inverse of \mathbf{AB} is \mathbf{B^{-1}}\mathbf{A^{-1}}, we would need to prove \mathbf{\left(AB\right)\left(B^{-1}A^{-1}\right)}=\mathbf{I} and \mathbf{\left(B^{-1}A^{-1}\right)\left(AB\right)}=\mathbf{I}
    • \mathbf{\left(AB\right)\left(B^{-1}A^{-1}\right)}
      • =\mathbf{A\left(BB^{-1}\right)A^{-1}} (using the associative laws to change the brackets WITHOUT changing the order of the arrangement of the matrices)
      • =\mathbf{AIA^{-1}}
      • =\mathbf{AA^{-1}}
      • =\mathbf{I}
      • Take note that as tempting as it is to just write =\mathbf{AA^{-1}BB^{-1}}=\mathbf{II}=\mathbf{I}, that is WRONG as it uses (wrongly) Commutative Law.
    • \mathbf{\left(B^{-1}A^{-1}\right)\left(AB\right)} Try this on your own first, then only compare with below
      • \begin{align} &= \mathbf{B^{-1}\left(A^{-1}A\right)B} \\ &= \mathbf{B^{-1}IB}\\ &= \mathbf{B^{-1}B}\\ &= \mathbf{I}\\ \end{align}
    • \therefore \mathbf{\left(AB\right)^{-1}}=\mathbf{B^{-1}}\mathbf{A^{-1}}

G) Scalars

k\mathbf{\left(A+B\right)} =k\mathbf{A}+k\mathbf{B}

\left(k\mathbf{A}\right)\left(m\mathbf{B}\right) =km\left(\mathbf{AB}\right) Note that the arrangement of the matrices \mathbf{A} and \mathbf{B} are NOT changed in relation to each other.

  • In other words, while multiplication of matrices do not abide Commutative Law, scalars do. This allows us to leave the multiplication with scalar as the last step, to avoid multiplying matrices with large or fractional values.

Manipulating Matrix Equations

The fact that there are no division in matrices and that \mathbf{AB}\neq\mathbf{BA} puts serious limitations to how we can manipulate a matrix equation compared to normal algebra. Part of the difficulty is unlearning things we are so used to in normal algebra when dealing with matrix.

  • If \mathbf{X+A}=\mathbf{B}, \therefore \mathbf{X} =\mathbf{B}-\mathbf{A}
  • However, if \mathbf{AX}=\mathbf{B}, how can we solve for \mathbf{X}?
  • We take note that \mathbf{AA^{-1}}=\mathbf{I}=\mathbf{A^{-1}A} and \mathbf{AI}=\mathbf{A}=\mathbf{IA}
  • Meaning, any matrix multiplied with it's inverse (if it exist) will give us the identity matrix, which itself will disappear when multiplied with any other matrix.
  • Thus, to get rid of a matrix from one side of the equation, we need to multiply it with its inverse
  • IMPORTANT NOTE : Either
    • Multiply BOTH sides at the FRONT
    • Multiply BOTH sides at the END
    • NO mixing (one front, one end), and certainly NO multiplying anywhere else (like in the middle). These are all because \mathbf{AB}\neq\mathbf{BA}.

Examples

Express \mathbf{X} in terms of the other matrices. (All matrices are non-singular)

  • \mathbf{AX}=\mathbf{B}
  • Analyze :
    • We need to get rid of \mathbf{A}, so we multiply BOTH sides with \mathbf{A^{-1}}, and we will multiply it in front ( so that we can get the \mathbf{A} and \mathbf{A^{-1}} together)
    • \mathbf{AX}=\mathbf{B} (you still should copy back the original equation first)
    • {\color{Red}\mathbf{A^{-1}}}\mathbf{AX}={\color{Red}\mathbf{A^{-1}}}\mathbf{B}
    • \mathbf{IX}=\mathbf{A^{-1}B}
    • \mathbf{X}=\mathbf{A^{-1}B}
    • Note : It is WRONG to write
      • {\color{Red}\mathbf{A^{-1}}}\mathbf{AX}=\mathbf{B}{\color{Red}\mathbf{A^{-1}}} (one front, one end)
      • \mathbf{A}{\color{Red}\mathbf{A^{-1}}}\mathbf{X}={\color{Red}\mathbf{A^{-1}}}\mathbf{B}(in the middle)

  • \mathbf{XA}=\mathbf{B}
  • Try it first : \mathbf{X}=\mathbf{BA^{-1}}
    • \begin{align} \mathbf{XA}&=\mathbf{B}\\ \mathbf{XAA^{-1}}&=\mathbf{BA^{-1}} \\ \mathbf{XI}&=\mathbf{BA^{-1}} \\ \mathbf{X}&=\mathbf{BA^{-1}} \\ \end{align}

Given \mathbf{AXA^{-1}}=\mathbf{B}, where \mathbf{A}=\begin{pmatrix} 2 & -1\\ -2 & 3 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} and \mathbf{X} is a 2 \times 2 matrix. Find \mathbf{X}.

  • Analyze : We should be needing \mathbf{A^{-1}}, so might as well find it first
    • \mathbf{A^{-1}}=\frac{1}{6-2}\begin{pmatrix} 3 & 1\\ 2 & 2 \end{pmatrix} =\frac{1}{4}\begin{pmatrix} 3 & 1\\ 2 & 2 \end{pmatrix} (It is not needed to multiply in the scalar as question did not require us to state \mathbf{A^{-1}})
    • Analyze : We need to get rid of the front \mathbf{A} and also the back \mathbf{A^{-1}}
    • Thus
      • Multiply \mathbf{A^{-1}} in front of both sides
      • AND multiply \mathbf{A} at the end of both sides
    • \mathbf{AXA^{-1}}=\mathbf{B}
      • {\color{Red}\mathbf{A^{-1}}}\mathbf{AXA^{-1}}{\color{Blue}\mathbf{A}}={\color{Red}\mathbf{A^{-1}}}\mathbf{B}{\color{Blue}\mathbf{A}}
      • \mathbf{IXI}=\frac{1}{4}\begin{pmatrix} 3 & 1\\ 2 & 2 \end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & -1\\ -2 & 3 \end{pmatrix}
      • \mathbf{IXI}=\frac{1}{4}\begin{pmatrix} 3 & 1\\ 2 & 2 \end{pmatrix}\begin{pmatrix} 0 & 2\\ -2 & 3 \end{pmatrix}
      • \mathbf{X}=\frac{1}{4}\begin{pmatrix} -2 & 9\\ -4 & 10 \end{pmatrix}
      • \mathbf{X}=\begin{pmatrix} -\frac{1}{2} & \frac{9}{4}\\ -1 & \frac{5}{2} \end{pmatrix} Check using calculator (substitute into given equation)

To Find Inverse from Equation

IMPORTANT : When the question gives/ask us to prove an equation, AND THEN ask us to find the inverse of some matrix, we are NOT ALLOWED to find the inverse using the adjoin method. We won't be given any marks, and will just waste precious time during the exam.

For example, we might be given/asked to prove \mathbf{AB}=k\mathbf{I}

  • To find \mathbf{A^{-1}}, multiply both sides with \mathbf{A^{-1}}
  • To find \mathbf{B^{-1}}, multiply both sides with \mathbf{B^{-1}}

Note: Practice showing the full working, even though it is not always required in marking scheme.

Examples

Find \mathbf{A^{-1}} and \mathbf{B^{-1}} if

  • \mathbf{AB}=\mathbf{I}
    • Lets find \mathbf{A^{-1}} first
      • multiply both sides with \mathbf{A^{-1}}
      • \mathbf{AB}=\mathbf{I}
      • {\color{Red}\mathbf{A^{-1}}}\mathbf{AB}={\color{Red}\mathbf{A^{-1}}}\mathbf{I}
      • Simplify, showing full working
      • \mathbf{IB}=\mathbf{A^{-1}}
      • \mathbf{B}=\mathbf{A^{-1}}
    • Try finding \mathbf{B^{-1}} on your own first
      • \begin{align} \mathbf{AB}&=\mathbf{I}\\ \mathbf{AB}{\color{Red}\mathbf{B^{-1}}}&=\mathbf{I}{\color{Red}\mathbf{B^{-1}}}\\ \mathbf{AI}&=\mathbf{B^{-1}}\\ \mathbf{A}&=\mathbf{B^{-1}} \end{align}

  • \mathbf{AB}=9\mathbf{I}
      • \mathbf{AB}=9\mathbf{I}
      • \mathbf{A^{-1}}\mathbf{AB}=\mathbf{A^{-1}}\left(9\mathbf{I}\right) Note that the 9\, is a scalar, not a matrix, and thus we can move it in front, that is \mathbf{A^{-1}}\left(9\mathbf{I}\right)=9\mathbf{A^{-1}I}=9\mathbf{A^{-1}}
      • \mathbf{IB}=9\mathbf{A^{-1}}
      • \mathbf{A^{-1}}=\frac{1}{9}\mathbf{B} DON'T write as =\frac{\mathbf{B}}{9}
    • \begin{align} \mathbf{AB}&=9\mathbf{I}\\ \mathbf{AB}\mathbf{B^{-1}}&=\left(9\mathbf{I}\right)\mathbf{B^{-1}}\\ \mathbf{AI}&=9\mathbf{B^{-1}}\\ \mathbf{B^{-1}}&=\frac{1}{9}\mathbf{A} \end{align}

Find \mathbf{A^{-1}} if

  • \mathbf{A}^{2}+3\mathbf{A}=2\mathbf{I}
    • Might not look obvious at first, but the usual method of multiplying with \mathbf{A^{-1}} will work
      • \mathbf{A}^{2}+3\mathbf{A}=2\mathbf{I}
      • \mathbf{A^{-1}}\left(\mathbf{A}^{2}+3\mathbf{A}\right)=\mathbf{A^{-1}}\left(2\mathbf{I}\right) Multiply in front is easier, but behind works too.
      • \mathbf{A}+3\mathbf{I}=2\mathbf{A^{-1}} Note that 3\mathbf{A^{-1}}\mathbf{A}=3\mathbf{I}, NOT 3\,
      • \mathbf{A^{-1}}=\frac{1}{2}\left(\mathbf{A}+3\mathbf{I}\right)
    • Alternatively
      • \mathbf{A}^{2}+3\mathbf{A}=2\mathbf{I}
      • \mathbf{A}\left(\mathbf{A}+3\mathbf{I}\right)=2\mathbf{I}
      • \mathbf{A}\left[\frac{1}{2}\left(\mathbf{A}+3\mathbf{I}\right)\right]=\mathbf{I}
      • \therefore \mathbf{A^{-1}}=\frac{1}{2}\left(\mathbf{A}+3\mathbf{I}\right)

  • \mathbf{A}^{3}-2\mathbf{A}^{2}+3\mathbf{A}+5\mathbf{I}=\mathbf{0}
    • Comparing with the top one, probably better to bring the 5\mathbf{I} to the other side first
      • Try it on your own first. \mathbf{A^{-1}}=-\frac{1}{5}\left(\mathbf{A^{2}}-2\mathbf{A}+3\mathbf{I}\right)
      • \begin{align} \mathbf{A}^{3}-2\mathbf{A}^{2}+3\mathbf{A}+5\mathbf{I}=\mathbf{0}\\ \mathbf{A}^{3}-2\mathbf{A}^{2}+3\mathbf{A}=-5\mathbf{I}\\ \mathbf{A^{-1}}\left(\mathbf{A}^{3}-2\mathbf{A}^{2}+3\mathbf{A}\right)=\mathbf{A^{-1}}\left(-5\mathbf{I}\right)\\ \mathbf{A}^{2}-2\mathbf{A}+3\mathbf{I}=-5\mathbf{A^{-1}}\\ \mathbf{A^{-1}}=-\frac{1}{5}\left(\mathbf{A}^{2}-2\mathbf{A}+3\mathbf{I}\right) \end{align}
      • Alternatively
        • \begin{align} \mathbf{A}^{3}-2\mathbf{A}^{2}+3\mathbf{A}+5\mathbf{I}=\mathbf{0}\\ \mathbf{A}^{3}-2\mathbf{A}^{2}+3\mathbf{A}=-5\mathbf{I}\\ \mathbf{A}\left(\mathbf{A}^{2}-2\mathbf{A}+3\mathbf{I}\right)=-5\mathbf{I}\\ \mathbf{A}\left[-\frac{1}{5}\left(\mathbf{A}^{2}-2\mathbf{A}+3\mathbf{I}\right)\right]=\mathbf{I}\\ \therefore \mathbf{A^{-1}}=-\frac{1}{5}\left(\mathbf{A}^{2}-2\mathbf{A}+3\mathbf{I}\right) \end{align}


Given \mathbf{A}=\begin{pmatrix} -1 & 1 & -2\\ 2 & 2 & 0 \\ 1 & 2 & -2 \end{pmatrix} \mathbf{B}=\begin{pmatrix} -4 & -2 & 4\\ 4 & 4 & -4 \\ 2& 3 & -4 \end{pmatrix}. Find \mathbf{AB}. Hence, find \mathbf{A^{-1}} and \mathbf{B^{-1}}.

  • This is a pretty standard type of question but there are many pitfalls to be careful of.
    • \mathbf{AB}=\begin{pmatrix} -1 & 1 & -2\\ 2 & 2 & 0 \\ 1 & 2 & -2 \end{pmatrix}\begin{pmatrix} -4 & -2 & 4\\ 4 & 4 & -4 \\ 2& 3 & -4 \end{pmatrix}
    • =\begin{pmatrix} 4 & 0 & 0\\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix} Note : Since the question ask us to find \mathbf{AB} (and NOT prove \mathbf{AB}=4\mathbf{I}), do not write it as 4\mathbf{I} or 4\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} for THIS part.
    • Since \mathbf{AB}=\begin{pmatrix} 4 & 0 & 0\\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{pmatrix}=4\mathbf{I} (We make it clear here we are starting the second part)
      • \begin{align} \mathbf{A^{-1}}\mathbf{AB}&=\mathbf{A^{-1}}\left(4\mathbf{I}\right)\\ \mathbf{IB}&=4\mathbf{A^{-1}}\\ \mathbf{A^{-1}}&=\frac{1}{4}\mathbf{B}\\ &=\frac{1}{4}\begin{pmatrix} -4 & -2 & 4\\ 4 & 4 & -4 \\ 2& 3 & -4 \end{pmatrix}\\ &=\begin{pmatrix} -1 & -\frac{1}{2} & 1\\ 1 & 1 & -1 \\ \frac{1}{2} & \frac{3}{4} & -1 \end{pmatrix} \end{align} Note: Final answer MUST be written as a SINGLE matrix
      • \begin{align} \mathbf{AB}\mathbf{B^{-1}}&=\left(4\mathbf{I}\right)\mathbf{B^{-1}}\\ \mathbf{AI}&=4\mathbf{B^{-1}}\\ \mathbf{B^{-1}}&=\frac{1}{4}\mathbf{A}\\ &=\frac{1}{4}\begin{pmatrix} -1 & 1 & -2\\ 2 & 2 & 0 \\ 1 & 2 & -2 \end{pmatrix}\\ &=\begin{pmatrix} -\frac{1}{4} & \frac{1}{4} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} & 0 \\ -\frac{1}{4} & \frac{1}{2} & -\frac{1}{2} \end{pmatrix} \end{align}

Other Examples

  • Given \mathbf{A}^{2}-\mathbf{A}+2\mathbf{I}=\mathbf{0}. Prove that \mathbf{A}^{4}=2\mathbf{I}-3\mathbf{A}.
  • Analyze:
    • What does \mathbf{A}^{4} means?
      • \mathbf{A}^{4}=\mathbf{A}\mathbf{A}\mathbf{A}\mathbf{A}
      • But of course we would relate it to \mathbf{A}^{2}-\mathbf{A}+2\mathbf{I}=\mathbf{0}, which has \mathbf{A}^{2} (and also \mathbf{A})
      • Thus, it makes more sense we start with \mathbf{A}^{4}=\mathbf{A}^{2}\mathbf{A}^{2}
      • Since we we would need \mathbf{A}^{2}, we first rearrange the first line
    • \mathbf{A}^{2}-\mathbf{A}+2\mathbf{I}=\mathbf{0}
      • \mathbf{A}^{2}=\mathbf{A}-2\mathbf{I}
    • \mathbf{A}^{4}=\mathbf{A}^{2}\mathbf{A}^{2}
      • =\left(\mathbf{A}-2\mathbf{I}\right)\left(\mathbf{A}-2\mathbf{I}\right)
      • =\mathbf{A}^{2}-2\mathbf{AI}-2\mathbf{IA}+4\mathbf{I}^{2} Note that \mathbf{IA}=\mathbf{AI}=\mathbf{A} and \mathbf{I}^{2}=\mathbf{I}
      • =\mathbf{A}^{2}-4\mathbf{A}+4\mathbf{I}
      • Hmmm... what are we supposed to do next?
        • Whenever we are stuck in a prove question, compare what we have and what we need to prove
        • in this case, compare \mathbf{A}^{2}-4\mathbf{A}+4\mathbf{I} and 2\mathbf{I}-3\mathbf{A}
        • We will see that we need to get rid of \mathbf{A}^{2}
        • But what is \mathbf{A}^{2}?
        • Look carefully...
        • Can see it?
        • Right at the beginning, \mathbf{A}^{2}=\mathbf{A}-2\mathbf{I}
        • This is second key step in this question, it can be a bit difficult to see since it's weird in a way (like going round in a circle)
      • =\left(\mathbf{A}-2\mathbf{I}\right)-4\mathbf{A}+4\mathbf{I}
      • =2\mathbf{I}-3\mathbf{A}\quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved}
    • Further analysis
      • This shows an interesting property of matrices where we have expressed \mathbf{A}^{4} which is a multiplication of matrices into a LINEAR function of \mathbf{A}. This is because the first line allows us the express \mathbf{A}^{2} in terms of \mathbf{A}, thus if we keep on using it, all indices of \mathbf{A} will end up as 1. It will also work for other indices and even inverse.
      • But why would we want to do that? Probably because it's easier (in a way) to calculate that linear function.
      • We MUST use the above way to prove the equation. We can't, for example, substitute the matrix to RHS and LHS and show its the same matrix.


  • \mathbf{A}=\begin{pmatrix} 1 & 2\\ 0 & -3 \end{pmatrix}. If \mathbf{A}^{2}+m\mathbf{A}=n\mathbf{I}, find m\, and n\,. Hence, find \mathbf{A}^{3} & \mathbf{A^{-1}}.
  • Analyze
    • The first part should be pretty obvious, the second part is Hence, thus we will use a must use the values of m\, and n\, found.
      • \begin{align} \mathbf{A}^2& =\begin{pmatrix} 1 & 2\\ 0 & -3 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 0 & -3 \end{pmatrix}\\ & = \begin{pmatrix} 1 & -4\\ 0 & 9 \end{pmatrix} \end{align}
      • \mathbf{A}^{2}+m\mathbf{A}=n\mathbf{I}
      • \therefore \begin{pmatrix} 1 & -4\\ 0 & 9 \end{pmatrix}+m \begin{pmatrix} 1 & 2\\ 0 & -3 \end{pmatrix}=n\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}
      • \begin{array}{lll} \therefore & -4+2m=0 \qquad & 1+m=n \\ & m=2 & n=3 \end{array} Though we only used two of the 4 available equations, we really should use the other two to check, since a mistake in this part will nullify everything in the second part.
    • Because of the Hence, we must now substitute back the values into the equation, and then use the equation to find the wanted matrix.
      • We first substitute the values
        • \mathbf{A}^{2}+2\mathbf{A}=3\mathbf{I}
      • Let's start with \mathbf{A}^{3}, remember that we must express it UNTIL we get a LINEAR function in \mathbf{A}
        • From experience from the first example, we know we will need \mathbf{A}^{2} in terms of \mathbf{A}
        • \mathbf{A}^{2}=3\mathbf{I}-2\mathbf{A}
        • \begin{align} \mathbf{A}^{3} &=\mathbf{A}^{2}\mathbf{A}\\ & =\left(3\mathbf{I}-2\mathbf{A}\right)\mathbf{A}\\ & = 3\mathbf{A}-2\mathbf{A}^{2}\\ \end{align} Done? No. Even though we have calculated \mathbf{A}^{2}, we are NOT allowed to substitute it here. We must use the LINEAR FORM
        • \begin{align} & = 3\mathbf{A}-2\left(3\mathbf{I}-2\mathbf{A}\right)\\ & = 7\mathbf{A}-6\mathbf{I} \end{align} Done? No. The question did ask us to find the matrix itself (and not just express it)
        • \begin{align} & = 7\begin{pmatrix} 1 & 2\\ 0 & -3 \end{pmatrix}-6\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} \\ & =\begin{pmatrix} 1 & 14\\ 0 & -27 \end{pmatrix} \end{align}
      • As for \mathbf{A}^{-1}, we do it like before (refer to earlier part about finding inverse from equation)
        • \begin{align} \mathbf{A}^{2}+2\mathbf{A}& =3\mathbf{I}\\ \mathbf{A}^{-1}\left(\mathbf{A}^{2}+2\mathbf{A}\right)& =\mathbf{A}^{-1}\left(3\mathbf{I}\right)\\ \mathbf{A}+2\mathbf{I}&=3\mathbf{A}^{-1}\\ \mathbf{A}^{-1}&=\frac{1}{3}\left(\mathbf{A}+2\mathbf{I}\right)\\ &=\frac{1}{3}\left[\begin{pmatrix} 1 & 2\\ 0 & -3 \end{pmatrix}+2\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\right]\\ &=\frac{1}{3}\begin{pmatrix} 3 & 2\\ 0 & -1\end{pmatrix}\\ &=\begin{pmatrix} 1 & \frac{2}{3}\\ 0 & -\frac{1}{3} \end{pmatrix} \end{align}
    • Both answers can be confirmed by multiplying \mathbf{A} three times by hand and finding \mathbf{A}^{-1} using formula, or by using calculator.
    • Further analysis (IMPORTANT)
      • It is VERY important to understand the method used here. We must use the given equation to find whatever matrix needed to be in a LINEAR form of \mathbf{A}, and if we are asked to find the matrix itself, then use that LINEAR form to find the value.
      • DO NOT substitute any matrix prematurely (substituting \mathbf{A}^{2} for example) until we get the LINEAR form
      • Any other method is NOT allowed. We can(should), however, use the alternative methods to check.


  • If \mathbf{A^{-1}PA}=\mathbf{B}, find \mathbf{P}^{2} & \mathbf{P}^{3}.
  • Analyze
    • We want \mathbf{P}^{2}, which is \mathbf{PP}, thus we will first need \mathbf{P}
    • We must do this carefully, keeping in mind we CANNOT simply rearrange the order of the matrices
    • \begin{align} \mathbf{A^{-1}PA}&=\mathbf{B}\\ \mathbf{AA^{-1}PAA^{-1}}&=\mathbf{ABA^{-1}}\\ \mathbf{IPI}&=\mathbf{ABA^{-1}}\\ \mathbf{P}&=\mathbf{ABA^{-1}}\\ \end{align}
    • \mathbf{P^{2}}=\mathbf{PP} Let's substitute the \mathbf{P}
      • =\left(\mathbf{ABA^{-1}}\right)\left(\mathbf{ABA^{-1}}\right) Remember that we CANNOT just simply cut away all the \mathbf{A} and \mathbf{A}^{-1}. However, we do see that the middle two in this case is already next to each other
      • =\mathbf{AB\left(A^{-1}A\right)BA^{-1}}
      • =\mathbf{ABIBA^{-1}}
      • =\mathbf{ABBA^{-1}}
      • =\mathbf{AB^{2}A^{-1}}
      • Can't be simplified anymore
    • Try \mathbf{P^{3}} on your own first
      • \begin{align} \mathbf{P^{3}}& =\mathbf{P^{2}P}\\ &=\left(\mathbf{AB^{2}A^{-1}}\right)\left(\mathbf{ABA^{-1}}\right)\\ &=\mathbf{AB^{2}\left(A^{-1}A\right)BA^{-1}}\\ &=\mathbf{AB^{2}IBA^{-1}}\\ &=\mathbf{AB^{2}BA^{-1}}\\ &=\mathbf{AB^{3}A^{-1}}\\ \end{align}


  • If \mathbf{A}^{2}=\mathbf{A}, find \mathbf{A}^{21}.
  • Analyze
    • The answer is pretty obvious, since \mathbf{A} multiply with \mathbf{A} gives us back \mathbf{A}, even if we keep multiplying, it will still give us \mathbf{A} However, the marks is in the working.
    • We could, well, do it one by one
      • \begin{align} \mathbf{A}^{21}&=\mathbf{AAAAAAAAAAAAAAAAAAAAA}\\ &=\mathbf{AAAAAAAAAAAAAAAAAAAA}^{2}\\ &=\mathbf{AAAAAAAAAAAAAAAAAAAA}\\ &=\mathbf{AAAAAAAAAAAAAAAAAAA}^{2}\\ &=\mathbf{AAAAAAAAAAAAAAAAAAA}\\ & \quad .\\ & \quad .\\ & \quad .\\ &=\mathbf{A}^{2}\\ &=\mathbf{A} \end{align}
      • But of course, we are not going to do that. Even if you are able to write everything, and the answer is technically correct, you still might not be given marks as it doesn't satisfy the marking scheme.
    • However, we still did learn something, where we can see each \mathbf{A}^{2} simplified into \mathbf{A}, so why not pair all of them up in the first place.
      • \begin{align} \mathbf{A}^{21}&=\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}\\ &=\mathbf{AAAAAAAAAAA}\\ &=\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}\\ &=\mathbf{AAAAAA}\\ &=\mathbf{A}^{2}\mathbf{A}^{2}\mathbf{A}^{2}\\ &=\mathbf{AAA}\\ &=\mathbf{A}^{2}\mathbf{A}\\ &=\mathbf{AA}\\ &=\mathbf{A}^{2}\\ &=\mathbf{A} \end{align}
      • Better, but still won't satisfy the marking scheme (I think)
    • But we learn more here. We were able to pair them, and then pair the resulting \mathbf{A} into a single \mathbf{A}. The effect is turning four\mathbf{A} into one \mathbf{A}, and so on. What we need is a systematic way to show it.
      • Instead of starting directly with the question, we are going to simplify lower indices first, and then build up from there
      • Starting from what is given
        • \mathbf{A}^{2}=\mathbf{A}
      • We know we can pair two \mathbf{A}^{2} to get one \mathbf{A}. Two \mathbf{A}^{2} is \mathbf{A}^{4}, so we the next term we simplify is \mathbf{A}^{4}
        • \mathbf{A}^{4}=\mathbf{A}^{2}\mathbf{A}^{2} But each \mathbf{A}^{2} is \mathbf{A}
        • =\mathbf{AA}
        • =\mathbf{A}^{2}
        • =\mathbf{A}
      • The next step we will pair two \mathbf{A}^{4}, giving us \mathbf{A}^{8}
        • \begin{align} \mathbf{A}^{8}&=\mathbf{A}^{4}\mathbf{A}^{4}\\ &=\mathbf{AA}\\ &=\mathbf{A}^{2}\\ &=\mathbf{A} \end{align}
      • And the next
        • \begin{align} \mathbf{A}^{16}&=\mathbf{A}^{8}\mathbf{A}^{8}\\ &=\mathbf{AA}\\ &=\mathbf{A}^{2}\\ &=\mathbf{A} \end{align}
      • And the next
        • No. Since it will be \mathbf{A}^{32}, which will be bigger than what we need.
      • So what about \mathbf{A}^{21}? The biggest we have is \mathbf{A}^{16},which will still leave \mathbf{A}^{5}. However, we also have \mathbf{A}^{4}. In other words, express what is needed into the matrices we have already simplified above.
        • \mathbf{A}^{21}=\mathbf{A}^{16}\mathbf{A}^{4}\mathbf{A}
        • =\mathbf{AAA}
        • =\mathbf{A}^2\mathbf{A}
        • =\mathbf{AA}
        • =\mathbf{A}^{2}
        • =\mathbf{A} Phew. Done.
    • Further analyisis
      • Make sure you understand why in this question we jumped from \mathbf{A}^{2} to \mathbf{A}^{4} to \mathbf{A}^{8} and so on...


  • If \mathbf{A}^{3}=\mathbf{A}, find \mathbf{A}^{20}.
  • Try it on your own first! Answer is \mathbf{A}^{2}
    • \mathbf{A}^{3}=\mathbf{A}
    • The next step is \mathbf{A}^{9}. Do note that we need THREE \mathbf{A}^{3}, so it will be \mathbf{A}^{9}, not \mathbf{A}^{6}. \mathbf{A}^{6}=\mathbf{A}^{3}\mathbf{A}^{3}=\mathbf{AA} which can't be simplified further
    • \begin{align} \mathbf{A}^{9}&=\mathbf{A}^{3}\mathbf{A}^{3}\mathbf{A}^{3}\\ &=\mathbf{AAA}\\ &=\mathbf{A}^{3}\\ &=\mathbf{A} \end{align}
    • The next term is \mathbf{A}^{27}, which is not needed.
    • \mathbf{A}^{20}= \mathbf{A}^{9}\mathbf{A}^{9}\mathbf{AA}
    • =\mathbf{AAAA}
    • =\mathbf{A}^{3}\mathbf{A}
    • =\mathbf{AA} or \mathbf{A}^{2}

Exercise 3

1) \mathbf{A}=\begin{pmatrix} 1 & 0\\ 3 & 4 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 2 & 5\\ -1 & -4 \end{pmatrix} and \mathbf{X} is a 2 \times 2 matrix. Find matrix \mathbf{X} if

  • a) \mathbf{BX}=\mathbf{A} \mathbf{X}=\begin{pmatrix} \frac{19}{3} & \frac{20}{3}\\ -\frac{7}{3} & -\frac{8}{3} \end{pmatrix}
    • \begin{align} \mathbf{BX}&=\mathbf{A}\\ \mathbf{B^{-1}BX}&=\mathbf{B^{-1}A}\\ \mathbf{IX}&=\frac{1}{-3}\begin{pmatrix} -4 & -5\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 0\\ 3 & 4 \end{pmatrix}\\ \mathbf{X}&=-\frac{1}{3}\begin{pmatrix} -19 & -20\\ 7 & 8 \end{pmatrix}\\ &=\begin{pmatrix} \frac{19}{3} & \frac{20}{3}\\ -\frac{7}{3} & -\frac{8}{3} \end{pmatrix} \end{align}

  • b) \mathbf{XA}=\mathbf{B^{-1}} \mathbf{X}=\begin{pmatrix} \frac{1}{12} & \frac{5}{12}\\ \frac{1}{6} & -\frac{1}{6} \end{pmatrix}
    • \begin{align} \mathbf{XA}&=\mathbf{B^{-1}}\\ \mathbf{XAA^{-1}}&=\mathbf{B^{-1}A^{-1}}\\ \mathbf{XI}&=\frac{1}{-3}\begin{pmatrix} -4 & -5\\ 1 & 2 \end{pmatrix}\frac{1}{4}\begin{pmatrix} 4 & 0\\ -3 & 1 \end{pmatrix}\\ \mathbf{X}&=-\frac{1}{12}\begin{pmatrix} -1 & -5\\ -2 & 2 \end{pmatrix}\\ &=\begin{pmatrix} \frac{1}{12} & \frac{5}{12}\\ \frac{1}{6} & -\frac{1}{6} \end{pmatrix} \end{align}

  • c) \mathbf{XA^{-1}}=\mathbf{A} \mathbf{X}=\begin{pmatrix} 1 & 0 \\ 15 & 16 \end{pmatrix}
    • \begin{align} \mathbf{XA^{-1}}&=\mathbf{A}\\ \mathbf{XA^{-1}A}&=\mathbf{AA}\\ \mathbf{XI}&=\begin{pmatrix} 1 & 0\\ 3 & 4 \end{pmatrix}\begin{pmatrix} 1 & 0\\ 3 & 4 \end{pmatrix}\\ \mathbf{X}&=\begin{pmatrix} 1 & 0 \\ 15 & 16 \end{pmatrix}\\ \end{align}

  • d) \mathbf{XA}=\mathbf{AB} \mathbf{X}=\begin{pmatrix} -\frac{7}{4} & \frac{5}{4}\\ \frac{11}{4} & -\frac{1}{4} \end{pmatrix}
    • \begin{align} \mathbf{XA}&=\mathbf{AB}\\ \mathbf{XAA^{-1}}&=\mathbf{ABA^{-1}}\\ \mathbf{XI}&=\begin{pmatrix} 1 & 0\\ 3 & 4 \end{pmatrix}\begin{pmatrix} 2 & 5\\ -1 & -4 \end{pmatrix}\frac{1}{4}\begin{pmatrix} 4 & 0\\ -3 & 1 \end{pmatrix}\\ \mathbf{X}&=\frac{1}{4}\begin{pmatrix} 1 & 0\\ 3 & 4 \end{pmatrix}\begin{pmatrix} -7 & 5 \\ 8 & -4 \end{pmatrix}\\ &=\frac{1}{4}\begin{pmatrix} -7 & 5\\ 11 & -1 \end{pmatrix}\\ &=\begin{pmatrix} -\frac{7}{4} & \frac{5}{4}\\ \frac{11}{4} & -\frac{1}{4} \end{pmatrix} \end{align}

  • e) \mathbf{BXB^{-1}}=\mathbf{A} \mathbf{X}=\begin{pmatrix} 6 & 5 \\ -2 & -1 \end{pmatrix}
    • \begin{align} \mathbf{BXB^{-1}}&=\mathbf{A}\\ \mathbf{B^{-1}BXB^{-1}B}&=\mathbf{B^{-1}AB}\\ \mathbf{IXI}&=-\frac{1}{3}\begin{pmatrix} -4 & -5\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 & 0\\ 3 & 4 \end{pmatrix}\begin{pmatrix} 2 & 5\\ -1 & -4 \end{pmatrix}\\ \mathbf{X}&=-\frac{1}{3}\begin{pmatrix} -4 & -5\\ 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 2 & -1 \end{pmatrix}\\ &=-\frac{1}{3}\begin{pmatrix} -18 & -15\\ 6 & 3 \end{pmatrix}\\ &=\begin{pmatrix} 6 & 5\\ -2 & -1 \end{pmatrix} \end{align}

  • f) \mathbf{XAB}=\mathbf{B^{-1}} \mathbf{X}=\begin{pmatrix} \frac{7}{18} & \frac{5}{18}\\ -\frac{5}{36} & -\frac{1}{36} \end{pmatrix}
    • \begin{align} \mathbf{XAB}&=\mathbf{B^{-1}}\\ \mathbf{XABB^{-1}}&=\mathbf{B^{-1}B^{-1}}\\ \mathbf{XAI}&=\mathbf{B^{-1}B^{-1}}\\ \mathbf{XAA^{-1}}&=\mathbf{B^{-1}B^{-1}A^{-1}}\\ \mathbf{XI}&=-\frac{1}{3}\begin{pmatrix} -4 & -5\\ 1 & 2 \end{pmatrix}\left[-\frac{1}{3}\begin{pmatrix} -4 & -5\\ 1 & 2 \end{pmatrix}\right]\frac{1}{4}\begin{pmatrix} 4 & 0\\ -3 & 1 \end{pmatrix}\\ &=\frac{1}{36}\begin{pmatrix} -4 & -5\\ 1 & 2 \end{pmatrix}\begin{pmatrix} -1 & -5\\ -2 & 2 \end{pmatrix}\\ &=\frac{1}{36}\begin{pmatrix} 14 & 10\\ -5 & -1 \end{pmatrix}\\ &=\begin{pmatrix} \frac{7}{18} & \frac{5}{18}\\ -\frac{5}{36} & -\frac{1}{36} \end{pmatrix} \end{align}


2) \mathbf{A}=\begin{pmatrix} -4 & 3\\ -1 & 2 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 2 & -3\\ 1 & -4 \end{pmatrix}, \mathbf{P}=\begin{pmatrix} -2 & 0 & -3\\ 4 & 2 & -1 \\ 1 & -3 & 0 \end{pmatrix}, \mathbf{Q}=\begin{pmatrix} -3 & 9 & 6\\ -1 & 3 & -14 \\ -14 & -6 & -4 \end{pmatrix}

Show that \mathbf{AB}+5\mathbf{I}=\mathbf{0} and find \mathbf{PQ}. Hence, find \mathbf{A^{-1}} and \mathbf{Q^{-1}}.

  • \mathbf{PQ}=\begin{pmatrix} 48 & 0 & 0\\ 0 &48 & 0 \\ 0 & 0 & 48 \end{pmatrix},\mathbf{A^{-1}}=\begin{pmatrix} -\frac{2}{5} & \frac{3}{5}\\ -\frac{1}{5} & \frac{4}{5} \end{pmatrix}, \mathbf{Q^{-1}}=\begin{pmatrix} -\frac{1}{24} & 0 & -\frac{1}{16} \\ \frac{1}{12} &\frac{1}{24} & -\frac{1}{48} \\ \frac{1}{48} & -\frac{1}{16} & 0 \end{pmatrix}
    • \begin{align} \mathbf{AB}&=\begin{pmatrix} -4 & 3\\ -1 & 2 \end{pmatrix}\begin{pmatrix} 2 & -3\\ 1 & -4 \end{pmatrix}\\ &= \begin{pmatrix} -5 & 0\\ 0 & -5 \end{pmatrix}\\ &=-5\mathbf{I} \end{align}
    • \therefore \mathbf{AB}+5\mathbf{I}=\mathbf{0}
    • Alternatively, \mathbf{AB}+5\mathbf{I}=\dots=\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}=\mathbf{0}
    • \begin{align} \mathbf{PQ}&=\begin{pmatrix} -2 & 0 & -3\\ 4 & 2 & -1 \\ 1 & -3 & 0 \end{pmatrix}\begin{pmatrix} -3 & 9 & 6\\ -1 & 3 & -14 \\ -14 & -6 & -4 \end{pmatrix}\\ &=\begin{pmatrix} 48 & 0 & 0\\ 0 &48 & 0 \\ 0 & 0 & 48 \end{pmatrix} \end{align}
    • Since \mathbf{AB}+5\mathbf{I}=\mathbf{0}
    • \begin{align} \mathbf{AB}&=-5\mathbf{I}\\ \mathbf{A^{-1}}\mathbf{AB}&=\mathbf{A^{-1}}\left(-5\mathbf{I}\right)\\ \mathbf{IB}&=-5\mathbf{A^{-1}}\\ \mathbf{A^{-1}}&=-\frac{1}{5}\mathbf{B}\\ &=-\frac{1}{5}\begin{pmatrix} 2 & -3\\ 1 & -4 \end{pmatrix}\\ &=\begin{pmatrix} -\frac{2}{5} & \frac{3}{5}\\ -\frac{1}{5} & \frac{4}{5} \end{pmatrix} \end{align}
    • Since \mathbf{PQ}=\begin{pmatrix} 48 & 0 & 0\\ 0 &48 & 0 \\ 0 & 0 & 48 \end{pmatrix}=48\mathbf{I}
    • \begin{align} \mathbf{PQ}&=48\mathbf{I}\\ \mathbf{PQ}\mathbf{Q^{-1}}&=\left(48\mathbf{I}\right)\mathbf{Q^{-1}}\\ \mathbf{PI}&=48\mathbf{Q^{-1}}\\ \mathbf{Q^{-1}}&=\frac{1}{48}\mathbf{P}\\ &=\frac{1}{48}\begin{pmatrix} -2 & 0 & -3\\ 4 & 2 & -1 \\ 1 & -3 & 0 \end{pmatrix}\\ &=\begin{pmatrix} -\frac{1}{24} & 0 & -\frac{1}{16} \\ \frac{1}{12} &\frac{1}{24} & -\frac{1}{48} \\ \frac{1}{48} & -\frac{1}{16} & 0 \end{pmatrix} \end{align}


3) \mathbf{A}=\begin{pmatrix} 2 & 3\\ -1 & 0 \end{pmatrix}. If \mathbf{A}^{2}+m\mathbf{A}+n\mathbf{I}=\mathbf{0}, find m\, and n\,. Hence, prove \mathbf{A}^{3}=\mathbf{A}-6\mathbf{I}. Find also \mathbf{A}^{4} & \mathbf{A^{-1}}.

  • n=3, m=-2, \mathbf{A}^{3}=\begin{pmatrix} -4 & 3\\ -1 & -6 \end{pmatrix}, \mathbf{A}^{4}=\begin{pmatrix} -11 & -12\\ 4 & -3 \end{pmatrix}, \mathbf{A}^{-1}=\begin{pmatrix} 0 & -1\\ \frac{1}{3} & \frac{2}{3} \end{pmatrix}
    • \begin{align} \mathbf{A}^{2}&=\begin{pmatrix} 2 & 3\\ -1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 3\\ -1 & 0 \end{pmatrix}\\ &=\begin{pmatrix} 1 & 6\\ -2 & -3 \end{pmatrix} \end{align}
    • \mathbf{A}^{2}+m\mathbf{A}+n\mathbf{I}=\mathbf{0}
    • \begin{pmatrix} 1 & 6\\ -2 & -3 \end{pmatrix}+m\begin{pmatrix} 2 & 3\\ -1 & 0 \end{pmatrix}+n\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}
    • \begin{array}{lll} \therefore & 6+3m=0 & -3m+n=0\\ & m=-2 & n=3 \end{array}
    • \therefore \mathbf{A}^{2}-2\mathbf{A}+3\mathbf{I}=\mathbf{0}
    • \therefore \mathbf{A}^{2}=2\mathbf{A}-3\mathbf{I}
    • \begin{align} \mathbf{A}^{3}&=\mathbf{A}^{2}\mathbf{A}\\ &=\left(2\mathbf{A}-3\mathbf{I}\right)\mathbf{A}\\ &=2\mathbf{A}^{2}-3\mathbf{A}\\ &=2\left(2\mathbf{A}-3\mathbf{I}\right)-3\mathbf{A}\\ &=\mathbf{A}-6\mathbf{I}\quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \end{align}
    • \begin{align} \therefore \mathbf{A}^{3}&=\begin{pmatrix} 2 & 3\\ -1 & 0 \end{pmatrix}-6\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} -4 & 3\\ -1 & -6 \end{pmatrix} \end{align}
    • \begin{align} \mathbf{A}^{4}&=\mathbf{A}^{2}\mathbf{A}^{2}\\ &=\left(2\mathbf{A}-3\mathbf{I}\right)\left(2\mathbf{A}-3\mathbf{I}\right)\\ &=4\mathbf{A}^{2}-6\mathbf{AI}-6\mathbf{IA}+9\mathbf{I}^{2}\\ &=4\mathbf{A}^{2}-12\mathbf{A}+9\mathbf{I}\\ &=4\left(2\mathbf{A}-3\mathbf{I}\right)-12\mathbf{A}+9\mathbf{I}\\ &=-4\mathbf{A}-3\mathbf{I}\\ &=-4\begin{pmatrix} 2 & 3\\ -1 & 0 \end{pmatrix}-3\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} -11 & -12\\ 4 & -3 \end{pmatrix} \end{align}
    • \begin{align} \mathbf{A}^{2}-2\mathbf{A}+3\mathbf{I}&=\mathbf{0}\\ \mathbf{A}^{2}-2\mathbf{A}&=-3\mathbf{I}\\ \mathbf{A^{-1}}\left(\mathbf{A}^{2}-2\mathbf{A}\right)&=\mathbf{A^{-1}}\left(-3\mathbf{I}\right)\\ \mathbf{A}-2\mathbf{I}&=-3\mathbf{A^{-1}}\\ \mathbf{A^{-1}}&=\frac{1}{3}\left(2\mathbf{I}-\mathbf{A}\right)\\ &=\frac{1}{3}\left[2\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}-\begin{pmatrix} 2 & 3\\ -1 & 0 \end{pmatrix}\right]\\ &=\frac{1}{3}\begin{pmatrix} 0 & -3\\ 1 & 2 \end{pmatrix}\\ &=\begin{pmatrix} 0 & -1\\ \frac{1}{3} & \frac{2}{3} \end{pmatrix} \end{align}


4) If \mathbf{AP}=\mathbf{BA}, prove that \mathbf{P}^{2}=\mathbf{A^{-1}}\mathbf{B}^{2}\mathbf{A}. Find \mathbf{P}^{3}, hence deduce \mathbf{P}^{n}, where n\, is a natural number.

  • \mathbf{P}^{3}=\mathbf{A^{-1}}\mathbf{B}^{3}\mathbf{A},\mathbf{P}^{n}=\mathbf{A^{-1}}\mathbf{B}^{n}\mathbf{A}
    • \begin{align} \mathbf{AP}&=\mathbf{BA}\\ \mathbf{A^{-1}AP}&=\mathbf{A^{-1}BA}\\ \mathbf{IP}&=\mathbf{A^{-1}BA}\\ \mathbf{P}&=\mathbf{A^{-1}BA}\\ \end{align}
    • \begin{align} \mathbf{P^{2}}&=\mathbf{PP}\\ &=\left(\mathbf{A^{-1}BA}\right)\left(\mathbf{A^{-1}BA}\right)\\ &=\mathbf{A^{-1}B\left(AA^{-1}\right)BA}\\ &=\mathbf{A^{-1}BIBA}\\ &=\mathbf{A^{-1}BBA}\\ &=\mathbf{A^{-1}B^{2}A}\quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \\ \end{align}
    • \begin{align} \mathbf{P^{3}}&=\mathbf{P^{2}P}\\ &=\left(\mathbf{A^{-1}B^{2}A}\right)\left(\mathbf{A^{-1}BA}\right)\\ &=\mathbf{A^{-1}B^{2}\left(AA^{-1}\right)BA}\\ &=\mathbf{A^{-1}B^{2}IBA}\\ &=\mathbf{A^{-1}B^{2}BA}\\ &=\mathbf{A^{-1}B^{3}A}\\ \end{align}
    • \therefore \mathbf{P}^{n}=\mathbf{A^{-1}B}^{n}\mathbf{A}


5) \mathbf{A}=\begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} -1 & 3\\ 0 & 0 \end{pmatrix} prove that \mathbf{A}^{2}=\mathbf{A} and find \mathbf{B}^{3}. Hence, find \mathbf{A}^{27} and \mathbf{B}^{40}.

  • \mathbf{A}^{27}=\begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix}, \mathbf{B}^{3}=\begin{pmatrix} -1 & 3\\ 0 & 0 \end{pmatrix}, \mathbf{B}^{40}=\begin{pmatrix} 1 & -3\\ 0 & 0 \end{pmatrix}
    • \begin{align} \mathbf{A}^{2}&=\begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix}\begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix}\\ &=\mathbf{A}\quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \end{align}
    • \begin{align} \mathbf{A}^{2}&=\mathbf{A}\\ \mathbf{A}^{4}&=\mathbf{A}^{2}\mathbf{A}^{2}=\mathbf{AA}=\mathbf{A}^{2}=\mathbf{A}\\ \mathbf{A}^{8}&=\mathbf{A}^{4}\mathbf{A}^{4}=\mathbf{AA}=\mathbf{A}^{2}=\mathbf{A}\\ \mathbf{A}^{16}&=\mathbf{A}^{8}\mathbf{A}^{8}=\mathbf{AA}=\mathbf{A}^{2}=\mathbf{A}\\ \mathbf{A}^{27}&=\mathbf{A}^{16}\mathbf{A}^{8}\mathbf{A}^{2}\mathbf{A}\\ &=\mathbf{AAAA}\\ &=\mathbf{A}^{4}\\ &=\mathbf{A}\\ &=\begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix} \end{align}
    • \begin{align} \mathbf{B}^{3}&=\begin{pmatrix} -1 & 3\\ 0 & 0 \end{pmatrix}\begin{pmatrix} -1 & 3\\ 0 & 0 \end{pmatrix}\begin{pmatrix} -1 & 3\\ 0 & 0 \end{pmatrix}\\ &=\begin{pmatrix} -1 & 3\\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & -3\\ 0 & 0 \end{pmatrix}\\ &=\begin{pmatrix} -1 & 3\\ 0 & 0 \end{pmatrix} \end{align}
    • \begin{align} \mathbf{B}^{3}&=\mathbf{B}\\ \mathbf{B}^{9}&=\mathbf{B}^{3}\mathbf{B}^{3}\mathbf{B}^{3}=\mathbf{BBB}=\mathbf{B}^{3}=\mathbf{B}\\ \mathbf{B}^{27}&=\mathbf{B}^{9}\mathbf{B}^{9}\mathbf{B}^{9}=\mathbf{BBB}=\mathbf{B}^{3}=\mathbf{B}\\ \mathbf{B}^{40}&=\mathbf{B}^{27}\mathbf{B}^{9}\mathbf{B}^{3}\mathbf{B}\\ &=\mathbf{BBBB}\\ &=\mathbf{B}^{3}\mathbf{B}\\ &=\mathbf{BB}\\ &=\begin{pmatrix} 1 & -3\\ 0 & 0 \end{pmatrix} \end{align}

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