Matrices Past Year

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Contents

Preparation

  • Make sure you have have mastered all the materials in the previous parts.
  • If you are using the material here to learn this topic (as opposed to just a revision to supplement your school lessons), its' best you take a day (or more) off after learning the previous parts.
  • In fact, try take a week off, and then revise back all the materials before trying the following questions. This will be a good practice for how you are going to revise for the actual STPM.
  • Find a comfortable place & comfortable time.
  • DO NOT do this while you are half-awake, or directly after a long day in school. Else, you will frustrate yourself. Trust me.
  • Saving trees is a good thing, but DO NOT do this (in fact, any of the exercises) on rough paper/recycled paper.
  • "ROUGH PAPER = ROUGH WORK = CARELESS MISTAKES = LOSS OF MARKS"
  • Know that you will face questions that you have NEVER SEEN which will require you to adapt on the spot.
  • Keep a clock or watch handy and give yourself the suggested time to complete it. Check your answers too during that time limit.
  • After finishing, check the answers given. If you made mistakes or couldn't find the solution, you can refer to the answers/answers with guidance.

Questions

Estimated time :

1) Determine the value of k\, such that the determinant of matrix \begin{pmatrix} k & 1 & 3\\ 2k+1 & -3 & 2 \\ 0& k &2 \end{pmatrix} is 0.


2) Determine the value of a, b, c \, and so that matrix \begin{pmatrix} 2b-1 & a^{2} & b^{2}\\ 2a-1 & a & bc \\ b& b+c & 2c-1 \end{pmatrix} is a symmetrical matrix.


3) Matrix \mathbf{A}is given by \begin{pmatrix} 3 & 3 & 4\\ 5 & 4 & 1 \\ 1& 2 & 3 \end{pmatrix} . Find the adjoin of \mathbf{A}. Hence, find \mathbf{A}^{-1}.


4) The matrices \mathbf{P} and \mathbf{Q}, where \mathbf{PQ}=\mathbf{QP} are given by, \mathbf{P}=\begin{pmatrix} 2 & -2 & 0\\ 0 & 0 & 2 \\ a& b & c \end{pmatrix} \mathbf{Q}=\begin{pmatrix} -1 & 1 & 0\\ 0 & 0 & -1 \\ 0& -2 & 2 \end{pmatrix}. Determine the values of a, b\, and c\,. Find the real numbers m\, and n\, for which \mathbf{P}=m\mathbf{Q}+n\mathbf{I}, where \mathbf{I} is the 3 \times 3 identity matrix.


5) If \mathbf{A}=\begin{pmatrix} -1 & 2\\ -3 & 4 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 1 & 2\\ 1 & 3 \end{pmatrix}, find \mathbf{C} so that \mathbf{A}=\mathbf{BCB}^{-1}.


6) If \mathbf{M}=\begin{pmatrix} m & 2\\ 6 & m-1 \end{pmatrix}, find the set values of m\, for which the inverse of \mathbf{M} exist.


7) If \mathbf{A}=\begin{pmatrix} -1 & 4\\ 2 & 1 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 0 & -1\\ 3 & 2 \end{pmatrix} and \mathbf{C}=\begin{pmatrix} 3 & 4\\ 21 & 19 \end{pmatrix}, find matrix \mathbf{X} such that\mathbf{AXB}=\mathbf{C}.


8) The matrix A is given by \mathbf{A}=\begin{pmatrix} 1 & 2 & -3\\ 3 & 1 & 1 \\ 0 & 1 & -2 \end{pmatrix}

i) Find the matrix \mathbf{B} such that \mathbf{B}=\mathbf{A}^{2}-10\mathbf{I}, \mathbf{I} is the 3\times3 identity matrix.

ii) Find \left( \mathbf{A}+\mathbf{I}\right)\mathbf{B}, and hence find \left(\mathbf{A}+\mathbf{I}\right)^{21}\mathbf{B}.


9) \mathbf{A}, \mathbf{B} and \mathbf{C} are square matrices such that \mathbf{BA}=\mathbf{B}^{-1} and \mathbf{ABC}=\left(\mathbf{AB}\right)^{-1}. Show that \mathbf{A}^{-1}=\mathbf{B}^{2}=\mathbf{C}.

If \mathbf{B}=\begin{pmatrix} 1 & 2 & 0\\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix}, find \mathbf{C} and \mathbf{A}.


10) Given \mathbf{M}=\begin{pmatrix} 2 & 3 & 1\\ -1 & 0 & 4 \\ 1 & -1 & 1 \end{pmatrix}. Show that \mathbf{M}^{3}-3\mathbf{M}^{2}+8\mathbf{M}-24\mathbf{I}_{3}=\mathbf{0}. Deduce \mathbf{M}^{-1}.


11) If \mathbf{A}=\begin{pmatrix} 1 & -1 & 1\\ 2 & 0 & 1 \\ 0 & 2 & -1 \end{pmatrix} Show that \mathbf{A}^{3}=\mathbf{A}. Hence, find \mathbf{A}^{40}.


12) Show that the matrix \mathbf{A}=\begin{pmatrix} 0 & -3\\ -1 & 1 \end{pmatrix} satisfies the equation \mathbf{A}^{2}-\mathbf{A}-3\mathbf{I}=\mathbf{0} . Hence, without evaluating \mathbf{A}^{3} or \mathbf{A}^{4}, show that \mathbf{A}^{4}=7\mathbf{A}+12\mathbf{I}.


13) \mathbf{A}=\begin{pmatrix} 2 & 3\\ -1 & 1 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} x & y\\ 1 & -1 \end{pmatrix}. Find the diagonal matrix \mathbf{D} such that \mathbf{ADA}^{-1}=\mathbf{B}.


14) The matrices \mathbf{A} and \mathbf{B} are given by \mathbf{A}=\begin{pmatrix} -1 & 2 & 1\\ -3 & 1 & 4 \\ 0 & 1 & 2 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} -35 & 19 & 18\\ -27 & -13 & 45 \\ -3 & 12 & 5 \end{pmatrix}. Find the matrix \mathbf{A}^{2}\mathbf{B} and deduce the inverse of \mathbf{A} . Hence, solve the system of linear equations

\begin{array}{lll} x-2y-z=-8\\ 3x-y-4z=-15\\ y+2z=4 \end{array}


15) If \mathbf{P}=\begin{pmatrix} 5 & 2 & 3\\ 1 & -4 & 3 \\ 3 & 1 & 2 \end{pmatrix} and \mathbf{Q}=\begin{pmatrix} a & 1 & -18\\ b & -1 & 12 \\ -13 & -1 & c \end{pmatrix}, and \mathbf{PQ}=2\mathbf{I}, determine the value of a\,, b\, and c\,. Hence, find \mathbf{P}^{-1}.

Two groups of workers have their drinks at a stall. The first group comprising ten workers have five cups of tea, two cups of coffee an three glasses of fruit juice at a total cost of RM 11.80. The second group of six workers have three cups of tea, a cup of coffee and two glasses of fruit juice at a total cost of RM 7.10. The cost of a cup of tea and three glasses of fruit juice is the same as the cost of four cups of coffee. The cost of a cup of tea, a cup of coffee and a glass of fruit juice are RM x\,, RMy\, and RMz\, respectively, obtain a matrix equation to represent the above information. Hence, determine the cost of each drink.


16) \mathbf{M}=\begin{pmatrix} -10 & 4 & 9\\ 15 & -4 & -14 \\ -5 & 1 & 6 \end{pmatrix} and \mathbf{N}=\begin{pmatrix} 2 & 3 & 4\\ 4 & 3 & 1 \\ 1 & 2 & 4 \end{pmatrix}. Find \mathbf{MN} and deduce \mathbf{N}^{-1}.

Products X, Y and Z are assembled from three components A, B and C according to different proportions. Each product X consists of two components of A, four components of B and one component of C; each product of Y consists of three components of A, three components of B and two component of C; and each product of Z consists of four components of A, one components of B and four component of C. A total of 750 components of A, 1 000 components of B and 500 components of C are used. With x\,, y\, and z\, representing the numbers of products of X, Y and Z assembled, obtain a matrix equation representing the information given. Hence, find the numbers of products of X, Y and Z assembled.


17) The matrices \mathbf{A} and \mathbf{B} are given as \mathbf{A}=\begin{pmatrix} 6 & 4 & 3\\ 3 & 3 & 7 \\ 1 & 3 & 0 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 21 & -9 & -19\\ -7 & 3 & 33 \\ -6 & 14 & -6 \end{pmatrix}. Find \mathbf{AB} and \mathbf{A}^{-1}.

A company produces three types of instant coffee under the brands Jerai, Ledang and Mulu which contain the percentages (according to mass) of coffee powder, sugar and powdered milk as shown in the following table. Brand of instant coffee Composition by percentage

Brand of instant coffee Composition by percentage
Coffee powder Sugar Powdered milk
Jerai 60 30 10
Ledang 40 30 30
Mulu 30 70 0

The company mixes the coffee Jerai, Ledang and Mulu to yield a new instant coffee under the brand Jelemu in 50 g packet containing 44% coffee powder, 38% sugar and 18% powdered milk. If packet of the Jelemu coffee contains x\,g of the coffee Jerai, y\,g of the coffee Ledang and z\,g of the coffee Mulu, show that \mathbf{A}\begin{pmatrix} x \\ y\\ z \end{pmatrix}=\begin{pmatrix} 220 \\ 190\\ 90 \end{pmatrix}. Hence, determine the mass of the coffee Jerai, Ledang and Mulu in each 50 g of packet of coffee Jelemu.


18)a) \mathbf{A}=\begin{pmatrix} 1 & 1 & 2\\ 0 & 2 & 2 \\ -1 & 1 & 3 \end{pmatrix}. Find \mathbf{A}^{2}-6\mathbf{A}+11\mathbf{I}. Show that \mathbf{A}\left(\mathbf{A}^{2}-6\mathbf{A}+11\mathbf{I}\right)=6\mathbf{I} and deduce \mathbf{A}^{-1}.

b) A factory produces three types of nuts namely kacang kuda, kacang botak and kacang parang. The profit from 1 kg of kacang kuda, 1 kg of kacang botak and 2 kg of kacang parang is RM 9. The profit from 1 kg of kacang botak and 1 kg of kacang parang is RM 3. The profit from 1 kg of kacang botak and 3 kg of kacang parang is equal to the profit from 1 kg of kacang kuda. If x\,, y\, and z\, represent the profit from 1 kg of kacang kuda, 1 kg of kacang botak and 1 kg of kacang parang respectively, write a matrix equation to represent the above information. Hence, determine the profit from 1 kg of kacang kuda, 1 kg of kacang botak and 1 kg of kacang parang.


19)\mathbf{A}=\begin{pmatrix} 5 & -5 & 0\\ -5 & 0 & 5 \\ 0 & 10 & -5 \end{pmatrix}, \mathbf{A}=\begin{pmatrix} 2 & 1 & 1\\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{pmatrix}. Find \mathbf{AB} and \mathbf{B}^{-1}.

The following table shows the wholesale price in RM per kg, of three commodities, namely red chilies, long beans and cucumber, in the three towns of Kuala Terengganu, Kuala Lumpur and Johor Bahru.

Town Commodities
Red chilies Long Beans Cucumber
Kuala Terengganu 4 2 2
Kuala Lumpur 2 2 2
Johor Bahru 4 4 2

A company with branch offices located in each of the three towns has been awarded a contact worth RM 5000 in each town to supply x\, kg of red chilies, y\, kg of long bean and z\, kg of cucumbers in each town to the local retailer. The vegetables are obtain form the town itself. The profit earn by the branches in Kuala Terengganu, Kuala Lumpur and Johor Bahru are RM 2000, RM 3000 and RM 1500 respectively. Write a matrix equation in x\,, y\, and z\, to represent the above information. Hence, determine the quantities of red chilies, long beans and cucumbers supplied in each town.

Answers

Estimated time :

1) Determine the value of k\, such that the determinant of matrix \begin{pmatrix} k & 1 & 3\\ 2k+1 & -3 & 2 \\ 0& k &2 \end{pmatrix} is 0.

  • k=-\frac{1}{4} or k=2\,
  • \begin{array}{rcl} \begin{vmatrix} k & 1 & 3 \\ 2k+1 & -3 & 2 \\ 0 & k & 2 \end{vmatrix} & = & k\begin{vmatrix} -3 & 2 \\ k & 2 \end{vmatrix} -1\begin{vmatrix} 2k+1 & 2 \\ 0 & 2 \end{vmatrix} +3\begin{vmatrix} 2k+1 & -3 \\ 0 & k \end{vmatrix} \\ \therefore 0 & = & k\left(-6-2k\right)-\left(4k+2\right)+3\left(2k^2+k\right)\\ 0 & = & -6k-2k^{2}-4k-2+6k^2+3k\\ 0 & = & 4k^2-7k-2\\ 0 & = & \left(4k+1\right)\left(k-2\right)\\ k & =& -\frac{1}{4} \qquad \mathrm{or} \qquad k=2 \end{array}


2) Determine the value of a, b, c \, and so that matrix \begin{pmatrix} 2b-1 & a^{2} & b^{2}\\ 2a-1 & a & bc \\ b& b+c & 2c-1 \end{pmatrix} is a symmetrical matrix.

  • a=1,b=0,c=0\,
  • Since the matrix is symmetrical
    • \begin{array}{lll} 2a-1=a^2 \qquad & b=b^2 & b+c=bc \\ a^2-2a+1 & b^2-b=0 & \mathrm{when}\quad b=0, \\ \left(a-1\right)^2=0 & b\left(b-1\right)=0 & c=0 \\ a=1 & b=0 \quad\mathrm{or}\quad b=1 \qquad & \mathrm{when}\quad b=1, \\ & & 1+c=c\qquad\mathrm{(impossible)}\qquad \end{array}
    • \therefore a=1,b=0,c=0


3) Matrix \mathbf{A}is given by \begin{pmatrix} 3 & 3 & 4\\ 5 & 4 & 1 \\ 1& 2 & 3 \end{pmatrix} . Find the adjoin of \mathbf{A}. Hence, find \mathbf{A}^{-1}.

  • adj \mathbf{A}= \begin{pmatrix} 10 & -1 & -13\\ -14 & 5 & 17 \\ 6 & -3 & -3 \end{pmatrix} \mathbf{A}^{-1}=\begin{pmatrix} \frac{5}{6} & -\frac{1}{12} & -\frac{13}{12}\\ -\frac{7}{6} & \frac{5}{12} & \frac{17}{12} \\ \frac{1}{2} & -\frac{1}{4} & -\frac{1}{4} \end{pmatrix}
  • \mathbf{A}^T= \begin{pmatrix} 3 & 5 & 1\\ 3 & 4 & 2 \\ 4 & 1 & 3 \end{pmatrix}
    • adj \mathbf{A}=\begin{pmatrix} \begin{vmatrix} 4 & 2 \\ 1 & 3 \\ \end{vmatrix} & -\begin{vmatrix} 3 & 2 \\ 4 & 3 \\ \end{vmatrix} & \begin{vmatrix} 3 & 4 \\ 4 & 1 \\ \end{vmatrix} \\ -\begin{vmatrix} 5 & 1 \\ 1 & 3 \\ \end{vmatrix} & \begin{vmatrix} 3 & 1 \\ 4 & 3 \\ \end{vmatrix} & -\begin{vmatrix} 3 & 5 \\ 4 & 1 \\ \end{vmatrix} \\ \begin{vmatrix} 5 & 1 \\ 4 & 2 \\ \end{vmatrix} & -\begin{vmatrix} 3 & 1 \\ 3 & 2 \\ \end{vmatrix} & \begin{vmatrix} 3 & 5 \\ 3 & 4 \\ \end{vmatrix} \end{pmatrix}
    • =\begin{pmatrix} 10 & -1 & -13\\ -14 & 5 & 17 \\ 6 & -3 & -3 \end{pmatrix}
    • \begin{align} \left|\mathbf{A}\right| & =3\begin{vmatrix} 4 & 1 \\ 2 & 3 \\ \end{vmatrix}-3\begin{vmatrix} 5 & 1 \\ 1 & 3 \\ \end{vmatrix}+4\begin{vmatrix} 5 & 4 \\ 1 & 2 \\ \end{vmatrix}\\ & = 3\left(10\right)-3\left(14\right)+4\left(6\right)\\ & = 12 \end{align}
    • \begin{align} \mathbf{A}^{-1} & = \frac{1}{\left|\mathbf{A}\right|}adj \mathbf{A}\\ & = \frac{1}{12}\begin{pmatrix} 10 & -1 & -13\\ -14 & 5 & 17 \\ 6 & -3 & -3 \end{pmatrix}\\ & =\begin{pmatrix} \frac{5}{6} & -\frac{1}{12} & -\frac{13}{12}\\ -\frac{7}{6} & \frac{5}{12} & \frac{17}{12} \\ \frac{1}{2} & -\frac{1}{4} & -\frac{1}{4} \end{pmatrix} \end{align}


4) The matrices \mathbf{P} and \mathbf{Q}, where \mathbf{PQ}=\mathbf{QP} are given by, \mathbf{P}=\begin{pmatrix} 2 & -2 & 0\\ 0 & 0 & 2 \\ a& b & c \end{pmatrix} \mathbf{Q}=\begin{pmatrix} -1 & 1 & 0\\ 0 & 0 & -1 \\ 0& -2 & 2 \end{pmatrix}. Determine the values of a, b\, and c\,. Find the real numbers m\, and n\, for which \mathbf{P}=m\mathbf{Q}+n\mathbf{I}, where \mathbf{I} is the 3 \times 3 identity matrix.

  • a=0,b=4,c=-4;m=-2,n=0\,
    • \begin{array}{rcl} \mathbf{PQ}&=&\mathbf{QP}\\ \begin{pmatrix} 2 & -2 & 0\\ 0 & 0 & 2 \\ a& b & c \end{pmatrix}\begin{pmatrix} -1 & 1 & 0\\ 0 & 0 & -1 \\ 0& -2 & 2 \end{pmatrix}&=&\begin{pmatrix} -1 & 1 & 0\\ 0 & 0 & -1 \\ 0& -2 & 2 \end{pmatrix}\begin{pmatrix} 2 & -2 & 0\\ 0 & 0 & 2 \\ a& b & c \end{pmatrix} \end{array}
    • \begin{array}{lll} -a=2a \qquad & a-2c = 2b \qquad& -b+2c=-4+2c \\ a=0 & \therefore -2c=2b & \therefore b=4 \\ & b=-c & \therefore c=-4 \\ \end{array}
    • \therefore a=0,b=4,c=-4
    • \begin{align} \mathbf{P}&=m\mathbf{Q}+n\mathbf{I}\\ \begin{pmatrix} 2 & -2 & 0\\ 0 & 0 & 2 \\ 0 & 4 & -4 \end{pmatrix}& =m\begin{pmatrix} -1 & 1 & 0\\ 0 & 0 & -1 \\ 0& -2 & 2 \end{pmatrix}+n\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\\ & =\begin{pmatrix} -m+n & m & 0\\ 0 & n & -m \\ 0& -2m & 2m+n \end{pmatrix} \end{align}
    • \therefore m=-2,n=0


5) If \mathbf{A}=\begin{pmatrix} -1 & 2\\ -3 & 4 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 1 & 2\\ 1 & 3 \end{pmatrix}, find \mathbf{C} so that \mathbf{A}=\mathbf{BCB}^{-1}.

  • \mathbf{C}=\begin{pmatrix} 1 & 0\\ 0 & 2 \end{pmatrix}
  • \begin{align} \mathbf{A}&=\mathbf{BCB}^{-1}\\ \mathbf{B}^{-1}\mathbf{BCB}^{-1}\mathbf{B}&=\mathbf{B}^{-1}\mathbf{AB}\\ \mathbf{ICI}&=\mathbf{B}^{-1}\mathbf{AB}\\ \mathbf{C}&=\frac{1}{3-2}\begin{pmatrix} 3 & -2\\ -1 & 1 \end{pmatrix}\begin{pmatrix} -1 & 2\\ -3 & 4 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 1 & 3 \end{pmatrix}\\ &=\begin{pmatrix} 3 & -2\\ -1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 4\\ 1 & 6 \end{pmatrix}\\ &=\begin{pmatrix} 1 & 0\\ 0 & 2 \end{pmatrix} \end{align}


6) If \mathbf{M}=\begin{pmatrix} m & 2\\ 6 & m-1 \end{pmatrix}, find the set values of m\, for which the inverse of \mathbf{M} exist.

  • \left\{m: m\in \mathbb{R}, m\neq-3,m\neq4\right\}
    • \begin{align} \left|\mathbf{M}\right|&=m\left(m-1\right)-12\\ &=m^2-m-12 \end{align}
    • if \mathbf{M}^{-1} exists, \left|\mathbf{M}\right| \neq 0
    • Consider \left|\mathbf{M}\right|= 0
    • \begin{align} m^2-m-12=0\\ \left(m+3\right)\left(m-4\right)=0\\ m=-3, m=4 \end{align}
    • Since \left|\mathbf{M}\right| \neq 0, \therefore m\neq-3,m\neq4
    • \therefore \left\{m: m\in \mathbb{R}, m\neq-3,m\neq4\right\}


7) If \mathbf{A}=\begin{pmatrix} -1 & 4\\ 2 & 1 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 0 & -1\\ 3 & 2 \end{pmatrix} and \mathbf{C}=\begin{pmatrix} 3 & 4\\ 21 & 19 \end{pmatrix}, find matrix \mathbf{X} such that\mathbf{AXB}=\mathbf{C}.

  • \mathbf{X}=\begin{pmatrix} -2 & 3\\ -1 & 1 \end{pmatrix}
  • \begin{align} \mathbf{AXB}&=\mathbf{C}\\ \mathbf{A}^{-1}\mathbf{AXBB}^{-1}&=\mathbf{A}^{-1}\mathbf{CB}^{-1}\\ \mathbf{IXI}&=\mathbf{A}^{-1}\mathbf{CB}^{-1}\\ \mathbf{X}&= -\frac{1}{9}\begin{pmatrix} 1 & -4\\ -2 & -1 \end{pmatrix}\begin{pmatrix} 3 & 4\\ 21 & 19 \end{pmatrix}\left[\frac{1}{3}\begin{pmatrix} 2 & 1\\ -3 & 0 \end{pmatrix}\right]\\ & =-\frac{1}{27}\begin{pmatrix} 1 & -4\\ -2 & -1 \end{pmatrix}\begin{pmatrix} -6 & 3\\ -15 & 21 \end{pmatrix}\\ & =-\frac{1}{27}\begin{pmatrix} 54 & -81\\ 27 & -27 \end{pmatrix}\\ &= \begin{pmatrix} -2 & 3\\ -1 & 1 \end{pmatrix} \end{align}


8) The matrix A is given by \mathbf{A}=\begin{pmatrix} 1 & 2 & -3\\ 3 & 1 & 1 \\ 0 & 1 & -2 \end{pmatrix}

i) Find the matrix \mathbf{B} such that \mathbf{B}=\mathbf{A}^{2}-10\mathbf{I}, \mathbf{I} is the 3\times3 identity matrix.

  • \mathbf{B}=\begin{pmatrix} -3 & 1 & 5\\ 6 & -2 & -10 \\ 3 & -1 & -5 \end{pmatrix}
  • \begin{align} \mathbf{B}&=\mathbf{A}^{2}-10\mathbf{I}\\ &=\begin{pmatrix} 1 & 2 & -3\\ 3 & 1 & 1 \\ 0 & 1 & -2 \end{pmatrix}\begin{pmatrix} 1 & 2 & -3\\ 3 & 1 & 1 \\ 0 & 1 & -2 \end{pmatrix}-10\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 7 & 1 & 5\\ 6 & 8 & -10 \\ 3 & -1 & 5 \end{pmatrix}-\begin{pmatrix} 10 & 0 & 0\\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{pmatrix}\\ &=\begin{pmatrix} -3 & 1 & 5\\ 6 & -2 & -10 \\ 3 & -1 & -5 \end{pmatrix} \end{align}

ii) Find \left(\mathbf{A}+\mathbf{I}\right)\mathbf{B}, and hence find \left(\mathbf{A}+\mathbf{I}\right)^{21}\mathbf{B}.

  • \left(\mathbf{A}+\mathbf{I}\right)\mathbf{B}=\begin{pmatrix} -3 & 1 & 5\\ 6 & -2 & -10 \\ 3 & -1 & -5 \end{pmatrix},\left(\mathbf{A}+\mathbf{I}\right)^{21}\mathbf{B}=\begin{pmatrix} -3 & 1 & 5\\ 6 & -2 & -10 \\ 3 & -1 & -5 \end{pmatrix}
  • \begin{align} \left(\mathbf{A}+\mathbf{I}\right)\mathbf{B} &=\left[\begin{pmatrix} 1 & 2 & -3\\ 3 & 1 & 1 \\ 0 & 1 & -2 \end{pmatrix}+\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\right]\begin{pmatrix} -3 & 1 & 5\\ 6 & -2 & -10 \\ 3 & -1 & -5 \end{pmatrix}\\ &=\begin{pmatrix} 2 & 2 & -3\\ 3 & 2 & 1 \\ 0 & 1 & -1 \end{pmatrix}\begin{pmatrix} -3 & 1 & 5\\ 6 & -2 & -10 \\ 3 & -1 & -5 \end{pmatrix}\\ &=\begin{pmatrix} -3 & 1 & 5\\ 6 & -2 & -10 \\ 3 & -1 & -5 \end{pmatrix} \end{align}
  • \begin{align} \left(\mathbf{A}+\mathbf{I}\right)\mathbf{B}&=\mathbf{B}\\ \left(\mathbf{A}+\mathbf{I}\right)^2\mathbf{B}&=\left(\mathbf{A}+\mathbf{I}\right)\left[\left(\mathbf{A}+\mathbf{I}\right)\mathbf{B}\right]\\ &=\left(\mathbf{A}+\mathbf{I}\right)\mathbf{B}\\ &=\mathbf{B}\\ \left(\mathbf{A}+\mathbf{I}\right)^4\mathbf{B}&=\left(\mathbf{A}+\mathbf{I}\right)^2\left[\left(\mathbf{A}+\mathbf{I}\right)^2\mathbf{B}\right]\\ &=\left(\mathbf{A}+\mathbf{I}\right)^2\mathbf{B}\\ &=\mathbf{B}\\ \left(\mathbf{A}+\mathbf{I}\right)^8\mathbf{B}&=\left(\mathbf{A}+\mathbf{I}\right)^4\left[\left(\mathbf{A}+\mathbf{I}\right)^4\mathbf{B}\right]\\ &=\left(\mathbf{A}+\mathbf{I}\right)^4\mathbf{B}\\ &=\mathbf{B}\\ \left(\mathbf{A}+\mathbf{I}\right)^{16}\mathbf{B}&=\left(\mathbf{A}+\mathbf{I}\right)^8\left[\left(\mathbf{A}+\mathbf{I}\right)^8\mathbf{B}\right]\\ &=\left(\mathbf{A}+\mathbf{I}\right)^8\mathbf{B}\\ &=\mathbf{B}\\ \left(\mathbf{A}+\mathbf{I}\right)^{21}\mathbf{B}&=\left(\mathbf{A}+\mathbf{I}\right)^{16}\left(\mathbf{A}+\mathbf{I}\right)^4\left[\left(\mathbf{A}+\mathbf{I}\right)\mathbf{B}\right]\\ &=\left(\mathbf{A}+\mathbf{I}\right)^{16}\left[\left(\mathbf{A}+\mathbf{I}\right)^4\mathbf{B}\right]\\ &=\left(\mathbf{A}+\mathbf{I}\right)^{16}\mathbf{B}\\ &=\mathbf{B}\\ &=\begin{pmatrix} -3 & 1 & 5\\ 6 & -2 & -10 \\ 3 & -1 & -5 \end{pmatrix} \end{align}


9) \mathbf{A}, \mathbf{B} and \mathbf{C} are square matrices such that \mathbf{BA}=\mathbf{B}^{-1} and \mathbf{ABC}=\left(\mathbf{AB}\right)^{-1}. Show that \mathbf{A^{-1}}=\mathbf{B}^{2}=\mathbf{C}.

    • \begin{align} \mathbf{BA}&=\mathbf{B}^{-1}\\ \mathbf{BAA}^{-1}&=\mathbf{B}^{-1}\mathbf{A}^{-1}\\ \mathbf{BI}&=\mathbf{B}^{-1}\mathbf{A}^{-1}\\ \mathbf{BB}&=\mathbf{BB}^{-1}\mathbf{A}^{-1}\\ \mathbf{B^2}&=\mathbf{IA}^{-1}\\ \mathbf{A}^{-1}&=\mathbf{B}^{2}\\ \end{align}
    • \begin{align} \mathbf{ABC}&=\left(\mathbf{AB}\right)^{-1}\\ \mathbf{ABC}&=\mathbf{B}^{-1}\mathbf{A}^{-1}\\ \mathbf{ABC}&=\mathbf{B}^{-1}\mathbf{B}^{2}\\ \mathbf{ABC}&=\mathbf{B}\\ \mathbf{A}^{-1}\mathbf{ABC}&=\mathbf{A}^{-1}\mathbf{B}\\ \mathbf{IBC}&=\mathbf{B}^{2}\mathbf{B}\\ \mathbf{B}^{-1}\mathbf{BC}&=\mathbf{B}^{-1}\mathbf{B}^{3}\\ \mathbf{IC}&=\mathbf{B}^{2}\\ \mathbf{C}&=\mathbf{B}^{2}\\ \end{align}
    • \therefore \mathbf{A}^{-1}=\mathbf{B}^{2}=\mathbf{C}

If \mathbf{B}=\begin{pmatrix} 1 & 2 & 0\\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix}, find \mathbf{C} and \mathbf{A}.

  • \mathbf{C}=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 2 & 2 & 1 \end{pmatrix}, \mathbf{A}= \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix}
    • \begin{align} \mathbf{C}&=\mathbf{B}^{2}\\ &=\begin{pmatrix} 1 & 2 & 0\\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 & 0\\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 2 & 2 & 1 \end{pmatrix} \end{align}
    • \begin{align} \mathbf{A}^{-1}&=\mathbf{C}\\ \mathbf{AA}^{-1}&=\mathbf{AC}\\ \mathbf{I}&=\mathbf{AC}\\ \mathbf{IC}^{-1}&=\mathbf{ACC}^{-1}\\ \mathbf{C}^{-1}&=\mathbf{AI}\\ \mathbf{A}&=\mathbf{C}^{-1}\\ \end{align}
    • \begin{align} \left|\mathbf{C}\right| & =1\begin{vmatrix} 1 & 0 \\ 2 & 1 \\ \end{vmatrix}-0\begin{vmatrix} 0 & 0 \\ 2 & 1 \\ \end{vmatrix}+0\begin{vmatrix} 0 & 1 \\ 2 & 2 \\ \end{vmatrix}\\ & = 1 \end{align}
    • \mathbf{C}^T= \begin{pmatrix} 1 & 0 & 2\\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}
    • adj \mathbf{C}=\begin{pmatrix} \begin{vmatrix} 1 & 2 \\ 0 & 1 \\ \end{vmatrix} & -\begin{vmatrix} 0 & 2 \\ 0 & 1 \\ \end{vmatrix} & \begin{vmatrix} 0 & 1 \\ 0 & 0 \\ \end{vmatrix} \\ -\begin{vmatrix} 0 & 2 \\ 0 & 1 \\ \end{vmatrix} & \begin{vmatrix} 1 & 2 \\ 0 & 1 \\ \end{vmatrix} & -\begin{vmatrix} 1 & 0 \\ 0 & 0 \\ \end{vmatrix} \\ \begin{vmatrix} 0 & 2 \\ 1 & 2 \\ \end{vmatrix} & -\begin{vmatrix} 1& 2 \\ 0 & 2 \\ \end{vmatrix} & \begin{vmatrix} 1 & 0 \\ 0 & 1 \\ \end{vmatrix} \end{pmatrix}
    • =\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix}
    • \begin{align} \mathbf{A}&=\mathbf{C}^{-1} \\ & = \frac{1}{\left|\mathbf{C}\right|}adj \mathbf{C}\\ & = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix} \end{align}


10) Given \mathbf{M}=\begin{pmatrix} 2 & 3 & 1\\ -1 & 0 & 4 \\ 1 & -1 & 1 \end{pmatrix}. Show that \mathbf{M}^{3}-3\mathbf{M}^{2}+8\mathbf{M}-24\mathbf{I}_{3}=\mathbf{0}. Deduce \mathbf{M}^{-1}.

  • \mathbf{M}^{-1}=\begin{pmatrix} \frac{1}{6} & -\frac{1}{6} & \frac{1}{2}\\ \frac{5}{24} & \frac{1}{24} & -\frac{3}{8} \\ \frac{1}{24} & \frac{5}{24} & \frac{1}{8} \end{pmatrix}
    • \begin{align} \mathbf{M}^{2}&=\begin{pmatrix} 2 & 3 & 1\\ -1 & 0 & 4 \\ 1 & -1 & 1 \end{pmatrix}\begin{pmatrix} 2 & 3 & 1\\ -1 & 0 & 4 \\ 1 & -1 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 2 & 5 & 15\\ 2 & -7 & 3 \\ 4 & 2 & -2 \end{pmatrix}\\ \mathbf{M}^{3}&=\mathbf{M}^{2}\mathbf{M}\\ &=\begin{pmatrix} 2 & 5 & 15\\ 2 & -7 & 3 \\ 4 & 2 & -2 \end{pmatrix}\begin{pmatrix} 2 & 3 & 1\\ -1 & 0 & 4 \\ 1 & -1 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 14 & -9 & 37\\ 14 & 3 & -23 \\ 4 & 14 & 10 \end{pmatrix} \end{align}
    • \begin{align} &\mathbf{M}^{3}-3\mathbf{M}^{2}+8\mathbf{M}-24\mathbf{I}_{3}\\ &=\begin{pmatrix} 14 & -9 & 37\\ 14 & 3 & -23 \\ 4 & 14 & 10 \end{pmatrix}-3\begin{pmatrix} 2 & 5 & 15\\ 2 & -7 & 3 \\ 4 & 2 & -2 \end{pmatrix}+8\begin{pmatrix} 2 & 3 & 1\\ -1 & 0 & 4 \\ 1 & -1 & 1 \end{pmatrix}-24\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 14 & -9 & 37\\ 14 & 3 & -23 \\ 4 & 14 & 10 \end{pmatrix}+\begin{pmatrix} -6 & -15 & -45\\ -6 & 21 & -9 \\ -12 & -6 & 6 \end{pmatrix}+\begin{pmatrix} 16 & 24 & 8\\ -8 & 0 & 32 \\ 8 & -8 & 8 \end{pmatrix}+\begin{pmatrix} -24 & 0 & 0\\ 0 & -24 & 0 \\ 0 & 0 & 24 \end{pmatrix}\\ &=\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\\ &=\mathbf{0}\quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \end{align}
    • \begin{align} \mathbf{M}^{3}-3\mathbf{M}^{2}+8\mathbf{M}-24\mathbf{I}_{3}&=\mathbf{0}\\ \mathbf{M}^{3}-3\mathbf{M}^{2}+8\mathbf{M}&=24\mathbf{I}_{3}\\ \mathbf{M}^{-1}\left(\mathbf{M}^{3}-3\mathbf{M}^{2}+8\mathbf{M}\right)&=\mathbf{M}^{-1}\left(24\mathbf{I}_{3}\right)\\ \mathbf{M}^{2}-3\mathbf{M}+8\mathbf{I}&=24\mathbf{M}^{-1}\\ \mathbf{M}^{-1}&=\frac{1}{24}\left(\mathbf{M}^{2}-3\mathbf{M}+8\mathbf{I}\right)\\ &=\frac{1}{24}\left[\begin{pmatrix} 2 & 5 & 15\\ 2 & -7 & 3 \\ 4 & 2 & -2 \end{pmatrix}-3\begin{pmatrix} 2 & 3 & 1\\ -1 & 0 & 4 \\ 1 & -1 & 1 \end{pmatrix}+8\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\right]\\ &=\frac{1}{24}\begin{pmatrix} 4 & -4 & 12\\ 5 & 1 & -9 \\ 1 & 5 & 3 \end{pmatrix}\\ &=\begin{pmatrix} \frac{1}{6} & -\frac{1}{6} & \frac{1}{2}\\ \frac{5}{24} & \frac{1}{24} & -\frac{3}{8} \\ \frac{1}{24} & \frac{5}{24} & \frac{1}{8} \end{pmatrix}\\ \end{align}


11) If \mathbf{A}=\begin{pmatrix} 1 & -1 & 1\\ 2 & 0 & 1 \\ 0 & 2 & -1 \end{pmatrix} Show that \mathbf{A}^{3}=\mathbf{A}. Hence, find \mathbf{A}^{40}.

  • \mathbf{A}=\begin{pmatrix} -1 & 1 & -1\\ 2 & 0 & 1 \\ 4 & -2 & 3 \end{pmatrix}
    • \begin{align} \mathbf{A}^{3}&=\begin{pmatrix} 1 & -1 & 1\\ 2 & 0 & 1 \\ 0 & 2 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 & 1\\ 2 & 0 & 1 \\ 0 & 2 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 & 1\\ 2 & 0 & 1 \\ 0 & 2 & -1 \end{pmatrix}\\ &=\begin{pmatrix} 1 & -1 & 1\\ 2 & 0 & 1 \\ 0 & 2 & -1 \end{pmatrix}\begin{pmatrix} -1 & 1 & -1\\ 2 & 0 & 1 \\ 4 & -2 & 3 \end{pmatrix}\\ &=\begin{pmatrix} 1 & -1 & 1\\ 2 & 0 & 1 \\ 0 & 2 & -1 \end{pmatrix}\\ &=\mathbf{A}\quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \end{align}
    • \begin{align} \mathbf{A}^{3}&=\mathbf{A}\\ \mathbf{A}^{9}&=\mathbf{A}^{3}\mathbf{A}^{3}\mathbf{A}^{3}=\mathbf{AAA}=\mathbf{A}^{3}=\mathbf{A}\\ \mathbf{A}^{27}&=\mathbf{A}^{9}\mathbf{A}^{9}\mathbf{A}^{9}=\mathbf{AAA}=\mathbf{A}^{3}=\mathbf{A}\\ \mathbf{A}^{40}&=\mathbf{A}^{27}\mathbf{A}^{9}\mathbf{A}^{3}\mathbf{A}\\ &=\mathbf{AAAA}\\ &=\mathbf{A}^{3}\mathbf{A}\\ &=\mathbf{AA}\\ &=\begin{pmatrix} -1 & 1 & -1\\ 2 & 0 & 1 \\ 4 & -2 & 3 \end{pmatrix} \end{align}


12) Show that the matrix \mathbf{A}=\begin{pmatrix} 0 & -3\\ -1 & 1 \end{pmatrix} satisfies the equation \mathbf{A}^{2}-\mathbf{A}-3\mathbf{I}=\mathbf{0} . Hence, without evaluating \mathbf{A}^{3} or \mathbf{A}^{4}, show that \mathbf{A}^{4}=7\mathbf{A}+12\mathbf{I}.

    • \begin{align} \mathbf{A}^{2}-\mathbf{A}-3\mathbf{I} &=\begin{pmatrix} 0 & -3\\ -1 & 1 \end{pmatrix}\begin{pmatrix} 0 & -3\\ -1 & 1 \end{pmatrix}-\begin{pmatrix} 0 & -3\\ -1 & 1 \end{pmatrix}-3\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} 3 & -3\\ -1 & 4 \end{pmatrix}-\begin{pmatrix} 0 & -3\\ -1 & 1 \end{pmatrix}-\begin{pmatrix} 3 & 0\\ 0 & 3 \end{pmatrix}\\ &=\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}\\ &=\mathbf{0} \quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \end{align}
    • \begin{align} \mathbf{A}^{2}-\mathbf{A}-3\mathbf{I}=\mathbf{0}\\ \mathbf{A}^{2}=\mathbf{A}+3\mathbf{I} \end{align}
    • \begin{align} \mathbf{A}^{4}&=\mathbf{A}^{2}\mathbf{A}^{2}\\ &=\left(\mathbf{A}+3\mathbf{I}\right)\left(\mathbf{A}+3\mathbf{I}\right)\\ &=\mathbf{A}^{2}+3\mathbf{AI}+3\mathbf{IA}+\mathbf{I}^{2}\\ &=\mathbf{A}^{2}+6\mathbf{A}+\mathbf{I}\\ &=\left(\mathbf{A}+3\mathbf{I}\right)+6\mathbf{A}+\mathbf{I}\\ &=7\mathbf{A}+4\mathbf{I}\quad \frac{ \qquad }{ \qquad} \quad \mathrm{proved} \end{align}


13) \mathbf{A}=\begin{pmatrix} 2 & 3\\ -1 & 1 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} x & y\\ 1 & -1 \end{pmatrix}. Find the diagonal matrix \mathbf{D} such that \mathbf{ADA}^{-1}=\mathbf{B}.

  • \mathbf{D}=\begin{pmatrix} -3 & 0\\ 0 & 2 \end{pmatrix}
    • Let \mathbf{D}=\begin{pmatrix} a & 0\\ 0 & b \end{pmatrix}
    • \begin{align} \mathbf{ADA}^{-1}&=\mathbf{B}\\ \mathbf{ADA}^{-1}\mathbf{A}&=\mathbf{BA}\\ \mathbf{AD}&=\mathbf{BA}\\ \begin{pmatrix} 2 & 3\\ -1 & 1 \end{pmatrix}\begin{pmatrix} a & 0\\ 0 & b \end{pmatrix}&=\begin{pmatrix} x & y\\ 1 & -1 \end{pmatrix}\begin{pmatrix} 2 & 3\\ -1 & 1 \end{pmatrix}\\ \end{align}
    • \begin{array}{l} \therefore -a=3 \qquad b=2\\ a=-3\\ \therefore \mathbf{D}=\begin{pmatrix} -3 & 0\\ 0 & 2 \end{pmatrix} \end{array}


14) The matrices \mathbf{A} and \mathbf{B} are given by \mathbf{A}=\begin{pmatrix} -1 & 2 & 1\\ -3 & 1 & 4 \\ 0 & 1 & 2 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} -35 & 19 & 18\\ -27 & -13 & 45 \\ -3 & 12 & 5 \end{pmatrix}. Find the matrix \mathbf{A}^{2}\mathbf{B} and deduce the inverse of \mathbf{A} .

  • \mathbf{A}^{2}\mathbf{B}=\begin{pmatrix} 121 & 0 & 0\\ 0 & 121 & 0 \\ 0 & 0 & 121 \end{pmatrix}, \mathbf{A}^{-1}=\begin{pmatrix} -\frac{2}{11} & -\frac{3}{11} & \frac{7}{11}\\ \frac{6}{11} & -\frac{2}{11} & \frac{1}{11} \\ -\frac{3}{11} & \frac{1}{11} & \frac{5}{11} \end{pmatrix}
    • \begin{align} \mathbf{A}^{2}&=\begin{pmatrix} -1 & 2 & 1\\ -3 & 1 & 4 \\ 0 & 1 & 2 \end{pmatrix}\begin{pmatrix} -1 & 2 & 1\\ -3 & 1 & 4 \\ 0 & 1 & 2 \end{pmatrix}\\ &=\begin{pmatrix} -5 & 1 & 9\\ 0 & -1 & 9 \\ -3 & 3 & 8 \end{pmatrix}\\ \mathbf{A}^{2}\mathbf{B}&=\begin{pmatrix} -5 & 1 & 9\\ 0 & -1 & 9 \\ -3 & 3 & 8 \end{pmatrix}\begin{pmatrix} -35 & 19 & 18\\ -27 & -13 & 45 \\ -3 & 12 & 5 \end{pmatrix}\\ &=\begin{pmatrix} 121 & 0 & 0\\ 0 & 121 & 0 \\ 0 & 0 & 121 \end{pmatrix} \end{align}
    • \begin{align} \mbox{Since}\quad\mathbf{A}^{2}\mathbf{B}&=121\mathbf{I}\\ \mathbf{A}^{-1}\mathbf{A}^{2}\mathbf{B}&=\mathbf{A}^{-1}\left(121\mathbf{I}\right)\\ \mathbf{AB}&=121\mathbf{A}^{-1}\\ \mathbf{A}^{-1}&=\frac{1}{121}\mathbf{AB}\\ &=\frac{1}{121}\begin{pmatrix} -1 & 2 & 1\\ -3 & 1 & 4 \\ 0 & 1 & 2 \end{pmatrix}\begin{pmatrix} -35 & 19 & 18\\ -27 & -13 & 45 \\ -3 & 12 & 5 \end{pmatrix}\\ &=\frac{1}{121}\begin{pmatrix} -22 & -33 & 77\\ 66 & -22 & 11 \\ -33 & 11 & 55 \end{pmatrix}\\ &=\begin{pmatrix} -\frac{2}{11} & -\frac{3}{11} & \frac{7}{11}\\ \frac{6}{11} & -\frac{2}{11} & \frac{1}{11} \\ -\frac{3}{11} & \frac{1}{11} & \frac{5}{11} \end{pmatrix} \end{align}

Hence, solve the system of linear equations

\begin{array}{l} x-2y-z=-8\\ 3x-y-4z=-15\\ y+2z=4 \end{array}

  • x=-3, y=2, z=1\,
    • \begin{array}{llll} -x & +2y & +z &=8\\ -3x& +y & +4z &=15\\ & y & +2z &=4 \end{array}
    • \begin{align} \therefore \begin{pmatrix} -1 & 2 & 1\\ -3 & 1 & 4 \\ 0 & 1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \\ x \end{pmatrix}&=\begin{pmatrix} 8 \\ 15 \\ 4 \end{pmatrix}\\ \mathbf{A}\begin{pmatrix} x \\ y \\ x \end{pmatrix}&=\begin{pmatrix} 8 \\ 15 \\ 4 \end{pmatrix}\\ \mathbf{A}^{-1}\mathbf{A}\begin{pmatrix} x \\ y \\ x \end{pmatrix}&=\mathbf{A}^{-1}\begin{pmatrix} 8 \\ 15 \\ 4 \end{pmatrix}\\ \mathbf{I}\begin{pmatrix} x \\ y \\ x \end{pmatrix}&=\frac{1}{121}\begin{pmatrix} -22 & -33 & 77\\ 66 & -22 & 11 \\ -33 & 11 & 55 \end{pmatrix}\begin{pmatrix} 8 \\ 15 \\ 4 \end{pmatrix}\\\\ \begin{pmatrix} x \\ y \\ x \end{pmatrix}&=\frac{1}{11}\begin{pmatrix} -2 & -3 & 7\\ 6 & -2 & 1 \\ -3 & 1 & 5 \end{pmatrix}\begin{pmatrix} 8 \\ 15 \\ 4 \end{pmatrix}\\ &=\frac{1}{11}\begin{pmatrix} -33 \\ 22 \\ 11 \end{pmatrix}\\ &=\begin{pmatrix} -3 \\ 2 \\ 1 \end{pmatrix}\\ \end{align}
    • \therefore x=-3, y=2, z=1


15) If \mathbf{P}=\begin{pmatrix} 5 & 2 & 3\\ 1 & -4 & 3 \\ 3 & 1 & 2 \end{pmatrix} and \mathbf{Q}=\begin{pmatrix} a & 1 & -18\\ b & -1 & 12 \\ -13 & -1 & c \end{pmatrix}, and \mathbf{PQ}=2\mathbf{I}, determine the value of a\,, b\, and c\,. Hence, find \mathbf{P}^{-1}.

  • a=11, b=-7, c=22, \mathbf{P}^{-1}=\begin{pmatrix} \frac{11}{2} & \frac{1}{2} & -9\\ -\frac{7}{2} & -\frac{1}{2} & 6 \\ -\frac{13}{2} & -\frac{1}{2} & 11 \end{pmatrix}
    • \mathbf{PQ}=2\mathbf{I}
    • \begin{pmatrix} 5 & 2 & 3\\ 1 & -4 & 3 \\ 3 & 1 & 2 \end{pmatrix}\begin{pmatrix} a & 1 & -18\\ b & -1 & 12 \\ -13 & -1 & c \end{pmatrix}=\begin{pmatrix} 2 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}
    • \begin{array}{lll} \therefore a-4b-39=0\qquad & 3a+b-26=0\qquad\qquad & -90+24+3c=0 \\ a=4b+39 & 3\left(4b+39\right)+b=26 & c=22 \\ & 13b=-91 \\ & b=-7, a= 11 \end{array}
    • \therefore a=11, b=-7, c=22
    • \begin{align} \mathbf{PQ}&=2\mathbf{I}\\ \mathbf{P}^{-1}\mathbf{PQ}&=\mathbf{P}^{-1}\left(2\mathbf{I}\right)\\ \mathbf{IQ}&=2\mathbf{P}^{-1}\\ \mathbf{P}^{-1}&=\frac{1}{2}\mathbf{Q}\\ &=\frac{1}{2}\begin{pmatrix} 11 & 1 & -18\\ -7 & -1 & 12 \\ -13 & -1 & 22 \end{pmatrix}\\ &=\begin{pmatrix} \frac{11}{2} & \frac{1}{2} & -9\\ -\frac{7}{2} & -\frac{1}{2} & 6 \\ -\frac{13}{2} & -\frac{1}{2} & 11 \end{pmatrix} \end{align}

Two groups of workers have their drinks at a stall. The first group comprising ten workers have five cups of tea, two cups of coffee an three glasses of fruit juice at a total cost of RM 11.80. The second group of six workers have three cups of tea, a cup of coffee and two glasses of fruit juice at a total cost of RM 7.10. The cost of a cup of tea and three glasses of fruit juice is the same as the cost of four cups of coffee. The cost of a cup of tea, a cup of coffee and a glass of fruit juice are RM x\,, RMy\, and RMz\, respectively, obtain a matrix equation to represent the above information. Hence, determine the cost of each drink.

  • \begin{pmatrix} 5 & 2 & 3\\ 1 & -4 & 3 \\ 3 & 1 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \\ z \end{pmatrix}=\begin{pmatrix} 11.8\\ 0 \\ 7.1 \end{pmatrix} Price of tea = RM 1.00, Price of coffee = RM 1.30, Price of fruit juice = RM 1.40.
    • \begin{align} 5x+2y+3z &=11.80\\ 3x+y+2z &=7.10\\ x+3z &=4y \end{align}\begin{array}{lllll} & 5x & +2y & +3z & =11.8\\ \therefore & x & -4y & +3z & = 0 \\ & 3x & +y & +2z & =7.1 \end{array}
    • \therefore \begin{pmatrix} 5 & 2 & 3\\ 1 & -4 & 3 \\ 3 & 1 & 2 \end{pmatrix}\begin{pmatrix} x\\ y \\ z \end{pmatrix}=\begin{pmatrix} 11.8\\ 0 \\ 7.1 \end{pmatrix}
    • \begin{align} \therefore \mathbf{P}\begin{pmatrix} x\\ y \\ z \end{pmatrix}&=\begin{pmatrix} 11.8\\ 0 \\ 7.1 \end{pmatrix}\\ \mathbf{P}^{-1}\mathbf{P}\begin{pmatrix} x\\ y \\ z \end{pmatrix}&= \mathbf{P}^{-1}\begin{pmatrix} 11.8\\ 0 \\ 7.1 \end{pmatrix}\\ \mathbf{I}\begin{pmatrix} x\\ y \\ z \end{pmatrix}&=\frac{1}{2}\begin{pmatrix} 11 & 1 & -18\\ -7 & -1 & 12 \\ -13 & -1 & 22 \end{pmatrix}\begin{pmatrix} 11.8\\ 0 \\ 7.1 \end{pmatrix}\\ \begin{pmatrix} x\\ y \\ z \end{pmatrix}&=\frac{1}{2}\begin{pmatrix} 2 \\ 2.6 \\ 2.8 \end{pmatrix}\\ &=\begin{pmatrix} 1 \\ 1.3 \\ 1.4 \end{pmatrix} \end{align}
    • \therefore x=1, y=1.3, z=1.4\,
    • \therefore Price of tea = RM 1.00, Price of coffee = RM 1.30, Price of fruit juice = RM 1.40.


16) \mathbf{M}=\begin{pmatrix} -10 & 4 & 9\\ 15 & -4 & -14 \\ -5 & 1 & 6 \end{pmatrix} and \mathbf{N}=\begin{pmatrix} 2 & 3 & 4\\ 4 & 3 & 1 \\ 1 & 2 & 4 \end{pmatrix}. Find \mathbf{MN} and deduce \mathbf{N}^{-1}.

  • \mathbf{MN}\begin{pmatrix} 5 & 0 & 0\\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}, \mathbf{N}^{-1}= \begin{pmatrix} -2 & \frac{4}{5} & \frac{9}{5}\\ 3 & -\frac{4}{5} & -\frac{14}{5} \\ -1 & \frac{1}{5} & \frac{6}{5} \end{pmatrix}
    • \begin{align} \mathbf{MN}&=\begin{pmatrix} -10 & 4 & 9\\ 15 & -4 & -14 \\ -5 & 1 & 6 \end{pmatrix}\begin{pmatrix} 2 & 3 & 4\\ 4 & 3 & 1 \\ 1 & 2 & 4 \end{pmatrix}\\ &=\begin{pmatrix} 5 & 0 & 0\\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} \end{align}
    • \begin{align} \mbox{Since}\quad \mathbf{MN}&=\mathbf{5I}\\ \mathbf{MN}\mathbf{N}^{-1}&=\left(\mathbf{5I}\right)\mathbf{N}^{-1}\\ \mathbf{MI}&=5\mathbf{N}^{-1}\\ \mathbf{N}^{-1}&=\frac{1}{5}\mathbf{M}\\ &=\frac{1}{5}\begin{pmatrix} -10 & 4 & 9\\ 15 & -4 & -14 \\ -5 & 1 & 6 \end{pmatrix}\\ &=\begin{pmatrix} -2 & \frac{4}{5} & \frac{9}{5}\\ 3 & -\frac{4}{5} & -\frac{14}{5} \\ -1 & \frac{1}{5} & \frac{6}{5} \end{pmatrix} \end{align}

Products X, Y and Z are assembled from three components A, B and C according to different proportions. Each product X consists of two components of A, four components of B and one component of C; each product of Y consists of three components of A, three components of B and two component of C; and each product of Z consists of four components of A, one components of B and four component of C. A total of 750 components of A, 1 000 components of B and 500 components of C are used. With x\,, y\, and z\, representing the numbers of products of X, Y and Z assembled, obtain a matrix equation representing the information given. Hence, find the numbers of products of X, Y and Z assembled.

  • \begin{pmatrix}2 & 3 & 4\\ 4 & 3 & 1 \\ 1 & 2 & 4\end{pmatrix}\begin{pmatrix}x\\ y \\ z\end{pmatrix}=\begin{pmatrix} 750\\ 1000 \\ 500\end{pmatrix} Number of X produced = 200, Number of Y produced = 50, Number of Z produced = 50
    • \begin{array}{|c|c|c|c|} & A & B & C \\ \hline X & 2 & 4 & 1 \\ Y & 3 & 3 & 2 \\ Z & 4 & 1 & 4 \end{array}
    • Total A=750, Total B= 1000, Total C=500
    • \begin{align} \therefore 2x+3y+4z&=750\\ 4x+3y+z&=1000\\ x+2y=4z&=500 \end{align}
    • \therefore \begin{pmatrix}2 & 3 & 4\\ 4 & 3 & 1 \\ 1 & 2 & 4\end{pmatrix}\begin{pmatrix}x\\ y \\ z\end{pmatrix}=\begin{pmatrix} 750\\ 1000 \\ 500\end{pmatrix}
    • \begin{align} \therefore \mathbf{N}\begin{pmatrix}x\\ y \\ z\end{pmatrix}&=\begin{pmatrix} 750\\ 1000 \\ 500\end{pmatrix}\\ \mathbf{N}^{-1}\mathbf{N}\begin{pmatrix}x\\ y \\ z\end{pmatrix}&=\mathbf{N}^{-1}\begin{pmatrix} 750\\ 1000 \\ 500\end{pmatrix}\\ \mathbf{I}\begin{pmatrix}x\\ y \\ z\end{pmatrix}&=\frac{1}{5}\begin{pmatrix} -10 & 4 & 9\\ 15 & -4 & -14 \\ -5 & 1 & 6 \end{pmatrix}\begin{pmatrix} 750\\ 1000 \\ 500\end{pmatrix}\\ \begin{pmatrix}x\\ y \\ z\end{pmatrix}&=\frac{1}{5}\begin{pmatrix} 1000\\ 250 \\ 250\end{pmatrix}\\ &=\begin{pmatrix} 200\\ 50 \\ 50\end{pmatrix}\\ \end{align}
    • \therefore x=200, y=50, z=50
    • \therefore Number of X produced = 200, Number of Y produced = 50, Number of Z produced = 50


17) The matrices \mathbf{A} and \mathbf{B} are given as \mathbf{A}=\begin{pmatrix} 6 & 4 & 3\\ 3 & 3 & 7 \\ 1 & 3 & 0 \end{pmatrix}, \mathbf{B}=\begin{pmatrix} 21 & -9 & -19\\ -7 & 3 & 33 \\ -6 & 14 & -6 \end{pmatrix}. Find \mathbf{AB} and \mathbf{A}^{-1}.

A company produces three types of instant coffee under the brands Jerai, Ledang and Mulu which contain the percentages (according to mass) of coffee powder, sugar and powdered milk as shown in the following table. Brand of instant coffee Composition by percentage

Brand of instant coffee Composition by percentage
Coffee powder Sugar Powdered milk
Jerai 60 30 10
Ledang 40 30 30
Mulu 30 70 0

The company mixes the coffee Jerai, Ledang and Mulu to yield a new instant coffee under the brand Jelemu in 50 g packet containing 44% coffee powder, 38% sugar and 18% powdered milk. If packet of the Jelemu coffee contains x\,g of the coffee Jerai, y\,g of the coffee Ledang and z\,g of the coffee Mulu, show that \mathbf{A}\begin{pmatrix} x \\ y\\ z \end{pmatrix}=\begin{pmatrix} 220 \\ 190\\ 90 \end{pmatrix}. Hence, determine the mass of the coffee Jerai, Ledang and Mulu in each 50 g of packet of coffee Jelemu.


18)a) \mathbf{A}=\begin{pmatrix} 1 & 1 & 2\\ 0 & 2 & 2 \\ -1 & 1 & 3 \end{pmatrix}. Find \mathbf{A^{2}}-6\mathbf{A}+11\mathbf{I}. Show that \mathbf{A}\left(\mathbf{A^{2}}-6\mathbf{A}+11\mathbf{I}\right)=6\mathbf{I} and deduce \mathbf{A^{-1}}.

b) A factory produces three types of nuts namely kacang kuda, kacang botak and kacang parang. The profit from 1 kg of kacang kuda, 1 kg of kacang botak and 2 kg of kacang parang is RM 9. The profit from 1 kg of kacang botak and 1 kg of kacang parang is RM 3. The profit from 1 kg of kacang botak and 3 kg of kacang parang is equal to the profit from 1 kg of kacang kuda. If x\,, y\, and z\, represent the profit from 1 kg of kacang kuda, 1 kg of kacang botak and 1 kg of kacang parang respectively, write a matrix equation to represent the above information. Hence, determine the profit from 1 kg of kacang kuda, 1 kg of kacang botak and 1 kg of kacang parang.


19)\mathbf{A}=\begin{pmatrix} 5 & -5 & 0\\ -5 & 0 & 5 \\ 0 & 10 & -5 \end{pmatrix}, \mathbf{A}=\begin{pmatrix} 2 & 1 & 1\\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{pmatrix}. Find \mathbf{AB} and \mathbf{B^{-1}}.

The following table shows the wholesale price in RM per kg, of three commodities, namely red chilies, long beans and cucumber, in the three towns of Kuala Terengganu, Kuala Lumpur and Johor Bahru.

Town Commodities
Red chilies Long Beans Cucumber
Kuala Terengganu 4 2 2
Kuala Lumpur 2 2 2
Johor Bahru 4 4 2

A company with branch offices located in each of the three towns has been awarded a contact worth RM 5000 in each town to supply x\, kg of red chilies, y\, kg of long bean and z\, kg of cucumbers in each town to the local retailer. The vegetables are obtain form the town itself. The profit earn by the branches in Kuala Terengganu, Kuala Lumpur and Johor Bahru are RM 2000, RM 3000 and RM 1500 respectively. Write a matrix equation in x\,, y\, and z\, to represent the above information. Hence, determine the quantities of red chilies, long beans and cucumbers supplied in each town.

Answers(With guidance)

To Be Done

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