# 11 Matrices and Determinants Case Study 11.1 Matrices

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11 Matrices and Determinants Case Study 11.1 Matrices
11.3 Inverses of Square Matrices Chapter Summary

Case Study We received an order to produce three kinds of products. Teams X and Y will work together to finish this job.  Team X produce 500 pieces of product A, 200 pieces of product B and 350 pieces of product C  Team Y produce 200 pieces of product A, 400 pieces of product B and 450 pieces of product C That’s tedious! Great! Please calculate the total amount of materials needed by each team. Contents of  product A: 1.5 kg of copper, 0.2 kg of steel  product B: 0.6 kg of copper, 1.4 kg of steel  product C: 0.8 kg of copper, 1 kg of steel How to organize and calculate the total amount of copper and steel needed by each team? (1) Amount of copper needed by Team X ? (2) Amount of steel needed by Team X ? (3) Amount of copper needed by Team Y ? (4) Amount of steel needed by Team Y ?

Case Study Organization We can arrange the data in tabular form:
1st column 2nd column Product A Product B Product C Team X 500 200 350 Team Y 400 450 Copper (in kg) Steel (in kg) Product A 1.5 0.2 Product B 0.6 1.4 Product C 0.8 1 1st row Calculation (1) Amount of copper needed by Team X 1st column 2nd column (500  1.5  200  0.6  350  0.8) kg 730 kg 900 kg 1050 kg Copper Steel Team X 1150 kg Team Y  1150 kg 1st row (2) Amount of steel needed by Team X ? (3) Amount of copper needed by Team Y ? (4) Amount of steel needed by Team Y ?

11.1 Matrices A. Introduction
A rectangular array of numbers arranged in m rows and n columns is called a m  n matrix. An m  n matrix is represented in the form 3rd column nth column 2nd row or mth row A matrix with m rows and n columns is said to be a matrix of order m  n. The number aij in the ith row and the jth column of a matrix is called an element or entry. For example, in the 2  3 matrix , a12  4 and a23  7.

aij  bij for all i = 1, 2, 3, ... , m and j = 1, 2, 3, ... , n.
11.1 Matrices A. Introduction For an a m  n matrix, if m  1, it has only 1 row and is called a row matrix; if n  1, it has only 1 column and is called a column matrix. We should specify the row number first, then the column number. ( ) is a row matrix of order 1  3. is a column matrix of order 3  1. Two matrices are said to be equal if they satisfy the following definition: Equality of Matrices Two matrices A  (aij)m  n and B  (bij)m  n are equal if and only if they have the same order and their corresponding elements are equal, i.e., aij  bij for all i = 1, 2, 3, ... , m and j = 1, 2, 3, ... , n.

Example 11.1T 11.1 Matrices Solution: A. Introduction
If , find the values of w, x, y and z. Solution: From the definition, w  2, x  4, y  6 and z  10.

11.1 Matrices B. Special Types of Matrices Zero Matrix
A zero matrix, or a null matrix, is a matrix that all its elements are zero. For example, is a 2  3 matrix. Square Matrix A square matrix is a matrix with the same numbers of rows and columns. For example, is a square matrix of order 2. Notes: The order of a square matrix is denoted by its number of rows n.

11.1 Matrices B. Special Types of Matrices Identity Matrix
An identity matrix of order n, which is denoted by I, is an n  n square matrix with For example, is the identity matrix of order 3. An identity matrix is also called a unit matrix.

11.1 Matrices C. Operations of Matrices
Some rules on the operations of matrices: Addition of Matrices Suppose A  (aij)m  n and B  (bij)m  n are two matrices of order m  n. Then the sum of A and B is also an m  n matrix C  (cij)m  n with cij  aij  bij, for all i  1, 2, 3, ... , m and j  1, 2, 3, ... , n. For example, if and , then Note that the addition of matrices is defined only when the two matrices are of the same order.

11.1 Matrices C. Operations of Matrices Negative of Matrices
Let A  (aij)m  n be an m  n matrix. The negative of A, denoted by A, is the matrix whose elements are the negative of the corresponding elements of A, i.e., A  (aij)m  n, for all i  1, 2, 3, ... , m and j  1, 2, 3, ... , n. For example, if , then Subtraction of Matrices Suppose A  (aij)m  n and B  (bij)m  n are two matrices of order m  n. The difference of A and B is defined as A  B  A  (B).

Example 11.2T 11.1 Matrices Solution: C. Operations of Matrices
Suppose and Find the matrix Z such that Y  Z  X. Solution: ∵ Y  Z  X ∴ Z  X  Y When summing up matrices, we sum up each pair of the corresponding elements independently.

11.1 Matrices C. Operations of Matrices Properties of Matrix Addition
Let A  (aij)m  n, B  (bij)m  n and C  (cij)m  n be m  n matrices and 0 be the m  n zero matrix. Then we have: (a) A  B  B  A (Commutative Law) (b) (A  B)  C  A  (B  C) (Associative Law) (c) A  0  0  A  A (d) A  (A)  (A)  A  0 Proofs of (a) and (b): By the definition of addition of matrices, A  B  (aij)m  n  (bij)m  n (A  B)  C  (aij  bij)m  n  (cij)m  n  (aij  bij)m  n  [(aij  bij)  cij]m  n  (bij  aij)m  n  [aij  (bij  cij)]m  n  (bij)m  n  (aij)m  n  (aij)m  n  (bij  cij)m  n  B  A  A  (B  C)

11.1 Matrices C. Operations of Matrices
Scalar Multiplication of Matrices The scalar multiplication of an m  n matrix A  (aij)m  n and a real number k, which is denoted by kA, is an m  n matrix whose elements are the corresponding elements of A multiplied by k, i.e., kA  (kaij)m  n, for all i  1, 2, 3, ... , m and j  1, 2, 3, ... , n. For example, Properties of Scalar Multiplication Let A and B be two m  n matrices and h, k be two real numbers. We have (a) k(A  B)  kA  kB; (Distributive Law) (b) (h  k)A  hA  kA; (c) hkA  h(kA)  k(hA).

Example 11.3T 11.1 Matrices Solution: C. Operations of Matrices
Suppose and Evaluate 2X  3Y and 4Y  2X. Solution: 2X  3Y 4Y  2X

cij  ai1b1j  ai2b2j  ...  ainbnj  ,
11.1 Matrices C. Operations of Matrices Multiplication of Matrices Let A  (aij)m  n be an m  n matrix and B  (bij)n  p be an n  p matrix. The product AB is an m  p matrix C  (cij)m  p where cij  ai1b1j  ai2b2j  ...  ainbnj  , for all i  1, 2, 3, …, m and j  1, 2, 3, …, p. To understand the process of the multiplication of matrices, students may also refer to the Case Study at the beginning of this chapter. Notes: When calculating the product AB, the matrix A should be placed on the left while B is placed on the right. Multiplication of matrices is non-commutative, i.e., for two matrices A and B, AB  BA in general.

11.1 Matrices C. Operations of Matrices Suppose and .
Also consider the product BA. ∵ B is a 3  2 matrix and A is a 2  3 matrix. ∴ BA is a 3  3 matrix. ∴ AB  BA ∵ A is a 2  3 matrix and B is a 3  2 matrix. ∴ AB is a 2  2 matrix.

Example 11.4T 11.1 Matrices Solution: C. Operations of Matrices
For each of the following pairs of matrices X and Y, find XY and YX. (a) , (b) , Solution: (a) XY

Example 11.4T 11.1 Matrices Solution: C. Operations of Matrices
For each of the following pairs of matrices X and Y, find XY and YX. (a) , (b) , Solution: YX (a)

Example 11.4T 11.1 Matrices Solution: C. Operations of Matrices
For each of the following pairs of matrices X and Y, find XY and YX. (a) , (b) , Solution: (b) XY The number of columns of Y is not equal to the number of rows of X. YX is undefined.

11.1 Matrices C. Operations of Matrices Suppose , and .
Even though A  0 and B  0, we still have AB  0: ∴ AB  0 does not imply A  0 or B  0. The following shows AC  0: Consider AB  AC AB  AC  0 A(B  C)  0 ∵ A  0 and B  C. ∴ AB  AC does not imply A  0 or B  C  0.

Example 11.5T 11.1 Matrices Solution: C. Operations of Matrices
Let Find a non-zero square matrix B of order 2 such that (a) AB  0, (b) BA  0. Solution: Let , where a, b, c and d are some constants. (a) (b) ∵ AB  0 ∵ BA  0 ∴ c  d  0 ∴ b  d  0

11.1 Matrices C. Operations of Matrices
Properties of Matrix Multiplication Let h and k be real numbers and A, B and C be matrices such that the following matrix products are defined. We have: (a) (AB)C  A(BC); (Associative Law) (b) (i) A(B + C)  AB + AC; (ii) (A + B)C  AC + BC; (c) k(AB)  (kA)B  A(kB); (d) (hA)(kB)  (hk)AB; (e) A0  0A  0, where A is a square matrix and 0 is a zero square matrix; (f) AI  IA  A, where A is a square matrix and I is an identity matrix. (Distributive Law) Remarks: The proofs are left for students.

11.1 Matrices C. Operations of Matrices Power of Square Matrices
For any square matrix A and any positive integer n, we have For square matrices A and B of same order: 1. (A  B)2  (A  B)(A  B)  AA  AB  BA  BB  A2  AB  BA  B2 The expressions cannot be reduced to the form we learnt in junior form unless AB  BA. 2. (A  B)(A  B)  AA  AB  BA  BB  A2  AB  BA  B2 In general, (A  B)2  A2  2AB  B2 and (A  B)(A  B)  A2  B2.

Example 11.6T 11.1 Matrices Solution: C. Operations of Matrices Let .
(a) Find the matrix X 2. (b) Hence, find the matrix 3X 2  2X  4I, where I is the 3  3 identity matrix. Solution: X 2  XX. X 2 is also a 3  3 matrix. (a) X 2

Example 11.6T 11.1 Matrices Solution: C. Operations of Matrices Let .
(a) Find the matrix X 2. (b) Hence, find the matrix 3X 2  2X  4I, where I is the 3  3 identity matrix. Solution: (b) 3X 2  2X  4I

Example 11.7T 11.1 Matrices Solution: C. Operations of Matrices
If , show, by mathematical induction, that for all positive integers n. The following shows an outline of solution only. Students should show your workings clearly. Solution: For n  1, obviously L.H.S.  R.H.S. ∴ The proposition is true for n  1. When n  k  1, show that R.H.S Assume the proposition is true for some positive integers k, that is, When n  k  1, L.H.S.  X k  1  R.H.S. ∴ The proposition is true for n  k  1.

cij  aji for all i  1, 2, … n and j  1, 2, …, m.
11.1 Matrices C. Operations of Matrices Transpose of Matrix Let A  (aij)m  n be an m  n matrix. The transpose of matrix of A, denoted by At or AT, is an n  m matrix At  (cij)n  m such that cij  aji for all i  1, 2, … n and j  1, 2, …, m. The transpose of a matrix A is obtained by interchanging the rows and the columns in A, for examples:

11.1 Matrices C. Operations of Matrices Properties of Transposes
Let A and B be two m  n matrices, we have (a) (At)t  A; (b) (A  B)t  At  Bt; (c) (kA)t  kAt, where k is any constant. Let A be an m  n matrix and B be an n  p matrix, we have (d) (AB)t  BtAt. Remarks: The proofs are left for students.

Example 11.8T 11.1 Matrices Solution: C. Operations of Matrices
Given that and If (At)2  pAt  qI  0, find the values of p and q. Solution: ∵ ( At )2  pAt  qI  0 By comparing the corresponding elements of the matrices on both sides, we have p  9 and q  8 .

11.2 Determinants A. Introduction For an n  n square matrix A ,
Similar to matrices, only determinants of at most order 3 will de discussed. denote the determinant of A by Determinant of Order 2 For a 2  2 square matrix A  , the value of its determinant, which is denoted by | A| or det A, is defined by  a11a22  a12a21. a11a22  a12a21 is called the expansion of the determinant.

Example 11.9T 11.2 Determinants Solution: A. Introduction
If  5, find the value of x. Solution:

11.2 Determinants A. Introduction Determinant of Order 3
For a 3  3 square matrix A  , the value of its determinant is defined by  a11a22a33  a12a23a31  a13a21a32  a13a22a31  a11a23a32  a12a21a33. To memorize the expansion of the determinant: This rule is called the rule of Sarrus. Notes:       This rule is only applicable for determinants of order 3.

Example 11.10T 11.2 Determinants Solution: A. Introduction
Evaluate the following determinants. (a) (b) Solution: (a)  (2)(2)(1)  1(1)(4)  3(5)(0)  3(2)(4)  (2)(1)(0)  1(5)(1)  21 (b)  a(1)(0)  0(b)(1)  1(0)(c)  1(1)(1)  a(b)(c)  0(0)(0)  (1  abc)

Example 11.11T 11.2 Determinants Solution: A. Introduction
Let a, b, c, d and e be five distinct numbers. If , prove that c(ae  bd)  a  e  b  d. Solution:

11.2 Determinants B. Properties of Determinants
The following shows some of the properties of determinants, which are true for determinants of any order. For any square matrix A, the determinant of A is equal to that of the transpose of A, i.e., If any two rows (or columns) of a matrix are interchanged, the determinant changes sign but its absolute value remains unchanged. e.g., ; Remarks: These properties can be verified by expanding of the determinants.

11.2 Determinants B. Properties of Determinants
If all the elements in any one row (or column) of a matrix are multiplied by a factor, then the determinant is just the product of the original determinant and the factor. e.g., for any k. For example, if , then (i) , (ii)

11.2 Determinants B. Properties of Determinants When k  0, we have:
The determinant of a matrix is zero if all the elements in a row (or column) are zero, i.e., When all the elements are also multiplied by k, we have: If all the elements of an n  n square matrix are multiplied by the same factor, then the resulting determinant is the product of the original determinant and the nth power of the factor, i.e.,

11.2 Determinants B. Properties of Determinants
The determinant of a matrix is zero if the elements of a row (or a column) are proportional to those of another row (or another column), i.e., if In particular, we have: If any two rows (or columns) of a matrix are equal, the determinant is equal to zero, i.e.,

11.2 Determinants B. Properties of Determinants
Consider the result of addition of matrices, we have: If all the elements in any row (or column) of a matrix can be expressed as the sum of two terms, then the determinant can also be expressed as the sum of the two determinants, i.e., When p, q and r are proportional to the elements of the other row, we have: If all the elements in a row (or column) of a matrix is added or subtracted by multiples of the other row (or column), then the value of the determinant will remain unchanged, i.e., for any k.

11.2 Determinants B. Properties of Determinants
Finally, for the product of two square matrices, we have: For any n  n square matrices A and B, the product of their determinants is equal to the determinant of the matrix AB, i.e., . Verification: Let and Then L.H.S. R.H.S.

Example 11.12T 11.2 Determinants Solution:
B. Properties of Determinants Example 11.12T Without expanding the determinant, show that Solution: Take out the common factor 7 from C2 R1  R2  R1; R3  R2  R3 In order to show the value of the determinant equals to zero, we need to show that any two rows or columns are the same.

Example 11.13T 11.2 Determinants Solution:
B. Properties of Determinants Example 11.13T Without expanding the determinant, show that 3 is a factor of . (Given that the determinant is non-zero.) Solution: C2  C3  C2 Take out the common factor 3 from C2 Since all the elements in the determinant are integers, its value in an integer. ∴ 3 is a factor of the given determinant.

11.2 Determinants C. Evaluation of Determinants of Order 3
Consider the determinant The expansion of the determinant  aei  bfg  cdh  ceg  afh  bdi  a(ei  fh)  b( fg  di)  c(dh  eg) Group the a terms, the b terms and the c terms; Arrange in alphabetical order  a(ei  fh)  b(di  fg)  c(dh  eg) The / sign of each term is determined by the position of a, b, c as shown below:

11.2 Determinants C. Evaluation of Determinants of Order 3
Consider the determinant The expansion of the determinant  aei  bfg  cdh  ceg  afh  bdi  b( fg  di)  e(ai  cg)  h(cd  af ) Group the b terms, the e terms and the h terms; Arrange in alphabetical order  b(di  fg)  e(ai  cg)  h(af  cd) The / sign of each term is determined by the position of b, e, h as shown below:

11.2 Determinants C. Evaluation of Determinants of Order 3
Summarize the results as follows: The determinant of order 3 can be expanded along any row or column, i.e., or , etc. In a determinant of order 3: For each of the elements a, c, e, g and i, cofactor  minor; For each of the elements b, d, f and h, cofactor  (minor). Remarks: For each of the element,  minor corresponding determinant obtained  cofactor product of the minor and the sign of the term

Example 11.14T 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Example 11.14T Evaluate the determinant by expanding along (a) the first row, (b) the third column. Solution: (a) Value of the determinant  1(6  48)  7(15  72)  4(30  18)  615 (b) Value of the determinant  4(30  18)  8(6  63)  3(2  35)  615

Example 11.15T 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Example 11.15T Show that Hence evaluate Solution: R3  R1  R3 Expand along R3

Example 11.16T 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Example 11.16T Let and Find the determinant of AB. Solution: Students may try to find the matrix AB first, and then the determinant of AB. However, each of the determinants of A and B can be evaluated more easily in this case.

Example 11.17T 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Example 11.17T Factorize Solution:

Example 11.18T 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Example 11.18T Prove that Solution: C1  C2  C3  C1

Example 11.19T 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Example 11.19T Solve the equation Solution: , where x  0. When x  0, the determinant becomes R1  R3  R1

11.3 Inverses of Square Matrices
A. Introduction For a non-zero real number n, is a multiplicative inverse of n such that  n  1, i.e., n1n  1. For matrices, matrix division is not defined. We can try to find a matrix B such that BA  AB  I. Inverse of a Matrix If square matrices A and B of order n satisfy the relationship AB = BA = I, where I is the identity matrix of order n, then the matrix B is called the inverse of A and denoted by A1, i.e., AA1  A1A  I.

11.3 Inverses of Square Matrices
A. Introduction For example, consider and and ∴ B is the inverse of A and A is the inverse of B. In particular, the inverse of an identity matrix is the identity matrix itself. Note the following relation in real numbers: 11  1  1

11.3 Inverses of Square Matrices
A. Introduction Actually, not all square matrices have their corresponding inverses. Singular and Non-singular Matrices A square matrix A is said to be non-singular or invertible if and only if its inverse exists. Otherwise, it is said to be singular or non-invertible. If the inverse of a square matrix exists, then we have: Uniqueness of Inverse The inverse of a non-singular square matrix is unique. Proof (using contraction): Suppose B and C are two distinct inverse matrices of A, i.e., AB  BA  I and AC  CA  I. Then B  BI  BAC  IC  C, which contradicts to B  C.

transpose of cofactors of A
11.3 Inverses of Square Matrices A. Introduction Consider the matrix Suppose , i.e., By comparing the corresponding elements of the matrices on both sides, we have: transpose of cofactors of A For a 2  2 square matrix , the cofactors of a, b, c and d are d, c, b and a respectively. determinant of A

11.3 Inverses of Square Matrices
A. Introduction Inverse of a 2  2 matrix Let If A is non-singular, then the inverse of A is given by: The determinant of A is 1. For example, if , then

transpose of cofactors of A
11.3 Inverses of Square Matrices A. Introduction Inverse of a 3  3 matrix Let If A is non-singular, then the inverse of A is given by: transpose of cofactors of A cofactor

11.3 Inverses of Square Matrices
A. Introduction In general, the inverse of a non-singular matrix A contains the factor Thus if , the matrix is singular. Theorem A square matrix A is non-singular if an only if Proof: ‘if ’: If A is non-singular, then there exists a matrix B such that AB  BA  I. ∵ , and ‘only if ’: If , then we can find the inverse: ∴ A is non-singular.

Example 11.20T 11.3 Inverses of Square Matrices Solution:
A. Introduction Example 11.20T Find the inverses of the following matrices. (a) (b) Solution: (a)

Example 11.20T 11.3 Inverses of Square Matrices Solution:
A. Introduction Example 11.20T Find the inverses of the following matrices. (a) (b) Solution: (b)

Example 11.21T 11.3 Inverses of Square Matrices Solution:
A. Introduction Example 11.21T Let P be a square matrix such that 2I  P  P2  0. Prove that P is non-singular and find P1 in terms of P and I. Solution: 2I  P  P2  0 P  P2  2I P(I  P)  2I ∴ P is non-singular and

11.3 Inverses of Square Matrices
B. Properties of Inverses Properties of Inverses Let A and B be two non-singular square matrices of the same order, k be a non-zero real number, and n be a positive integer. Then (a) (A1)1  A; (b) (kA)1  k1A1; (c) (An)1  (A1)n; (d) (At)1  (A1)t; (e) ; (f) (AB)1  B1A1. Proof of (f): ∵ (AB)(B1A1)  A(BB1)A1  AIA1  AA1  I and (B1A1)(AB)  B1(A1A)B  B1IB  B1B  I ∴ By definition, (AB)1  B1A1.

Example 11.22T 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Example 11.22T Let and (a) Find A1 and B1. (b) Hence find (AB2)1 and [(AB)t]1. Solution: (a)

Example 11.22T 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Example 11.22T Let and (a) Find A1 and B1. (b) Hence find (AB2)1 and [(AB)t]1. Solution: (b) (AB2)1  (B2)1A1  (B1)2A1

Example 11.22T 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Example 11.22T Let and (a) Find A1 and B1. (b) Hence find (AB2)1 and [(AB)t]1. Solution: [(AB)t]1  [(AB)1]t  (B1A1)t (b)

Example 11.23T 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Example 11.23T Let (a) Find M 2. (b) Hence find M 1. Solution: (a) (b)

Example 11.24T 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Example 11.24T Let (a) Find X 1. (b) Hence find Y if YX  Solution: (a)

Example 11.24T 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Example 11.24T Let (a) Find X 1. (b) Hence find Y if YX  Solution: (b) Y  (YX)X 1

Example 11.25T 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Example 11.25T Let and (a) Find the matrix Y 1XY. (b) Hence find X 1000. Solution: The following shows an outline of solution only. Students should show your workings clearly. (a)

Example 11.25T 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Example 11.25T Let and (a) Find the matrix Y 1XY. (b) Hence find X 1000. Solution: (b) Consider (Y 1XY)1000  (Y 1XY)(Y 1XY)(Y 1…) … (… Y)(Y 1XY)  Y 1X(I) X(I) … (I) XY  Y 1X 1000Y ∴ Y(Y 1XY)1000Y 1  X 1000

Chapter Summary 11.1 Matrices 1. Definition
An m  n matrix is represented in the form An m  n matrix may also be represented by the symbol (aij)m  n or [aij]m  n.

Chapter Summary 11.1 Matrices 2. Operations of Matrices
Let A  (aij)m  n and B  (bij)m  n be two matrices and k be a real number. (a) Addition A  B  (aij  bij)m  n, for all i  1, 2, ... , m and j  1, 2, ... , n (b) Subtraction A  B  (aij  (1)bij)m  n, for all i  1, 2, ... , m and j  1, 2, ... , n (c) Scalar Multiplication kA  (kaij)m  n (d) Transpose At  (cij)n  m where cij  aji, for all i  1, 2, ... , n and j  1, 2, ... , m (e) Multiplication Let A  (aij)m  n, B  (bij)n  p and C  (cij)m  p. If AB  C, then cij  ai1b1j  ai2b2j  ...  ainbnj 

Chapter Summary 11.2 Determinants 1. Determinant of order 2
 a11a22  a12a21 2. Determinant of order 3  a11a22a33  a12a23a31  a13a21a32  a13a22a31  a11a23a32  a12a21a33

Chapter Summary 11.3 Inverses of Square Matrices 1. Definition
For a square matrix A, if there exists a matrix B such that AB  BA  I, then B is called the inverse of A and is denoted by A1. 2. Inverse of a 2  2 matrix: 3. Inverse of a 3  3 matrix:

Follow-up 11.1 11.1 Matrices Solution: A. Introduction
If ( )  ( i j k ), find the values of i, j and k. Solution: From the definition, i  3, j  5 and k  7.

Follow-up 11.2 11.1 Matrices Solution: C. Operations of Matrices
Suppose and Find the matrix C such that A  B  C. Solution: ∵ A  B  C ∴ C  A  B

Follow-up 11.3 11.1 Matrices Solution: C. Operations of Matrices
Suppose and Evaluate 2P  Q and P  4Q. Solution: 2P  Q P  4Q

Follow-up 11.4 11.1 Matrices Solution: C. Operations of Matrices
For each of the following pairs of matrices P and Q, find PQ and QP. (a) , (b) , Solution: (a) PQ

Follow-up 11.4 11.1 Matrices Solution: C. Operations of Matrices
For each of the following pairs of matrices P and Q, find PQ and QP. (a) , (b) , Solution: QP (a) In general, for two square matrices P and Q of the same order, PQ  QP.

Follow-up 11.4 11.1 Matrices Solution: C. Operations of Matrices
For each of the following pairs of matrices P and Q, find PQ and QP. (a) , (b) , Solution: (b) PQ The number of columns of Q is not equal to the number of rows of P. QP is undefined.

Follow-up 11.5 11.1 Matrices Solution: C. Operations of Matrices
Let Find a non-zero square matrix B of order 2 such that (a) AB  0, (b) BA  0. Solution: Let , where a, b, c and d are some constants. (a) (b) ∵ AB  0 ∵ BA  0 ∴ a  b  0 ∴ a  b  0 and c  d  0 b  a and d  c

Follow-up 11.6 11.1 Matrices Solution: C. Operations of Matrices Let .
(a) Find the matrices A2 and A4. (b) Hence find the matrix A4  2A2  3I, where I is the 2  2 identity matrix. Solution: (a) A2 A4  A2 A2

Follow-up 11.6 11.1 Matrices Solution: C. Operations of Matrices Let .
(a) Find the matrices A2 and A4. (b) Hence find the matrix A4  2A2  3I, where I is the 2  2 identity matrix. Solution: (b) A4  2A2  3I

Follow-up 11.7 11.1 Matrices Solution: C. Operations of Matrices
If , show, by mathematical induction, that for all positive integers n. Solution: R.H.S. For n  1, L.H.S. ∴ The proposition is true for n  1. Assume the proposition is true for some positive integers k, that is, .

Follow-up 11.7 11.1 Matrices Solution: C. Operations of Matrices
If , show, by mathematical induction, that for all positive integers n. Solution: When n  k  1, L.H.S.  M k  1 ∴ The proposition is true for n  k  1. ∴ By the principle of mathematical induction, the proposition is true for all positive integers n.

Follow-up 11.8 11.1 Matrices Solution: C. Operations of Matrices
Given that and If (At)2  pAt  qI  0, find the values of p and q. Solution: ∵ ( At )2  pAt  qI  0 By comparing the corresponding elements of the matrices on both sides, we have p  6 and q  1 .

Follow-up 11.9 11.2 Determinants Solution: A. Introduction
If  9, find the possible values of x. Solution: or

Follow-up 11.10 11.2 Determinants Solution: A. Introduction
Evaluate the following determinants. (a) (b) Solution: (a)  (1)(3)(5)  0(0)(4)  2(0)(0)  2(3)(4)  (1)(0)(0)  0(0)(5)  9 (b)  1(1)(1)  a(b)(c)  (c)(a)(b)  (c)(1)(c)  1(b)(b)  a(a)(1)  1  a2  b2  c2

Follow-up 11.11 11.2 Determinants Solution: A. Introduction
Let a, b, c and d be four distinct numbers. If , prove that (a  d)(c  b)  0. Solution:

Follow-up 11.12 11.2 Determinants Solution:
B. Properties of Determinants Follow-up 11.12 Without expanding the determinant, show that Solution: Take out the common factor 3 from C2 In order to show the value of the determinant equals to zero, we need to show that any two rows or columns are the same. C3  C2  C3

Follow-up 11.13 11.2 Determinants Solution:
B. Properties of Determinants Follow-up 11.13 Without expanding the determinant, show that is divisible by 25. Solution: C1  C3  C1 C1  C2  C1 Take out the common factor 5 from C1

Follow-up 11.13 11.2 Determinants Solution:
B. Properties of Determinants Follow-up 11.13 Without expanding the determinant, show that is divisible by 25. Solution: C3  C1  C3 integer ∴ The given determinant is divisible by 25.

Follow-up 11.14 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Follow-up 11.14 Evaluate the determinant by expanding along (a) the third column, (b) the second row. Solution: (a) Value of the determinant  2(63  12)  8(9  24)  5(3  42)  45 (b) Value of the determinant  7(30  18)  3(5  8)  8(9  24)  45

Follow-up 11.15 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Follow-up 11.15 Show that Hence evaluate Solution: R3  R1  R3; R2  4  R1  R2 Expand along R3

Follow-up 11.16 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Follow-up 11.16 Let and Find the determinant of AB. Solution:

Follow-up 11.17 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Follow-up 11.17 Factorize c2  (a  b)2  (c  a  b)(c  a  b) Solution: (b  c)2  a2  (b  c  a)(b  c  a) R2  R1  R2 ; R3  R1  R3 C1  C2  C3  C1

Follow-up 11.17 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Follow-up 11.17 Factorize Solution:

Follow-up 11.18 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Follow-up 11.18 Prove that Solution: R1  R2  R1; R2  R3  R2 Take out the common factors from R1 and R2 Expand (a  b  c)(c  a) on the R.H.S. to complete the proof.

Follow-up 11.19 11.2 Determinants Solution:
C. Evaluation of Determinants of Order 3 Follow-up 11.19 Solve the equation Solution: R1  R3  R1; R2  R3  R2 Take out the common factors from R1 and R2 or

Follow-up 11.20 11.3 Inverses of Square Matrices Solution:
A. Introduction Follow-up 11.20 Find the inverses of the following matrices. (a) (b) Solution: (a)

Follow-up 11.20 11.3 Inverses of Square Matrices Solution:
A. Introduction Follow-up 11.20 Find the inverses of the following matrices. (a) (b) Solution: (b)

Follow-up 11.21 11.3 Inverses of Square Matrices Solution:
A. Introduction Follow-up 11.21 Let X be a square matrix such that 2X 2  4X  5I  0. Prove that X is non-singular and find X 1 in terms of X and I. Solution: 2X 2  4X  5I  0 2X 2  4X  5I 2X (X  2I )  5I ∴ X is non-singular and

Follow-up 11.22 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Follow-up 11.22 Let and (a) Find M 1 and N 1. (b) Hence find (M 2N)1 and (NM t)1. Solution: (a)

Follow-up 11.22 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Follow-up 11.22 Let and (a) Find M 1 and N 1. (b) Hence find (M 2N)1 and (NM t)1. Solution: (b) (M 2N)1  N 1(M 2)1  N 1(M 1)2

Follow-up 11.22 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Follow-up 11.22 Let and (a) Find M 1 and N 1. (b) Hence find (M 2N)1 and (NM t)1. Solution: (b) (NM t)1  (M t)1N 1  (M 1)tN 1

Follow-up 11.23 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Follow-up 11.23 Let (a) Find P 2. (b) Hence find P1. Solution: (a) (b)

Follow-up 11.24 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Follow-up 11.24 Let (a) Find P 1. (b) Hence find Q if PQ  Solution: (a)

Follow-up 11.24 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Follow-up 11.24 Let (a) Find P 1. (b) Hence find Q if PQ  Solution: (b) Q  P 1(PQ)

Follow-up 11.25 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Follow-up 11.25 Let and (a) Show that (b) Hence find Q800. Solution: (a)

Follow-up 11.25 11.3 Inverses of Square Matrices Solution:
B. Properties of Inverses Follow-up 11.25 Let and (a) Show that (b) Hence find Q800. Solution: (b) Consider (P 1QP)800  (P 1QP)(P 1QP)(P 1…) … (… P)(P 1QP)  P 1Q(I) Q(I) … (I) QP  P 1Q800P ∴ P(P 1QP)800P 1  Q 800