4.10 SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 1:

It is given that matrix A =

(\begin{matrix} 3 & - 1 \\ 5 & - 2 \end{matrix})

(a) Find the inverse matrix of A.

(b) Write the following simultaneous linear equations as matrix equation:

3u – v = 9

5u – 2v = 13

Hence, using matrix method, calculate the value of u and v.

Solution:
(a)

\begin{array}{l} A^{- 1} = \frac{1}{3 (- 2) - (5) (- 1)} (\begin{matrix} - 2 & 1 \\ - 5 & 3 \end{matrix}) \\ = - 1 (\begin{matrix} - 2 & 1 \\ - 5 & 3 \end{matrix}) = (\begin{matrix} 2 & - 1 \\ 5 & - 3 \end{matrix}) \end{array}

(b)

\begin{array}{l} (\begin{matrix} 3 & - 1 \\ 5 & - 2 \end{matrix}) (\begin{array}{l} u \\ v \end{array}) = (\begin{array}{l} 9 \\ 13 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = - 1 (\begin{matrix} - 2 & 1 \\ - 5 & 3 \end{matrix}) (\begin{array}{l} 9 \\ 13 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = - 1 (\begin{array}{l} (- 2) (9) + (1) (13) \\ (- 5) (9) + (3) (13) \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = - 1 (\begin{array}{l} - 5 \\ - 6 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = (\begin{array}{l} 5 \\ 6 \end{array}) \\ ∴ u = 5, v = 6 \end{array}

Question 2:

It is given that matrix A =

(\begin{matrix} 2 & - 5 \\ 1 & 3 \end{matrix})

and matrix B =

m (\begin{matrix} 3 & k \\ - 1 & 2 \end{matrix})

such that AB =

(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

(a) Find the value of m and of k.

(b) Write the following simultaneous linear equations as matrix equation:

2u – 5v = –15

u + 3v = –2

Hence, using matrix method, calculate the value of u and v.

Solution:

(a) Since AB =

(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

, B is the inverse of A.

m = \frac{1}{(2) (3) - (- 5) (1)} = \frac{1}{11}

k = 5

(b)

\begin{array}{l} (\begin{matrix} 2 & - 5 \\ 1 & 3 \end{matrix}) (\begin{array}{l} u \\ v \end{array}) = (\begin{array}{l} - 15 \\ - 2 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = \frac{1}{11} (\begin{matrix} 3 & 5 \\ - 1 & 2 \end{matrix}) (\begin{array}{l} - 15 \\ - 2 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = \frac{1}{11} (\begin{array}{l} (3) (- 15) + (5) (- 2) \\ (- 1) (- 15) + (2) (- 2) \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = \frac{1}{11} (\begin{array}{l} - 55 \\ 11 \end{array}) \\ (\begin{array}{l} u \\ v \end{array}) = (\begin{array}{l} - 5 \\ 1 \end{array}) \\ ∴ u = - 5, v = 1 \end{array}

4.4 Multiplication of Matrix by a Number

Posted on April 24, 2020 by user

4.4 Multiplication of a Matrix by a Number

When matrix is multiplied by a number, every element in the matrix is multiplied by the number.

Example:

Given that

A = (\begin{matrix} - 2 & 4 \\ 5 & - 6 \end{matrix})

, find each of the following.
(a) 3A
(b) -2A

Solution:

\begin{array}{l} (a) 3 A = 3 (\begin{matrix} - 2 & 4 \\ 5 & - 6 \end{matrix}) \\ = (\begin{matrix} 3 \times (- 2) & 3 \times 4 \\ 3 \times 5 & 3 \times (- 6) \end{matrix}) \\ = (\begin{matrix} - 6 & 12 \\ 15 & - 18 \end{matrix}) \end{array}

\begin{array}{l} (b) - 2 A = - 2 (\begin{matrix} - 2 & 4 \\ 5 & - 6 \end{matrix}) \\ = (\begin{matrix} - 2 \times (- 2) & - 2 \times 4 \\ - 2 \times 5 & - 2 \times (- 6) \end{matrix}) \\ = (\begin{matrix} 4 & - 8 \\ - 10 & 12 \end{matrix}) \end{array}

4.4 Multiplication of Matrix by a Number (Sample Questions)

Posted on April 24, 2020 by user

Example 1:

Express the following as single matrix.

(a) 3 (\begin{matrix} 1 & - 4 \\ - 3 & 2 \end{matrix}) + 2 (\begin{matrix} - 3 & 1 \\ 3 & - 4 \end{matrix})

(b) 2 (\begin{matrix} - 1 & 0 \\ 9 & - 4 \end{matrix}) - 4 (\begin{matrix} - 2 & 2 \\ 3 & - 6 \end{matrix}) - \frac{1}{3} (\begin{matrix} - 9 & - 3 \\ - 6 & 15 \end{matrix})

Solution:

\begin{array}{l} (a) 3 (\begin{matrix} 1 & - 4 \\ - 3 & 2 \end{matrix}) + 2 (\begin{matrix} - 3 & 1 \\ 3 & - 4 \end{matrix}) \\ = (\begin{matrix} 3 \times 1 & 3 \times (- 4) \\ 3 \times (- 3) & 3 \times 2 \end{matrix}) + (\begin{matrix} 2 \times (- 3) & 2 \times 1 \\ 2 \times 3 & 2 \times (- 4) \end{matrix}) \\ = (\begin{matrix} 3 & - 12 \\ - 9 & 6 \end{matrix}) + (\begin{matrix} - 6 & 2 \\ 6 & - 8 \end{matrix}) \\ = (\begin{matrix} 3 + (- 6) & - 12 + 2 \\ - 9 + 6 & 6 - (- 8) \end{matrix}) \\ = (\begin{matrix} - 3 & - 10 \\ - 3 & 14 \end{matrix}) \end{array}

\begin{array}{l} (b) \\ 2 (\begin{matrix} - 1 & 0 \\ 9 & - 4 \end{matrix}) - 4 (\begin{matrix} - 2 & 2 \\ 3 & - 6 \end{matrix}) - \frac{1}{3} (\begin{matrix} - 9 & - 3 \\ - 6 & 15 \end{matrix}) \\ = (\begin{matrix} - 2 & 0 \\ 18 & - 8 \end{matrix}) - (\begin{matrix} - 8 & 8 \\ 12 & - 24 \end{matrix}) - (\begin{matrix} \frac{1}{3} \times (- 9) & \frac{1}{3} \times (- 3) \\ \frac{1}{3} \times (- 6) & \frac{1}{3} \times 15 \end{matrix}) \\ = (\begin{matrix} - 2 & 0 \\ 18 & - 8 \end{matrix}) - (\begin{matrix} - 8 & 8 \\ 12 & - 24 \end{matrix}) - (\begin{matrix} - 3 & - 1 \\ - 2 & 5 \end{matrix}) \\ = (\begin{matrix} - 2 - (- 8) - (- 3) & 0 - 8 - (- 1) \\ 18 - 12 - (- 2) & - 8 - (- 24) - 5 \end{matrix}) \\ = (\begin{matrix} 9 & - 7 \\ 8 & 11 \end{matrix}) \end{array}

4.5 Multiplication of Two Matrices

Posted on April 24, 2020 by user

4.5 Multiplication of Two Matrices

1. Two matrices can be multiplied when the number of columns for the first matrix is the same as the number of rows for the second matrix.

For instance, if A is a m × n matrix and B is a n × t matrix, then the product of P = AB. The order of matrix P is m × t

\begin{array}{l} Examples: \\ (a) (a b) (\begin{array}{l} c \\ d \end{array}) = (a c + b d) \\ 1 \times 22 \times 1 1 \times 1 \\ (b) (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{array}{l} e \\ f \end{array}) = (\begin{array}{l} a e + b f \\ c e + d f \end{array}) \\ 2 \times 22 \times 12 \times 1 \\ (c) (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} e & f \\ g & h \end{matrix}) = (\begin{matrix} a e + b g & a f + b h \\ c e + d g & c f + d h \end{matrix}) \\ 2 \times 22 \times 22 \times 2 \\ (d) (\begin{array}{l} a \\ b \end{array}) (c d) = (\begin{matrix} a c & a d \\ b c & b d \end{matrix}) \\ 2 \times 11 \times 22 \times 2 \\ (e) (a b c) (\begin{array}{l} d \\ e \\ f \end{array}) = (a d + b e + c f) \\ 1 \times 33 \times 11 \times 1 \\ (f) (\begin{matrix} a & b \\ \begin{array}{l} c \\ e \end{array} & \begin{array}{l} d \\ f \end{array} \end{matrix}) (\begin{array}{l} g \\ h \end{array}) = (\begin{array}{l} a g + b h \\ c g + d h \\ e g + f h \end{array}) \end{array}

Examples:

Determine whether the following pairs of matrices can be multiplied and state the order of the product of those that can be multiplied.

\begin{array}{l} (a) (\begin{matrix} 3 & - 5 \\ 1 & 2 \end{matrix}) (3 7) \\ (b) (\begin{matrix} 2 & 9 \\ 1 & 3 \end{matrix}) (\begin{array}{l} 8 \\ 6 \end{array}) \\ (c) (- 10 - 6) (\begin{array}{l} 7 \\ 2 \end{array}) \\ (d) (\begin{array}{l} 8 \\ 6 \end{array}) (\begin{matrix} 2 & 9 \\ 1 & 3 \end{matrix}) \\ (e) (\begin{array}{l} 7 \\ - 3 \end{array}) (2 10) \end{array}

Solution:

\begin{array}{l} (a) (\begin{matrix} 3 & - 5 \\ 1 & 2 \end{matrix}) (3 7) \\ 2 \times 2 1 \times 2 \leftarrow 2 \neq 1 ∴ Cannot be multiplied \\ (b) (\begin{matrix} 2 & 9 \\ 1 & 3 \end{matrix}) (\begin{array}{l} 8 \\ 6 \end{array}) \\ 2 \times 2 2 \times 1 \leftarrow 2 = 2 ∴ Can be multiplied . \\ Order of product = 2 \times 1 \\ (c) (- 10 - 6) (\begin{array}{l} 7 \\ 2 \end{array}) \\ 1 \times 2 2 \times 1 \leftarrow 2 = 2 ∴ Can be multiplied . \\ Order of product = 1 \times 1 \\ (d) (\begin{array}{l} 8 \\ 6 \end{array}) (\begin{matrix} 2 & 9 \\ 1 & 3 \end{matrix}) \\ 2 \times 1 2 \times 2 \leftarrow 1 \neq 2 ∴ Cannot be multiplied \\ (e) (\begin{array}{l} 7 \\ - 3 \end{array}) (2 10) \\ 2 \times 1 1 \times 2 \leftarrow 1 = 1 ∴ Can be multiplied . \\ Order of product = 2 \times 2 \end{array}

4.5 Multiplication of Two Matrices (Sample Question 1)

Posted on April 24, 2020 by user

Example 1:
Find the product of the following pairs of matrices.

\begin{array}{l} (a) (1 5 2) (\begin{array}{l} 2 \\ 4 \\ 3 \end{array}) \\ (b) (\begin{matrix} 2 & 8 \\ - 3 & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ 4 & - 2 \end{matrix}) \\ (c) (\begin{array}{l} - 3 \\ 5 \end{array}) (2 6) \\ (d) (\begin{matrix} 0 & 4 \\ - 1 & 3 \end{matrix}) (\begin{array}{l} 7 \\ - 2 \end{array}) \\ (e) (7 - 4) (\begin{matrix} - 2 & 0 \\ - 1 & 3 \end{matrix}) \end{array}

Solution:

\begin{array}{l} (a) \\ (1 5 2) (\begin{array}{l} 2 \\ 4 \\ 3 \end{array}) \leftarrow \begin{array}{l} Matrices analysis \\ 1 \times 3 and 3 \times 1 \\ ↓ ↓ \\ = 1 \times 1 matrix \end{array} \\ = (1 \times 2 \oplus 5 \times 4 \oplus 2 \times 3) \\ = (2 + 20 + 6) \\ = (28) \end{array}

(b)

\begin{array}{l} (\begin{matrix} 2 & 8 \\ - 3 & 1 \end{matrix}) (\begin{matrix} 1 & 0 \\ 4 & - 2 \end{matrix}) \leftarrow \begin{array}{l} Matrices analysis \\ 2 \times 2 and 2 \times 2 \\ ↓ ↓ \\ = 2 \times 2 matrix \end{array} \\ = (\begin{matrix} 2 \times 1 + 8 \times 4 & 2 \times 0 + 8 \times - 2 \\ - 3 \times 1 + 1 \times 4 & - 3 \times 0 + 1 \times - 2 \end{matrix}) \\ = (\begin{matrix} 34 & - 16 \\ 1 & - 2 \end{matrix}) \end{array}

(c)

\begin{array}{l} (\begin{array}{l} - 3 \\ 5 \end{array}) (2 6) \leftarrow \begin{array}{l} Matrices analysis \\ 2 \times 1 and 1 \times 2 \\ ↓ ↓ \\ = 2 \times 2 matrix \end{array} \\ = (\begin{matrix} - 3 \times 2 & - 3 \times 6 \\ 5 \times 2 & 5 \times 6 \end{matrix}) \\ = (\begin{matrix} - 6 & - 18 \\ 10 & 30 \end{matrix}) \end{array}

(d)

\begin{array}{l} (\begin{matrix} 0 & 4 \\ - 1 & 3 \end{matrix}) (\begin{array}{l} 7 \\ - 2 \end{array}) \leftarrow \begin{array}{l} Matrices analysis \\ 2 \times 2 and 2 \times 1 \\ ↓ ↓ \\ = 2 \times 1 matrix \end{array} \\ = (\begin{array}{l} 0 \times 7 + 4 \times - 2 \\ - 1 \times 7 + 3 \times - 2 \end{array}) \\ = (\begin{array}{l} - 8 \\ - 13 \end{array}) \end{array}

(e)

\begin{array}{l} (7 - 4) (\begin{matrix} - 2 & 0 \\ - 1 & 3 \end{matrix}) \leftarrow \begin{array}{l} Matrices analysis \\ 1 \times 2 and 2 \times 2 \\ ↓ ↓ \\ = 1 \times 2 matrix \end{array} \\ = (7 \times - 2 + (- 4 \times - 1) 7 \times 0 + (- 4 \times 3)) \\ = (- 14 + 4 0 - 12) \\ = (- 10 - 12) \end{array}

4.7 Inverse Matrix

Posted on April 24, 2020 by user

4.7 Inverse Matrix

1. If A is a square matrix, B is another square matrix and A × B = B × A = I, then matrix A is the inverse matrix of matrix B and vice versa. Matrix A is called the inverse matrix of B for multiplication and vice versa.

2. The symbol A^-1 denotes the inverse matrix of A.

3. Inverse matrices can only exist for square matrices but not all square matrices have inverse matrices.

4. If AB ≠ I or BA ≠ I, then A is not the inverse of B and B is not the inverse of A.

Example 1:

Determine whether matrix

A = (\begin{matrix} 2 & 9 \\ 1 & 5 \end{matrix})

is an inverse matrix of matrix

B = (\begin{matrix} 5 & - 9 \\ - 1 & 2 \end{matrix}) .

Solution:

\begin{array}{l} A B = (\begin{matrix} 2 & 9 \\ 1 & 5 \end{matrix}) (\begin{matrix} 5 & - 9 \\ - 1 & 2 \end{matrix}) \\ = (\begin{matrix} 2 \times 5 + 9 \times - 1 & 2 \times - 9 + 9 \times 2 \\ 1 \times 5 + 5 \times - 1 & 1 \times - 9 + 5 \times 2 \end{matrix}) \\ = (\begin{matrix} 10 + (- 9) & - 18 + 18 \\ 5 + (- 5) & - 9 + 10 \end{matrix}) \\ = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = I \\ A B = (\begin{matrix} 5 & - 9 \\ - 1 & 2 \end{matrix}) (\begin{matrix} 2 & 9 \\ 1 & 5 \end{matrix}) \\ = (\begin{matrix} 5 \times 2 + (- 9) \times 1 & 5 \times 9 + (- 9) \times 5 \\ - 1 \times 2 + 2 \times 1 & - 1 \times 9 + 2 \times 5 \end{matrix}) \\ = (\begin{matrix} 10 + (- 9) & 18 - 18 \\ - 2 + 2 & - 9 + 10 \end{matrix}) \\ = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) = I \\ A B = B A = I \\ ∴ A is the inverse matrix of B and vice versa . \end{array}

5. The inverse of a matrix may also be found using a formula.

A = (\begin{matrix} a & b \\ c & d \end{matrix})

, then the inverse matrix of A, A^-1, is given by the formula below.

A^{- 1} = \frac{1}{a d - b c} (\begin{matrix} d & - b \\ - c & a \end{matrix}), where a d - b c \neq 0

6. ad – bc is known as the determinant of matrix A.

7. If the determinant, ad – bc = 0, then the inverse matrix of A does not exist.

Example 2:

Find the inverse matrix of

A = (\begin{matrix} 6 & 1 \\ - 9 & - 1 \end{matrix})

using the formula.

Solution:

\begin{array}{l} A = (\begin{matrix} 6 & 1 \\ - 9 & - 1 \end{matrix}) \\ a = 6, b = 1, c = - 9, d = - 1 \\ A^{- 1} = \frac{1}{a d - b c} (\begin{matrix} d & - b \\ - c & a \end{matrix}) \\ A^{- 1} = \frac{1}{6 \times - 1 - (1 \times - 9)} (\begin{matrix} - 1 & - 1 \\ 9 & 6 \end{matrix}) \\ A^{- 1} = \frac{1}{- 6 + 9} (\begin{matrix} - 1 & - 1 \\ 9 & 6 \end{matrix}) \\ A^{- 1} = \frac{1}{3} (\begin{matrix} - 1 & - 1 \\ 9 & 6 \end{matrix}) = (\begin{matrix} - \frac{1}{3} & - \frac{1}{3} \\ 3 & 2 \end{matrix}) \end{array}

Example 3:

The inverse matrix of

(\begin{matrix} - 7 & 2 \\ - 9 & 2 \end{matrix}) is r (\begin{matrix} 2 & s \\ 9 & t \end{matrix}) .

Find the value of r, of s and of t.

Solution:

\begin{array}{l} Let A = (\begin{matrix} - 7 & 2 \\ - 9 & 2 \end{matrix}) \\ A^{- 1} = \frac{1}{- 7 \times 2 - (- 9) \times 2} (\begin{matrix} 2 & - 2 \\ 9 & - 7 \end{matrix}) \\ A^{- 1} = \frac{1}{4} (\begin{matrix} 2 & - 2 \\ 9 & - 7 \end{matrix}) \\ ∴ r (\begin{matrix} 2 & s \\ 9 & t \end{matrix}) = \frac{1}{4} (\begin{matrix} 2 & - 2 \\ 9 & - 7 \end{matrix}) \\ By comparison, \\ r = \frac{1}{4}, s = - 2, t = - 7. \end{array}

4.6 Identity Matrix

Posted on April 24, 2020 by user

4.6 Identity Matrix

1. Identity matrix is a square matrix, usually denoted by the letter I and is also known as unit matrix.

2. All the diagonal elements (from top left to bottom right) of an identity matrix are 1 and the rest of the elements are 0.

For example,

(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) and (\begin{matrix} 1 & 00 \\ \begin{array}{l} 0 \\ 0 \end{array} & \begin{array}{l} 10 \\ 01 \end{array} \end{matrix}) are identity matrices .

3. If I is the identity matrix of order n × n and A is a matrix of the same order, then IA = A and AI = A.

Example 1:

Determine whether each of the following is an identity matrix of

(\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) .

(a) (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) (b) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})

Solution:

\begin{array}{l} (a) (\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \\ = (\begin{matrix} - 2 \times 1 + 4 \times 0 & - 2 \times 0 + 4 \times 1 \\ 3 \times 1 + 7 \times 0 & 3 \times 0 + 7 \times 1 \end{matrix}) \\ = (\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) \\ Therefore, (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) is an identity matrix . \\ (b) (\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) \\ = (\begin{matrix} - 2 \times 0 + 4 \times 1 & - 2 \times 1 + 4 \times 0 \\ 3 \times 0 + 7 \times 1 & 3 \times 1 + 7 \times 0 \end{matrix}) \\ = (\begin{matrix} 4 & - 2 \\ 7 & 3 \end{matrix}) \\ \neq (\begin{matrix} - 2 & 4 \\ 3 & 7 \end{matrix}) \\ Therefore, (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) is not an identity matrix . \end{array}

Example 2:

Find the product of the following pairs of matrices and determine whether the given matrix is an identity matrix.

\begin{array}{l} (a) (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) and (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) \\ (b) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) and (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) \end{array}

Solution:

\begin{array}{l} (a) (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) \\ = (\begin{matrix} - 3 \times 1 + 2 \times 0 & - 3 \times 0 + 2 \times 1 \\ 5 \times 1 + 7 \times 0 & 5 \times 0 + 7 \times 1 \end{matrix}) = (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) \\ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) \\ = (\begin{matrix} 1 \times - 3 + 0 \times 5 & 1 \times 2 + 0 \times 7 \\ 0 \times - 3 + 1 \times 5 & 0 \times 2 + 1 \times 7 \end{matrix}) = (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) \\ ∴ (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}) is an identity matrix for (\begin{matrix} - 3 & 2 \\ 5 & 7 \end{matrix}) . \\ (b) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) \\ = (\begin{matrix} 0 \times 1 + 0 \times 5 & 0 \times 8 + 0 \times 3 \\ 1 \times 1 + 1 \times 5 & 1 \times 8 + 1 \times 3 \end{matrix}) = (\begin{matrix} 0 & 0 \\ 6 & 11 \end{matrix}) \\ (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) \\ = (\begin{matrix} 1 \times 0 + 8 \times 1 & 1 \times 0 + 8 \times 1 \\ 5 \times 0 + 3 \times 1 & 5 \times 0 + 3 \times 1 \end{matrix}) = (\begin{matrix} 8 & 8 \\ 3 & 3 \end{matrix}) \\ ∴ (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) is NOT an identity matrix for (\begin{matrix} 1 & 8 \\ 5 & 3 \end{matrix}) . \end{array}

4.2 Equal Matrices

Posted on April 24, 2020 by user

4.2 Equal Matrices

(A) Determining whether two matrices are equal

1. Two matrices are equal if they have the same order and their corresponding elements are equal.

For example,

(\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} e & f \\ g & h \end{matrix})

Therefore, a = e, b = f, c = g and d = h.

Example 1:

Determine whether the following pairs of matrices are equal.

\begin{array}{l} (a) A = (\begin{matrix} 10 & 8 \\ - 3 & 1 \end{matrix}) and B = (\begin{matrix} 10 & 8 \\ - 3 & 1 \end{matrix}) \\ (b) P = (\begin{array}{l} 2 \\ - 4 \\ 10 \end{array}) and Q = (\begin{array}{l} 2 \\ - 3 \\ 10 \end{array}) \\ (c) M = (\begin{array}{l} 3 \\ 5 \end{array}) and N = (4 7) \end{array}

Solution:

(a) Equal

(b) Not equal, because -4 ≠ -3.

(c) Not equal, because the orders of the matrices are not equal.

(B) Solving problem involving equal matrices

1. When matrices are equal, elements whose values are unknown can be determined.

Example 2:

State the value of the unknowns in the following pairs of equal matrix.

(\begin{array}{l} 2 x \\ x + 2 y \end{array}) = (\begin{array}{l} - 8 \\ 10 \end{array})

Solution:

(\begin{array}{l} 2 x \\ x + 2 y \end{array}) = (\begin{array}{l} - 8 \\ 10 \end{array})

2x = -8
x = -4

x + 2y = 10
(-4) + 2y = 10
2y = 10 + 4
2y = 14
y = 7

4.3 Addition and Subtraction of Matrices (Sample Questions)

Posted on April 24, 2020 by user

Example 1:

Find the addition of the following matrices.

\begin{array}{l} (a) (18 - 7) + (3 6) \\ (b) (\begin{matrix} - 13 & 0 \\ 7 & - 1 \end{matrix}) + (\begin{matrix} - 1 & 3 \\ 5 & 6 \end{matrix}) \end{array}

Solution:

\begin{array}{l} (a) \\ (18 - 7) + (3 6) \\ = (18 + 3 - 7 + 6) = (21 - 1) \\ (b) \\ (\begin{matrix} - 13 & 0 \\ 7 & - 1 \end{matrix}) + (\begin{matrix} - 1 & 3 \\ 5 & 6 \end{matrix}) \\ = (\begin{matrix} - 13 + (- 1) & 0 + 3 \\ 7 + 5 & - 1 + 6 \end{matrix}) = (\begin{matrix} - 14 & 3 \\ 12 & 5 \end{matrix}) \end{array}

Example 2:

Find the subtraction of the following matrices.

\begin{array}{l} (a) (\begin{array}{l} 9 \\ 6 \end{array}) - (\begin{array}{l} 7 \\ - 2 \end{array}) \\ (b) (\begin{matrix} 3 & - 4 \\ 0 & 5 \end{matrix}) - (\begin{matrix} - 7 & - 3 \\ - 6 & 1 \end{matrix}) \end{array}

Solution:

\begin{array}{l} (a) (\begin{array}{l} 9 \\ 6 \end{array}) - (\begin{array}{l} 7 \\ - 2 \end{array}) = (\begin{array}{l} 9 - 7 \\ 6 - (- 2) \end{array}) = (\begin{array}{l} 2 \\ 8 \end{array}) \\ (b) (\begin{matrix} 3 & - 4 \\ 0 & 5 \end{matrix}) - (\begin{matrix} - 7 & - 3 \\ - 6 & 1 \end{matrix}) \\ = (\begin{matrix} 3 - (- 7) & - 4 - (- 3) \\ 0 - (- 6) & 5 - 1 \end{matrix}) \\ = (\begin{matrix} 3 + 7 & - 4 + 3 \\ 0 + 6 & 5 - 1 \end{matrix}) = (\begin{matrix} 10 & - 1 \\ 6 & 4 \end{matrix}) \end{array}

Example 3:
Given that

(\begin{array}{l} p \\ q \end{array}) + (\begin{array}{l} 5 p \\ 9 \end{array}) = (\begin{array}{l} - 12 \\ 3 q + 1 \end{array})

, find the values of p and q.

Solution:

\begin{array}{l} (\begin{array}{l} p \\ q \end{array}) + (\begin{array}{l} 5 p \\ 9 \end{array}) = (\begin{array}{l} - 12 \\ 3 q + 1 \end{array}) \\ (\begin{array}{l} p + 5 p \\ q + 9 \end{array}) = (\begin{array}{l} - 12 \\ 3 q + 1 \end{array}) \end{array}

p + 5p = –12

6p = –12

p = –2

q + 9 = 3q + 1

q – 3q = 1 – 9

–2q = –8

q = 4

Example 4:

Find the values of m and n in the following matrix equation.

(\begin{matrix} 7 & 6 \\ n & 1 \end{matrix}) - (\begin{matrix} - 1 & m \\ - 4 & - 2 \end{matrix}) = (\begin{matrix} 5 & 12 \\ 6 & 11 \end{matrix})

Solution:

(\begin{matrix} 7 & 6 \\ n & 1 \end{matrix}) - (\begin{matrix} - 1 & m \\ - 4 & - 2 \end{matrix}) = (\begin{matrix} 5 & 12 \\ 6 & 11 \end{matrix})

6 – m = 12

–m = 6

m = –6

n – (–4) = 6

n + 4 = 6

n = 2

4.1 Matrix

Posted on April 24, 2020 by user

4.1 Matrices

A matrix is a rectangular array of numbers enclosed in large brackets.

For example (\begin{matrix} 2 & 0 \\ 3 & 1 \end{matrix}) is a matrix .

(A) Rows, Columns and Order of Matrices

1. A matrix which has m rows and n columns is known as a matrix of order m × n.

2. A row matrix is a matrix which has only one row.

Example:

\begin{array}{l} (4), (2 6), (3 8    5) \\ 1 \times 1 1 \times 2 1 \times 3 \\ Only 1 row \end{array}

3. A column matrix is a matrix which has only one column.

Example:

\begin{array}{l} (3), (\begin{array}{l} 2 \\ 6 \end{array}), (\begin{array}{l} - 5 \\ 7 \\ - 9 \end{array}) \\ 1 \times 12 \times 13 \times 1 \\ Only 1 column \end{array}

4. A square matrix is a matrix which has equal number of rows and columns.
Example:

\begin{array}{l} (3), (\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}), (\begin{matrix} 1 & 39 \\ \begin{array}{l} 0 \\ 6 \end{array} & \begin{array}{l} 41 \\ 35 \end{array} \end{matrix}) \\ 1 \times 12 \times 23 \times 3 \\ Number of rows = Number of columns \end{array}