9th class exercise 1.1 notes

Exercise 1.1 of Unit 1 Matrices and Determinants

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Exercise 1.1 Solution

Q.1- Find the order of matrices.

To find the order (dimensions) of a matrix, you simply count the number of rows and columns in the matrix. The order is represented as “m x n,” where “m” is the number of rows and “n” is the number of columns.

Let’s find the order of each given matrix:

Matrix A = [[2, 3], [-5, 6]]
Order of A = 2 x 2 (2 rows, 2 columns)

Matrix B = [[2, 0], [3, 5]]
Order of B = 2 x 2 (2 rows, 2 columns)

Matrix C = [[2, 4]]
Order of C = 1 x 2 (1 row, 2 columns)

Matrix D = [[4], [0], [6]]
Order of D = 3 x 1 (3 rows, 1 column)

Matrix E = [[a, d], [b, e], [c, f]]
Order of E = 3 x 2 (3 rows, 2 columns)

Matrix F = [2]
Order of F = 1 x 1 (1 rows, 1 columns)

Matrix G = [[2, 3, 0], [1, 2, 3], [2, 4, 5]]
Order of G = 3 x 3 (3 rows, 3 columns)

Matrix H = [[2, 3, 4], [1, 0, 6]]
Order of H = 2 x 3 (2 rows, 3 columns)

So, the orders of the given matrices are:

A: 2 x 2
B: 2 x 2
C: 1 x 2
D: 3 x 1
E: 3 x 2
F: 1 x 1
G: 3 x 3
H: 2 x 3

Q.2- Which matrices are equal?

To determine which matrices are equal, we need to compare each pair of matrices based on their elements. Two matrices are equal if and only if they have the same order (dimensions) and all their corresponding elements are equal.

Let’s compare the matrices:

A = [3]
B = [[3, 5]]
C = [5-2]
D = [[5, 3]]
E = [[4, 0], [6, 2]]
F = [[2], [6]]
G = [[3-1], [3+3]]
H = [[4, 0], [6, 2]]
I = [[3, 3+2]]
J = [[2 + 2, 2 – 2], [2 + 4, 2 + 0]]

Let’s compare them:

A: [3]
B: [[3, 5]]
C: [5-2] = [3]
D: [[5, 3]]
E: [[4, 0], [6, 2]]
F: [[2], [6]]
G: [[3-1], [3+3]] = [[2], [6]]
H: [[4, 0], [6, 2]]
I: [[3, 3+2]] = [[3, 5]]
J: [[2 + 2, 2 – 2], [2 + 4, 2 + 0]] = [[4, 0], [6, 2]]

So
A=C
E=H=J
B=I
F=G

Q.3- Find the values of a, b, c and d which satisfy the matrix equation

Given equation:
[[a + c, a + 2b], [c – 1, 4d – 6]] = [[0, -7], [3, 2d]]

Now, let’s equate the elements:

Equating the elements in the first row, first column:
a + c = 0

Equating the elements in the first row, second column:
a + 2b = -7

Equating the elements in the second row, first column:
c – 1 = 3

Equating the elements in the second row, second column:
4d – 6 = 2d

Now, let’s solve these equations:

From equation 1: a + c = 0
From equation 3: c – 1 = 3

Let’s find the value of c from the second equation:
c = 3 + 1
c = 4

Now, we have the value of c. Let’s substitute it into equations 1 and 2:

From equation 1 (a + c = 0):
a + 4 = 0
a = -4

From equation 2 (a + 2b = -7):
-4 + 2b = -7
2b = -7 + 4
2b = -3
b = -3/2

Now, let’s find the value of d from equation 4:

4d – 6 = 2d
4d – 2d = 6
2d = 6
d = 6/2
d = 3

So, the values of a, b, c, and d that satisfy the matrix equation are:
a = -4
b = -3/2
c = 4
d = 3

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