9th class exercise 1.4 notes

Exercise 1.4 of Unit 1 Matrices and Determinants

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Exercise 1.4 Solution

Find the product of following matrices;
(i) [[1, 2]] * [[4], [0]]
(ii )[1 2] [[5], [- 4]]
(iii) [-3 0] [[4], [0]]
(iv) [6 + 0] * [[4], [0]]
(v) [[1, 2], [- 3, 0], [6, – 1]] * [[4, 5], [0, – 4]]

Solution:
To find the product of two matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix. The resulting product will have the number of rows from the first matrix and the number of columns from the second matrix.

Let’s find the products of the given matrices:

(i) [[1, 2]] * [[4], [0]]

The first matrix is a 1×2 matrix, and the second matrix is a 2×1 matrix.
Product = 1 * 4 + 2 * 0 = 4

(ii) [1 2] * [[5], [-4]]

The first matrix is a 1×2 matrix, and the second matrix is a 2×1 matrix.
Product = (1 * 5) + (2 * -4) = 5 – 8 = -3

(iii) [-3 0] * [[4], [0]]

The first matrix is a 1×2 matrix, and the second matrix is a 2×1 matrix.
Product = (-3 * 4) + (0 * 0) = -12 + 0 = -12

(iv) [6 0] * [[4], [0]]

The first matrix is a 1×2 matrix, and the second matrix is a 2×1 matrix.
Product = (6 * 4) + (0 * 0) = 24 + 0 = 24

(v) [[1, 2], [-3, 0], [6, -1]] * [[4, 5], [0, -4]]

The first matrix is a 3×2 matrix, and the second matrix is a 2×2 matrix.
The resulting product will be a 3×2 matrix.

Product:

First element of the resulting matrix:
(1 * 4) + (2 * 0) = 4

Second element of the resulting matrix:
(1 * 5) + (2 * -4) = 5 – 8 = -3

Third element of the resulting matrix:
(-3 * 4) + (0 * 0) = -12 + 0 = -12

Fourth element of the resulting matrix:
(-3 * 5) + (0 * -4) = -15 + 0 = -15

Fifth element of the resulting matrix:
(6 * 4) + (-1 * 0) = 24 + 0 = 24

Sixth element of the resulting matrix:
(6 * 5) + (-1 * -4) = 30 + 4 = 34

The product of the matrices is:
[[4, -3], [-12, -15], [24, 34]]

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