9th class exercise 1.6 notes

Exercise 1.6 of Unit 1 Matrices and Determinants

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Exercise 1.6 Solution

Use matrices, if possible, to solve the following systems of linear equations (i) the matrix inversion method (ii) the Cramer’s rule
(i) 2x – 2y = 4
3x + 2y = 6
(ii) 2x + y = 3
6x + 5y = 1

(iii) 4x + 2y = 8
3x – y = – 1
(iv) 3x – 2y = – 6
5x – 2y = – 10
(v) 3x – 2y = 4
-6x + 4y = 7
(vii) 2x -2y = 4
-5x -2y = -10
(vi) 4x + y = 9
-3x – y = -5
(viii) 3x – 4y = 4 x + 2y = 8

Solution
To solve the given systems of linear equations using matrices, we can use two methods: (i) the matrix inversion method and (ii) Cramer’s rule.

(i) Matrix Inversion Method:
In this method, we express the systems of equations in matrix form (AX = B), where A is the coefficient matrix, X is the column matrix of variables, and B is the column matrix of constants. To solve for X, we can use the formula X = A^(-1) * B, where A^(-1) is the inverse of matrix A.

(ii) Cramer’s Rule:
Cramer’s rule can be used to solve systems of linear equations where the number of equations is the same as the number of variables. In this method, we find the determinants of the coefficient matrix (A) and the matrices formed by replacing each column of A with the constants matrix (B) one by one. The solutions for the variables are then given by x = det(Ax) / det(A), y = det(Ay) / det(A), and so on.

Let’s solve each system of equations using both methods:

(i) 2x – 2y = 4
3x + 2y = 6

Matrix form: AX = B
A = [[2, -2], [3, 2]]
X = [[x], [y]]
B = [[4], [6]]

Determinant of A: det(A) = (2 * 2) – (-2 * 3) = 4 + 6 = 10

Inverse of A: A^(-1) = (1/det(A)) * [[2, 2], [-3, 2]] = (1/10) * [[2, 2], [-3, 2]] = [[1/5, 1/5], [-3/10, 1/5]]

Solution using Matrix Inversion Method:
X = A^(-1) * B = [[1/5, 1/5], [-3/10, 1/5]] * [[4], [6]] = [[(1/5 * 4) + (1/5 * 6)], [(-3/10 * 4) + (1/5 * 6)]] = [[2], [1]]

Solution using Cramer’s Rule:
x = det(Ax) / det(A) = [[4, -2], [6, 2]] / 10 = (4 * 2) – (-2 * 6) / 10 = (8 + 12) / 10 = 20 / 10 = 2
y = det(Ay) / det(A) = [[2, 4], [3, 6]] / 10 = (2 * 6) – (4 * 3) / 10 = (12 – 12) / 10 = 0 / 10 = 0

The solutions are x = 2 and y = 0.

(ii) 2x + y = 3
6x + 5y = 1

Matrix form: AX = B
A = [[2, 1], [6, 5]]
X = [[x], [y]]
B = [[3], [1]]

Determinant of A: det(A) = (2 * 5) – (1 * 6) = 10 – 6 = 4

Inverse of A: A^(-1) = (1/det(A)) * [[5, -1], [-6, 2]] = (1/4) * [[5, -1], [-6, 2]] = [[5/4, -1/4], [-6/4, 2/4]] = [[5/4, -1/4], [-3/2, 1/2]]

Solution using Matrix Inversion Method:
X = A^(-1) * B = [[5/4, -1/4], [-3/2, 1/2]] * [[3], [1]] = [[(5/4 * 3) + (-1/4 * 1)], [(-3/2 * 3) + (1/2 * 1)]] = [[14/4], [-7/2]] = [[7/2], [-7/2]]

Solution using Cramer’s Rule:
x = det(Ax) / det(A) = [[3, 1], [1, 5]] / 4 = (3 * 5) – (1 * 1) / 4 = (15 – 1) / 4 = 14 / 4 = 7/2
y = det(Ay) / det(A) = [[2, 3], [6, 1]] / 4 = (2 * 1) – (3 * 6) / 4 = (2 – 18) / 4 = -16 / 4 = -4

The solutions are x = 7/2 and y = -4.

Using the Matrix Inversion Method:

(iii) 4x + 2y = 8
3x – y = -1

Step 1: Write the coefficient matrix and the constant matrix.

Coefficient matrix A:
A = [[4, 2], [3, -1]]

Constant matrix B:
B = [[8], [-1]]

Step 2: Calculate the determinant of matrix A (det(A)).

det(A) = (4 * -1) – (2 * 3) = -4 – 6 = -10

Step 3: Check if det(A) ≠ 0. If det(A) = 0, the system has either no solution or infinite solutions and cannot be solved using the matrix inversion method. Otherwise, it has a unique solution.

In this case, det(A) ≠ 0, so the system has a unique solution.

Step 4: Calculate the inverse of matrix A (A^(-1)).

A^(-1) = (1/det(A)) * adj(A)

where adj(A) is the adjugate (adjoint) of matrix A.

adj(A) = [[-1, -2], [-3, 4]]

A^(-1) = (1/(-10)) * [[-1, -2], [-3, 4]]
A^(-1) = [[1/10, 1/5], [3/10, -2/5]]

Step 5: Calculate the solution matrix X (X = A^(-1) * B).

X = A^(-1) * B
X = [[1/10, 1/5], [3/10, -2/5]] * [[8], [-1]]
X = [[(1/10 * 8) + (1/5 * -1)], [(3/10 * 8) + (-2/5 * -1)]]
X = [[(8/10) – (1/5)], [(24/10) + (2/5)]]
X = [[(4/5) – (1/5)], [(26/10)]]
X = [[3/5], [13/5]]

The solution to the system is x = 3/5 and y = 13/5.

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