Determinants Ex 4.6: Solving Systems of Linear Equations by Matrix Method

Understanding Systems of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. For instance, a system with two variables x and y might look like:

a₁x + b₁y = c₁
a₂x + b₂y = c₂

And for three variables x, y, z:

a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃

The matrix method allows us to represent such systems in a compact form: AX = B.

Here's what each part represents:

• A is the coefficient matrix, formed by the coefficients of the variables. For a 3x3 system, A = [[a₁, b₁, c₁], [a₂, b₂, c₂], [a₃, b₃, c₃]].
• X is the variable matrix (or column vector), containing the variables. For a 3x3 system, X = [[x], [y], [z]].
• B is the constant matrix (or column vector), containing the constants on the right-hand side of the equations. For a 3x3 system, B = [[d₁], [d₂], [d₃]].

This matrix representation simplifies the process of solving and analyzing the system significantly.

Solving Systems Using Matrix Inversion: A Step-by-Step Guide

The core idea behind the matrix inversion method is to isolate the variable matrix X. If we have AX = B and the inverse of matrix A (denoted as A⁻¹) exists, we can pre-multiply both sides by A⁻¹:

A⁻¹(AX) = A⁻¹B
(A⁻¹A)X = A⁻¹B (by associativity of matrix multiplication)
IX = A⁻¹B (since A⁻¹A = I, the identity matrix)
X = A⁻¹B

This formula, X = A⁻¹B, is the key! It tells us that if we can find A⁻¹, we can directly find the values of x, y, z by multiplying A⁻¹ with B.

When does A⁻¹ exist? The inverse of a matrix A exists if and only if its determinant, |A|, is non-zero (i.e., |A| ≠ 0). If |A| ≠ 0, the system has a unique solution.

Steps to solve a system of linear equations using the matrix method:

1. Write the system in matrix form AX = B. Identify matrices A, X, and B.
2. Calculate the determinant of the coefficient matrix A, i.e., |A|.
3. Check the value of |A|:

• If |A| ≠ 0, proceed to step 4. A unique solution exists.
• If |A| = 0, the system may have infinitely many solutions or no solution. You must proceed to the consistency analysis (covered in the next section).

1. Find the adjoint of matrix A, denoted as adj A.
2. **Calculate the inverse of matrix A, A⁻¹ = (1/|A|) * adj A**.
3. Calculate the variable matrix X by multiplying A⁻¹ with B: X = A⁻¹B.
4. Identify the values of x, y, z from the matrix X.

Consistency and Inconsistency of Systems of Equations

Understanding whether a system of linear equations has a solution (and how many) is crucial. This property is called consistency.

1. Consistent System: A system of linear equations is said to be consistent if it has at least one solution. This solution can be unique or there can be infinitely many solutions.

• Case 1: Unique Solution: This occurs when |A| ≠ 0. If the determinant of the coefficient matrix is non-zero, then A⁻¹ exists, and the system AX = B has a unique solution given by X = A⁻¹B. This is the most straightforward case.
• Case 2: Infinitely Many Solutions: This happens when |A| = 0 AND (adj A)B = 0 (where 0 denotes the zero matrix). In this scenario, the equations are dependent, meaning one equation can be derived from the others, leading to an infinite number of solutions. You usually express these solutions in terms of a parameter.

2. Inconsistent System: A system of linear equations is said to be inconsistent if it has no solution.

• Case: No Solution: This occurs when |A| = 0 AND (adj A)B ≠ 0 (where 0 denotes the zero matrix). When |A|=0, the matrix A is singular, meaning its rows/columns are linearly dependent. If (adj A)B is also non-zero, it signifies a contradiction within the system, making it impossible for any values of x, y, z to satisfy all equations simultaneously.

Summary for |A| = 0:

• Calculate (adj A)B.
• If (adj A)B = 0 (zero matrix), the system is consistent with infinitely many solutions.
• If (adj A)B ≠ 0 (non-zero matrix), the system is inconsistent with no solution.

Solved Examples: Step-by-Step Application

• Example 1: Solving a 2x2 system with a unique solution Solve the system of equations using the matrix method: 2x + 5y = 1 3x + 2y = 7 Step 1: Write the system in matrix form AX = B. A = [[2, 5], [3, 2]], X = [[x], [y]], B = [[1], [7]] Step 2: Calculate the determinant of A. |A| = (2 2) - (5 3) = 4 - 15 = -11 Step 3: Check |A|. Since |A| = -11 ≠ 0, a unique solution exists. Step 4: Find the adjoint of A. adj A = [[2, -5], [-3, 2]] (Swap diagonal elements, change signs of off-diagonal) Step 5: Calculate the inverse of A. A⁻¹ = (1/|A|) adj A = (1/-11) [[2, -5], [-3, 2]] = [[-2/11, 5/11], [3/11, -2/11]] Step 6: Calculate X = A⁻¹B. X = [[-2/11, 5/11], [3/11, -2/11]] [[1], [7]] X = [[(-2/11)1 + (5/11)7], [(3/11)1 + (-2/11)*7]] X = [[(-2 + 35)/11], [(3 - 14)/11]] X = [[33/11], [-11/11]] X = [[3], [-1]] Step 7: Identify the values of x and y. x = 3, y = -1 Final answer: The solution is x=3, y=-1.
• Example 2: Solving a 3x3 system with a unique solution Solve the system: x - y + z = 4 2x + y - 3z = 0 x + y + z = 2 Step 1: Write the system in matrix form AX = B. A = [[1, -1, 1], [2, 1, -3], [1, 1, 1]], X = [[x], [y], [z]], B = [[4], [0], [2]] Step 2: Calculate the determinant of A. |A| = 1(11 - (-3)1) - (-1)(21 - (-3)1) + 1(21 - 11) |A| = 1(1 + 3) + 1(2 + 3) + 1(2 - 1) |A| = 1(4) + 1(5) + 1(1) = 4 + 5 + 1 = 10 Step 3: Check |A|. Since |A| = 10 ≠ 0, a unique solution exists. Step 4: Find the adjoint of A (adj A). First, find the cofactor matrix C: C₁₁ = 1(1 - (-3)) = 4 C₁₂ = -1(2 - (-3)) = -5 C₁₃ = 1(2 - 1) = 1 C₂₁ = -1(-1 - 1) = 2 C₂₂ = 1(1 - 1) = 0 C₂₃ = -1(1 - (-1)) = -2 C₃₁ = 1(3 - 1) = 2 C₃₂ = -1(-3 - 2) = 5 C₃₃ = 1(1 - (-2)) = 3 C = [[4, -5, 1], [2, 0, -2], [2, 5, 3]] adj A = Cᵀ = [[4, 2, 2], [-5, 0, 5], [1, -2, 3]] Step 5: Calculate the inverse of A. A⁻¹ = (1/10) [[4, 2, 2], [-5, 0, 5], [1, -2, 3]] Step 6: Calculate X = A⁻¹B. X = (1/10) [[4, 2, 2], [-5, 0, 5], [1, -2, 3]] [[4], [0], [2]] X = (1/10) [[44 + 20 + 22], [-54 + 00 + 52], [14 + (-2)0 + 32]] X = (1/10) [[16 + 0 + 4], [-20 + 0 + 10], [4 + 0 + 6]] X = (1/10) * [[20], [-10], [10]] X = [[2], [-1], [1]] Step 7: Identify the values of x, y, and z. x = 2, y = -1, z = 1 Final answer: The solution is x=2, y=-1, z=1.
• Example 3: System with |A|=0 (Checking Consistency) Examine the consistency of the system: x + 2y = 5 2x + 4y = 10 Step 1: Write in matrix form AX = B. A = [[1, 2], [2, 4]], X = [[x], [y]], B = [[5], [10]] Step 2: Calculate |A|. |A| = (1 4) - (2 2) = 4 - 4 = 0 Step 3: Check |A|. Since |A| = 0, the system either has no solution or infinitely many solutions. We need to check (adj A)B. Step 4: Find adj A. adj A = [[4, -2], [-2, 1]] Step 5: Calculate (adj A)B. (adj A)B = [[4, -2], [-2, 1]] [[5], [10]] (adj A)B = [[45 + (-2)10], [(-2)5 + 1*10]] (adj A)B = [[20 - 20], [-10 + 10]] (adj A)B = [[0], [0]] (Zero matrix) Step 6: Conclude consistency. Since |A| = 0 and (adj A)B = 0, the system is consistent and has infinitely many solutions. Final answer: The system is consistent and has infinitely many solutions.

Exam Strategy and Common Pitfalls

To score well in questions related to Determinants Ex 4.6, pay attention to these critical points:

1. Careful Calculations: Matrix inversion involves calculating determinants, cofactors, and adjoints, followed by matrix multiplication. Each step is prone to sign errors or arithmetic mistakes. Double-check your calculations, especially when finding cofactors and their signs.
2. Order of Multiplication: Remember that in X = A⁻¹B, the order matters. It's A⁻¹ multiplied by B, not B by A⁻¹. Matrix multiplication is generally not commutative.
3. Don't Stop at |A|=0: A common mistake is to conclude 'no solution' immediately if |A|=0. This is only half the story! When |A|=0, you must calculate (adj A)B. Only if (adj A)B ≠ 0 is there no solution. If (adj A)B = 0, then infinitely many solutions exist.
4. Practice Cofactor Signs: The sign pattern for cofactors (-1)^(i+j) is crucial. Visualise [[+, -, +], [-, +, -], [+, -, +]] for a 3x3 matrix to avoid errors.
5. Present Your Work Clearly: For board exams, show every step: setting up AX=B, calculating |A|, adj A, A⁻¹, and finally X = A⁻¹B. Clearly state your conclusion about consistency and the solution set. This structured approach helps in fetching step-wise marks even if the final answer has a minor error.

Practice Questions with Solutions

• Q: Solve the following system of linear equations using the matrix method: 3x - 2y = 3 2x + y = 16 A: Step 1: Write in matrix form AX = B. A = [[3, -2], [2, 1]], X = [[x], [y]], B = [[3], [16]] Step 2: Calculate |A|. |A| = (3 1) - (-2 2) = 3 + 4 = 7 Step 3: Since |A| = 7 ≠ 0, a unique solution exists. Step 4: Find adj A. adj A = [[1, 2], [-2, 3]] Step 5: Calculate A⁻¹. A⁻¹ = (1/7) [[1, 2], [-2, 3]] Step 6: Calculate X = A⁻¹B. X = (1/7) [[1, 2], [-2, 3]] [[3], [16]] X = (1/7) [[13 + 216], [-23 + 316]] X = (1/7) [[3 + 32], [-6 + 48]] X = (1/7) [[35], [42]] X = [[5], [6]] Final answer: x = 5, y = 6.
• Q: Using the matrix method, solve the system of equations: x + y + z = 6 y + 3z = 11 x - 2y + z = 0 A: Step 1: Write in matrix form AX = B. A = [[1, 1, 1], [0, 1, 3], [1, -2, 1]], X = [[x], [y], [z]], B = [[6], [11], [0]] Step 2: Calculate |A|. |A| = 1(11 - 3(-2)) - 1(01 - 31) + 1(0(-2) - 11) |A| = 1(1 + 6) - 1(0 - 3) + 1(0 - 1) |A| = 7 + 3 - 1 = 9 Step 3: Since |A| = 9 ≠ 0, a unique solution exists. Step 4: Find adj A (cofactor matrix C first). C₁₁ = 7, C₁₂ = 3, C₁₃ = -1 C₂₁ = -(11 - 1(-2)) = -(1 + 2) = -3 C₂₂ = (11 - 11) = 0 C₂₃ = -(1(-2) - 11) = -(-2 - 1) = 3 C₃₁ = (13 - 11) = 2 C₃₂ = -(13 - 10) = -3 C₃₃ = (11 - 10) = 1 C = [[7, 3, -1], [-3, 0, 3], [2, -3, 1]] adj A = Cᵀ = [[7, -3, 2], [3, 0, -3], [-1, 3, 1]] Step 5: Calculate A⁻¹. A⁻¹ = (1/9) [[7, -3, 2], [3, 0, -3], [-1, 3, 1]] Step 6: Calculate X = A⁻¹B. X = (1/9) [[7, -3, 2], [3, 0, -3], [-1, 3, 1]] [[6], [11], [0]] X = (1/9) [[76 + (-3)11 + 20], [36 + 011 + (-3)0], [(-1)6 + 311 + 10]] X = (1/9) [[42 - 33 + 0], [18 + 0 + 0], [-6 + 33 + 0]] X = (1/9) * [[9], [18], [27]] X = [[1], [2], [3]] Final answer: x = 1, y = 2, z = 3.
• Q: Examine the consistency of the system of equations: 5x - y + 4z = 5 2x + 3y + 5z = 2 5x - 2y + 6z = -1 A: Step 1: Write in matrix form AX = B. A = [[5, -1, 4], [2, 3, 5], [5, -2, 6]], X = [[x], [y], [z]], B = [[5], [2], [-1]] Step 2: Calculate |A|. |A| = 5(36 - 5(-2)) - (-1)(26 - 55) + 4(2(-2) - 35) |A| = 5(18 + 10) + 1(12 - 25) + 4(-4 - 15) |A| = 5(28) + 1(-13) + 4(-19) |A| = 140 - 13 - 76 = 140 - 89 = 51 Step 3: Since |A| = 51 ≠ 0, the system is consistent and has a unique solution. Final answer: The system is consistent with a unique solution.
• Q: Examine the consistency of the system of equations: x + y + z = 1 2x + 3y + 2z = 2 ax + ay + 2az = 4 A: Step 1: Write in matrix form AX = B. A = [[1, 1, 1], [2, 3, 2], [a, a, 2a]], X = [[x], [y], [z]], B = [[1], [2], [4]] Step 2: Calculate |A|. |A| = 1(32a - 2a) - 1(22a - 2a) + 1(2a - 3a) |A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a) |A| = 4a - 2a - a = a Step 3: Check |A|. Case 1: If |A| ≠ 0, i.e., a ≠ 0, then the system is consistent with a unique solution. Case 2: If |A| = 0, i.e., a = 0. If a = 0, then A = [[1, 1, 1], [2, 3, 2], [0, 0, 0]]. |A|=0 is confirmed. Now we need to check (adj A)B when a=0. When a=0, the system becomes: x + y + z = 1 2x + 3y + 2z = 2 0x + 0y + 0z = 4 The last equation 0 = 4 is a contradiction. Alternatively, let's find adj A for a=0. C₁₁ = 0, C₁₂ = 0, C₁₃ = 0 (since last row is all zeros) C₂₁ = 0, C₂₂ = 0, C₂₃ = 0 C₃₁ = 1(2 - 3) = -1 C₃₂ = -(12 - 12) = 0 C₃₃ = 1(3 - 2) = 1 adj A = [[0, 0, -1], [0, 0, 0], [0, 0, 1]] (Mistake in calculation of C₃. If row is all zero, determinant of minor will be zero leading to Cofactors being zero for first 2 rows. But for 3rd row, Cofactors are not necessarily zero unless 2 rows are identical. Let's recalculate the cofactors for A = [[1, 1, 1], [2, 3, 2], [0, 0, 0]]. C₁₁ = 30 - 20 = 0 C₁₂ = -(20 - 20) = 0 C₁₃ = 20 - 30 = 0 C₂₁ = -(10 - 10) = 0 C₂₂ = 10 - 10 = 0 C₂₃ = -(10 - 10) = 0 C₃₁ = 12 - 13 = -1 C₃₂ = -(12 - 12) = 0 C₃₃ = 13 - 12 = 1 adj A = Cᵀ = [[0, 0, -1], [0, 0, 0], [0, 0, 1]] (Corrected C₂₁, C₂₂, C₂₃) Now (adj A)B = [[0, 0, -1], [0, 0, 0], [0, 0, 1]] [[1], [2], [4]] (adj A)B = [[01 + 02 + (-1)4], [01 + 02 + 04], [01 + 02 + 14]] (adj A)B = [[-4], [0], [4]] which is not the zero matrix. Since |A|=0 and (adj A)B ≠ 0 (specifically [[ -4 ], [ 0 ], [ 4 ]]), the system is inconsistent when a=0. Final answer: The system is consistent with a unique solution if a ≠ 0. The system is inconsistent (no solution) if a = 0.

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