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Algebra - Solving system of linear equations using matrix method

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Matrix Representation: A system of linear equations can be represented in the form AX=BAX = B, where AA is the coefficient matrix, XX is the column matrix of variables, and BB is the column matrix of constants. Visually, you can imagine the coefficients of each equation forming the horizontal rows of matrix AA.

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Consistency and Uniqueness: A system is consistent if it has at least one solution. If the determinant ∣A∣≠0|A| \neq 0, the system is non-singular and has a unique solution. Geometrically, in a 3D system, this represents three planes intersecting at exactly one specific point (x,y,z)(x, y, z).

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Singular Matrices and Solutions: If ∣A∣=0|A| = 0, the matrix AA is singular. In this case, the system may have either infinitely many solutions or no solution at all. This occurs visually when planes are parallel or overlap in a way that they don't meet at a single point.

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The Role of the Adjoint Matrix: To solve for variables, we use the adjoint of matrix AA, denoted as adj(A)adj(A), which is the transpose of the cofactor matrix. The adjoint helps 'undo' the effect of matrix AA on the variables XX.

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Matrix Inversion Method: The solution to the system is found using the formula X=A−1BX = A^{-1}B. This is only applicable when AA is non-singular. The process involves finding the determinant, then the cofactors, then the adjoint, and finally multiplying the inverse by matrix BB.

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Infinite Solutions vs. Inconsistency: When ∣A∣=0|A| = 0, we calculate (adjA)B(adj A)B. If (adjA)B=O(adj A)B = O (the zero matrix), the system has infinitely many solutions (planes coincide along a line or a plane). If (adjA)B≠O(adj A)B \neq O, the system is inconsistent and has no solution (planes are parallel and distinct).

📐Formulae

System Form: AX=BAX = B

Inverse of Matrix A: A−1=1∣A∣adj(A)A^{-1} = \frac{1}{|A|} adj(A) where ∣A∣≠0|A| \neq 0

Solution Formula: X=A−1BX = A^{-1}B

Determinant of 2x2 Matrix: det⁡(abcd)=ad−bc\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc

Condition for Unique Solution: ∣A∣≠0|A| \neq 0

Condition for Inconsistent System: ∣A∣=0|A| = 0 and (adjA)B≠O(adj A)B \neq O

💡Examples

Problem 1:

Solve the following system of equations using the matrix method: x+y+z=6x + y + z = 6 y+3z=11y + 3z = 11 x−2y+z=0x - 2y + z = 0

Solution:

Step 1: Write the system in matrix form AX=BAX = B. A=(1110131−21),X=(xyz),B=(6110)A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{pmatrix}, X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, B = \begin{pmatrix} 6 \\ 11 \\ 0 \end{pmatrix}

Step 2: Find the determinant ∣A∣|A|. ∣A∣=1(1−(−6))−1(0−3)+1(0−1)|A| = 1(1 - (-6)) - 1(0 - 3) + 1(0 - 1) ∣A∣=1(7)+3−1=9|A| = 1(7) + 3 - 1 = 9 Since ∣A∣≠0|A| \neq 0, a unique solution exists.

Step 3: Find the matrix of cofactors CijC_{ij} and adj(A)adj(A). C11=7,C12=3,C13=−1C_{11} = 7, C_{12} = 3, C_{13} = -1 C21=−3,C22=0,C23=3C_{21} = -3, C_{22} = 0, C_{23} = 3 C31=2,C32=−3,C33=1C_{31} = 2, C_{32} = -3, C_{33} = 1 adj(A)=(7−3230−3−131)adj(A) = \begin{pmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{pmatrix}

Step 4: Calculate X=A−1B=1∣A∣(adjA)BX = A^{-1}B = \frac{1}{|A|} (adj A)B. X=19(7−3230−3−131)(6110)X = \frac{1}{9} \begin{pmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{pmatrix} \begin{pmatrix} 6 \\ 11 \\ 0 \end{pmatrix} X=19((7)(6)+(−3)(11)+(2)(0)(3)(6)+(0)(11)+(−3)(0)(−1)(6)+(3)(11)+(1)(0))X = \frac{1}{9} \begin{pmatrix} (7)(6) + (-3)(11) + (2)(0) \\ (3)(6) + (0)(11) + (-3)(0) \\ (-1)(6) + (3)(11) + (1)(0) \end{pmatrix} X=19(42−3318−6+33)=19(91827)=(123)X = \frac{1}{9} \begin{pmatrix} 42 - 33 \\ 18 \\ -6 + 33 \end{pmatrix} = \frac{1}{9} \begin{pmatrix} 9 \\ 18 \\ 27 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} Therefore, x=1,y=2,z=3x = 1, y = 2, z = 3.

Explanation:

This solution follows the standard matrix inversion path. We first verify the system is solvable (non-singular), then calculate the adjoint via cofactors, and finally perform matrix multiplication to find the specific coordinates where the three planes intersect.