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Determinants - Solving system of linear equations using inverse of a matrix

Grade 12CBSE

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

🔑Concepts

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Matrix Representation of Linear Equations: A system of linear equations such as a1x+b1y+c1z=d1a_1x + b_1y + c_1z = d_1, a2x+b2y+c2z=d2a_2x + b_2y + c_2z = d_2, and a3x+b3y+c3z=d3a_3x + b_3y + c_3z = d_3 can be represented as a single matrix equation AX=BAX = B. Here, AA is the coefficient matrix, XX is the column matrix of variables, and BB is the column matrix of constants. Visually, this represents the intersection of three planes in 3D space.

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Non-Singular Matrix and Unique Solution: A square matrix AA is called non-singular if its determinant ∣A∣≠0|A| \neq 0. If AA is non-singular, the system of equations is consistent and possesses a unique solution, which is the single point where all planes or lines intersect.

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The Matrix Inverse Method: To solve for the variables in AX=BAX = B, we pre-multiply both sides by the inverse matrix A−1A^{-1} (if it exists). This results in A−1(AX)=A−1BA^{-1}(AX) = A^{-1}B, leading to IX=A−1BIX = A^{-1}B, or simply X=A−1BX = A^{-1}B. This is the primary tool for finding values of x,y,zx, y, z.

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Adjoint and Inverse Calculation: The inverse of a matrix is calculated using the formula A−1=1∣A∣adj(A)A^{-1} = \frac{1}{|A|} adj(A). The adjoint, adj(A)adj(A), is the transpose of the cofactor matrix. Visually, finding the adjoint involves calculating the minor for each element and applying a 'checkerboard' of signs (±)(\pm).

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Consistency and Inconsistency Criterion: If ∣A∣=0|A| = 0, the system is either inconsistent (no solution) or consistent with infinite solutions. To distinguish, calculate (adjA)B(adj A)B. If (adjA)B≠O(adj A)B \neq O, the system has no solution and is inconsistent, representing parallel or non-intersecting planes. If (adjA)B=O(adj A)B = O, the system has either infinitely many solutions (coincident planes) or no solution.

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Singular Matrix: A matrix AA is singular if ∣A∣=0|A| = 0. In this case, the inverse A−1A^{-1} does not exist. This happens when the rows or columns of the matrix are linearly dependent, meaning the equations do not provide enough independent information to locate a single point of intersection.

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Homogeneous Systems: A system where B=OB = O (all constant terms are zero) is called homogeneous (AX=OAX = O). Such a system always has at least the 'trivial solution' (x=0,y=0,z=0)(x=0, y=0, z=0). If ∣A∣≠0|A| \neq 0, the trivial solution is the only solution.

📐Formulae

AX=BAX = B

∣A∣=∑j=1na1jC1j|A| = \sum_{j=1}^{n} a_{1j} C_{1j} (Expansion along the first row)

Cij=(−1)i+jMijC_{ij} = (-1)^{i+j} M_{ij} (Cofactor formula where MM is the minor)

adj(A)=[Cij]Tadj(A) = [C_{ij}]^T

A−1=1∣A∣adj(A)A^{-1} = \frac{1}{|A|} adj(A)

X=A−1BX = A^{-1}B

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

Condition for No Solution: ∣A∣=0|A| = 0 and (adjA)B≠O(adj A)B \neq O

💡Examples

Problem 1:

Solve the following system of equations using matrix method: 5x+2y=45x + 2y = 4 and 7x+3y=57x + 3y = 5

Solution:

  1. Write in AX=BAX = B form: A=[5273]A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}, X=[xy]X = \begin{bmatrix} x \\ y \end{bmatrix}, B=[45]B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}.

  2. Find ∣A∣|A|: ∣A∣=(5)(3)−(7)(2)=15−14=1|A| = (5)(3) - (7)(2) = 15 - 14 = 1. Since ∣A∣≠0|A| \neq 0, A−1A^{-1} exists.

  3. Find adj(A)adj(A): For a 2×22 \times 2 matrix, swap diagonal elements and change signs of off-diagonal elements. adj(A)=[3−2−75]adj(A) = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}.

  4. Calculate A−1A^{-1}: A−1=11[3−2−75]=[3−2−75]A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}.

  5. Solve X=A−1BX = A^{-1}B: X=[3−2−75][45]=[(3)(4)+(−2)(5)(−7)(4)+(5)(5)]=[12−10−28+25]=[2−3]X = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} (3)(4) + (-2)(5) \\ (-7)(4) + (5)(5) \end{bmatrix} = \begin{bmatrix} 12 - 10 \\ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}.

Therefore, x=2,y=−3x = 2, y = -3.

Explanation:

This example demonstrates the step-by-step application of the matrix inverse method for a 2×22 \times 2 system. We first verify the existence of a solution by checking the determinant, calculate the adjoint, find the inverse, and multiply it by the constant matrix BB to find the variables.

Problem 2:

Solve the system: x−y+z=4x - y + z = 4, 2x+y−3z=02x + y - 3z = 0, x+y+z=2x + y + z = 2 using matrix inverse.

Solution:

  1. Matrix form: A=[1−1121−3111]A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}, X=[xyz]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B=[402]B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}.

  2. ∣A∣=1(1+3)−(−1)(2+3)+1(2−1)=4+5+1=10|A| = 1(1+3) - (-1)(2+3) + 1(2-1) = 4 + 5 + 1 = 10. (∣A∣≠0|A| \neq 0)

  3. Cofactors: A11=4,A12=−5,A13=1,A21=2,A22=0,A23=−2,A31=2,A32=5,A33=3A_{11}=4, A_{12}=-5, A_{13}=1, A_{21}=2, A_{22}=0, A_{23}=-2, A_{31}=2, A_{32}=5, A_{33}=3.

  4. adj(A)=[422−5051−23]adj(A) = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}.

  5. A−1=110[422−5051−23]A^{-1} = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}.

  6. X=A−1B=110[422−5051−23][402]=110[16+0+4−20+0+104+0+6]=110[20−1010]=[2−11]X = A^{-1}B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}.

Thus, x=2,y=−1,z=1x=2, y=-1, z=1.

Explanation:

For a 3×33 \times 3 system, we follow the same logical flow: establish the matrix equation, find the determinant to ensure a unique solution, calculate all nine cofactors to form the adjoint, and finally compute X=A−1BX = A^{-1}B through matrix multiplication.