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Number and Algebra - Systems of Linear Equations

Grade 12IB

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

πŸ”‘Concepts

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Linear Equations and Geometric Representation: A linear equation in two variables, such as ax+by=cax + by = c, represents a straight line in a 2D Cartesian plane. In three variables, ax+by+cz=dax + by + cz = d represents a flat, infinite plane in 3D space. A system of these equations represents the search for common intersection points between these lines or planes.

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Matrix Representation of Systems: A system of linear equations can be compactly represented as a matrix equation AX=BAX = B. Here, AA is the coefficient matrix containing the numbers multiplying the variables, XX is the column vector of variables (e.g., x,y,zx, y, z), and BB is the column vector of constants from the right side of the equations.

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The Role of the Determinant: For a system with the same number of equations and variables, a unique solution exists if and only if the determinant of the coefficient matrix is non-zero (det(A)β‰ 0det(A) \neq 0). Visually, this means the lines or planes intersect at exactly one specific point.

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Consistent vs. Inconsistent Systems: A system is 'consistent' if it has at least one solution and 'inconsistent' if it has no solutions. Geometrically, inconsistency in 2D occurs when lines are parallel and distinct. In 3D, it can occur when planes are parallel or when they intersect in a way that no single point is common to all three (forming a triangular prism shape).

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Dependent Systems and Infinite Solutions: If the equations are redundant (e.g., one is a multiple of another), the system is dependent and has infinitely many solutions. Geometrically, this means the lines are coincident (lie on top of each other) or the planes intersect along a common line or are the same plane. These solutions are often expressed using a parameter like Ξ»\lambda.

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Elementary Row Operations: These are tools used to simplify the augmented matrix [A∣B][A | B] without changing the solution set. They include swapping two rows (Ri↔RjR_i \leftrightarrow R_j), multiplying a row by a non-zero constant (kRiβ†’RikR_i \to R_i), and adding a multiple of one row to another (Ri+kRjβ†’RiR_i + kR_j \to R_i).

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Reduced Row Echelon Form (RREF): By applying row operations, a matrix is transformed into RREF, where the leading coefficient of each row is 11 and is the only non-zero entry in its column. Visually, this process is equivalent to 'straightening' the planes until they align with the axes, making the intersection point (x,y,z)(x, y, z) obvious.

πŸ“Formulae

ax+by+cz=dax + by + cz = d

AX=BAX = B

X=Aβˆ’1BΒ (ifΒ det(A)β‰ 0)X = A^{-1}B \text{ (if } det(A) \neq 0)

det(A)=adβˆ’bcΒ (forΒ 2x2Β matrix)det(A) = ad - bc \text{ (for 2x2 matrix)}

Ri↔Rj,kRiβ†’Ri,Ri+kRjβ†’RiR_i \leftrightarrow R_j, \quad kR_i \to R_i, \quad R_i + kR_j \to R_i

x=det(Ax)det(A),y=det(Ay)det(A)Β (Cramer’sΒ Rule)x = \frac{det(A_x)}{det(A)}, \quad y = \frac{det(A_y)}{det(A)} \text{ (Cramer's Rule)}

πŸ’‘Examples

Problem 1:

Solve the following system of linear equations: 2x+3y=82x + 3y = 8 xβˆ’y=βˆ’1x - y = -1

Solution:

Step 1: Use the substitution or elimination method. From the second equation, x=yβˆ’1x = y - 1. Step 2: Substitute this into the first equation: 2(yβˆ’1)+3y=82(y - 1) + 3y = 8. Step 3: Expand and simplify: 2yβˆ’2+3y=8β€…β€ŠβŸΉβ€…β€Š5y=10β€…β€ŠβŸΉβ€…β€Šy=22y - 2 + 3y = 8 \implies 5y = 10 \implies y = 2. Step 4: Solve for xx: x=2βˆ’1=1x = 2 - 1 = 1. The solution is (1,2)(1, 2).

Explanation:

This is a 2D system where two lines intersect at a unique point. The non-zero determinant of the coefficients det(231βˆ’1)=βˆ’2βˆ’3=βˆ’5det \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix} = -2 - 3 = -5 confirms a unique solution exists.

Problem 2:

Use an augmented matrix to solve: x+y+z=6x + y + z = 6 2xβˆ’y+z=32x - y + z = 3 x+2yβˆ’z=2x + 2y - z = 2

Solution:

Step 1: Write the augmented matrix: (111∣62βˆ’11∣312βˆ’1∣2)\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 2 & -1 & 1 & | & 3 \\ 1 & 2 & -1 & | & 2 \end{pmatrix} Step 2: Perform R2βˆ’2R1β†’R2R_2 - 2R_1 \to R_2 and R3βˆ’R1β†’R3R_3 - R_1 \to R_3: (111∣60βˆ’3βˆ’1βˆ£βˆ’901βˆ’2βˆ£βˆ’4)\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & -3 & -1 & | & -9 \\ 0 & 1 & -2 & | & -4 \end{pmatrix} Step 3: Swap R2R_2 and R3R_3, then perform R3+3R2β†’R3R_3 + 3R_2 \to R_3: (111∣601βˆ’2βˆ£βˆ’400βˆ’7βˆ£βˆ’21)\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & -2 & | & -4 \\ 0 & 0 & -7 & | & -21 \end{pmatrix} Step 4: Back-substitute: βˆ’7z=βˆ’21β€…β€ŠβŸΉβ€…β€Šz=3-7z = -21 \implies z = 3. From R2R_2, yβˆ’2(3)=βˆ’4β€…β€ŠβŸΉβ€…β€Šy=2y - 2(3) = -4 \implies y = 2. From R1R_1, x+2+3=6β€…β€ŠβŸΉβ€…β€Šx=1x + 2 + 3 = 6 \implies x = 1. The solution is (1,2,3)(1, 2, 3).

Explanation:

This example demonstrates Gaussian elimination. The 3D planes intersect at a single unique point (1,2,3)(1, 2, 3) because the row reduction resulted in three non-zero pivots.