Shortcut Tricks for Solving System of Linear Equation

Duration: 4 min

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  1. Paper-I | Mathematical Abilities: Ratio and Proportion (Ratios), Divisibility Rules, Data Interpretation, Mixture and Alligation, Number System, Speed Time and Distance, Arithmetic, Percentage, Average, Simple Interest and Compound Interest, Profit and Loss, Partnership, Time and Work, Algebra, Mensuration and Geometry, Trigonometry, Linear Algebra
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AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This educational video provides a tutorial on solving systems of linear equations using the Gauss Elimination Method. The instructor presents a system of three linear equations and explains the concept of an augmented matrix, combining coefficients and constants. The lecture involves performing elementary row operations to transform the coefficient matrix into a diagonal or echelon form. Through this process, the instructor systematically isolates variables, starting with z, then y, and finally x, to determine the unique solution. The video concludes by verifying the calculated solution against multiple-choice options written on the whiteboard.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The instructor introduces the problem by writing down three linear equations: x + 3y - 2z = 5, 3x + 5y + 6z = 7, and 2x + 4y + 3z = 8. He labels these as equations (1), (2), and (3). He defines the augmented matrix, noting that coefficients and constants are written together. He constructs the augmented matrix: [1 3 -2 | 5], [3 5 6 | 7], [2 4 3 | 8]. He states the goal is to convert the coefficient matrix into a diagonal matrix. He performs row operations: R2 <- R2 - 3R1, and R3 <- R3 - 2R1. This eliminates the x-term from the second and third rows, resulting in: [1 3 -2 | 5], [0 -4 12 | -8], [0 -2 7 | -2]. He then simplifies the second row by multiplying by -1/4, making the pivot 1. Next, he eliminates the y-term in the third row using R3 <- R3 + 2R2. This results in a matrix where the last row is [0 0 1 | 2], directly giving the value z = 2.

  2. 2:00 – 3:59 02:00-03:59

    With z determined, the instructor moves to back-substitution. He looks at the second row, which corresponds to the equation y - 3z = 2. Substituting z = 2, he calculates y - 6 = 2, leading to y = 8. He then moves to the first row equation x + 3y - 2z = 5. Substituting the known values y = 8 and z = 2, he gets x + 24 - 4 = 5, which simplifies to x + 20 = 5, yielding x = -15. He writes the final solution set x = -15, y = 8, z = 2 on the board. He points to the multiple-choice options on the right side of the whiteboard, specifically option (a) which reads x = -15, y = 8, z = 2. He confirms this matches his calculation, circles the option, and places a checkmark next to it. The video ends with a "Thanks for watching" graphic.

The lecture guides students through the Gauss Elimination method, demonstrating how to convert a system of equations into an augmented matrix, perform row operations to achieve echelon form, and use back-substitution to find the solution. The visual step-by-step breakdown ensures clarity in understanding matrix transformations and variable isolation.

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