Notes on Solving Systems of Equations and Matrices

Elimination and substitution strategies for systems

  • The goal of elimination is to find a variable with coefficients that are equal in magnitude but opposite in sign, so adding the equations cancels that variable.

  • When you have two equations with the same sign for one variable and opposite sign coefficients for the other, decide which variable to eliminate by agreement (e.g., voting among options).

  • General idea: multiply one equation by a constant so that the chosen variable has opposite coefficients in the two equations, then add (or subtract) the equations to eliminate that variable.

Worked example (elimination) to solve for (c, d)

  • Given a system with an elimination path leading to

    • First equation: c−2d=29c - 2d = 29

    • Second equation (after appropriate operation, e.g., multiply by -2 and add): the elimination yields a relation on c alone: −3c=51 ,-3c = 51\,,
      so c=−17.c = -17.

  • Substitute back into the first equation to solve for d:

    • c−2d=29c - 2d = 29 becomes −17−2d=29-17 - 2d = 29
      so −2d=46⇒d=−23.-2d = 46\Rightarrow d = -23.

  • Therefore, the ordered pair (c, d) = (−17,−23)(-17, -23) (alphabetical order by variable name: c, d).

Substitution method (illustrative path and decimal handling)

  • When solving by substitution, you may have a variable already isolated (e.g., in a solved-for form like y=frac32x+2y = frac{3}{2}x + 2).

  • To avoid decimals, you can clear them by multiplying both sides of the equation, and/or the entire system, by a common multiple (e.g., 10 if necessary) to obtain integer coefficients.

  • Example setup:

    • From a system: y=frac32x+2y = frac{3}{2}x + 2 and 6x−4y=46x - 4y = 4

    • Substituting y into the second equation gives: 6x−4(32x+2)=46x - 4\left(\tfrac{3}{2}x + 2\right) = 4 → 6x−6x−8=46x - 6x - 8 = 4 → −8=4-8 = 4, which is a contradiction.

    • Conclusion: no solution (the two lines are parallel and never meet).

  • Alternative method: clear decimals first, then substitute; either way, you can conclude whether there is a unique solution, no solution, or infinitely many solutions depending on whether the equations intersect, are parallel, or are the same line.

No solution vs. infinitely many solutions vs. a unique solution

  • No solution: the system corresponds to two parallel lines; there is no ordered pair that satisfies both equations.

  • Infinitely many solutions: the two equations represent the same line (one equation is a multiple of the other); every point on the line is a solution. This yields a whole line of ordered pairs (e.g., all (c, d) satisfying a single linear relation).

  • Unique solution: the two lines intersect at a single point; that point is the solution to the system.

Introducing matrices for systems

Quick definitions

  • A matrix is a rectangular array of numbers in rows and columns.

  • Dimensions: a matrix with r rows and s columns is called an r×s matrix.

  • Example dimensions:

    • A matrix with 2 rows and 3 columns is a 2×3 matrix.

    • A matrix with 3 rows and 2 columns is a 3×2 matrix.

  • Each entry is called an element (or entry).

Augmented matrices

  • A system of linear equations can be written as an augmented matrix, grouping the coefficients of the variables and the constants on the right of a dividing line (optional dotted line visualization).

  • Example: the system

    • 5x−2y=35x - 2y = 3

    • 4x+3y=104x + 3y = 10
      becomes the augmented matrix
      [5amp;−2amp;∣amp;3 4amp;3amp;∣amp;10]\begin{bmatrix}5 & -2 & | & 3 \ 4 & 3 & | & 10\end{bmatrix}
      which is often written more compactly as
      [5amp;−2amp;3 4amp;3amp;10]\begin{bmatrix}5 & -2 & 3 \ 4 & 3 & 10\end{bmatrix}

  • Another example from the video:

    • System: −x+5y=10-x + 5y = 10 and 3x+9y=−63x + 9y = -6

    • Augmented form: [−1amp;5amp;∣amp;10 3amp;9amp;∣amp;−6]\begin{bmatrix}-1 & 5 & | & 10 \ 3 & 9 & | & -6\end{bmatrix}

  • A fully solved system (when the left-hand side forms the identity matrix) yields the solution directly from the right-hand side, e.g.

    • [1amp;0amp;∣amp;2 0amp;1amp;∣amp;5]\begin{bmatrix}1 & 0 & | & 2 \ 0 & 1 & | & 5\end{bmatrix} implies x=2,y=5.x = 2, y = 5.

Reduced Row Echelon Form (RREF)

  • The left-hand side of an augmented matrix is in reduced row echelon form if:

    • The leading entry (the first 1 from the left, in each nonzero row) of every nonzero row is 1, and that 1 is the only nonzero entry in its column.

    • Each leading 1 is the only nonzero in its column (zeros above and below).

    • The leading 1s move to the right as you move down the rows (staircase pattern).

  • When an augmented matrix is in RREF, the solution to the system is immediately read off from the rightmost column for each variable.

  • Note: In two variables, solving by matrices is still possible with Gauss–Jordan elimination, but for simple two-equation-two-variable systems, substitution or elimination is often quicker by hand. Matrices shine as the system grows (e.g., three equations and three variables or more).

Row operations (the Gauss–Jordan toolkit)

  • We may perform these operations on rows (which do not change the solution set):

    • Swap two rows.

    • Multiply a row by a nonzero scalar.

    • Add a multiple of one row to another row (Rowi ← Rowi + k·Row_j).

  • These are the algebraic steps used to transform a system's augmented matrix into RREF.

  • It helps to keep a running log of row operations to avoid mistakes and to understand the path to the solution.

Worked matrix example (Gauss–Jordan elimination)

  • Start with the matrix from the video:

    • Left side: [−1amp;5 3amp;9],\begin{bmatrix}-1 & 5 \ 3 & 9\end{bmatrix},

    • Right side (constants): [10 −6].\begin{bmatrix}10 \ -6\end{bmatrix}.

    • Combined augmented form: [−1amp;5amp;∣amp;10 3amp;9amp;∣amp;−6].\begin{bmatrix}-1 & 5 & | & 10 \ 3 & 9 & | & -6\end{bmatrix}.

  • Step 1: Make the leading entry in the first row a 1 by multiplying Row 1 by -1:

    • Row 1 → -1·Row 1 gives [1,−5,∣,−10][1, -5, |, -10]; Row 2 stays [3,9,∣,−6][3, 9, |, -6].

  • Step 2: Eliminate the entry below the leading 1 in column 1 by Row2 ← Row2 − 3·Row1:

    • Row2 becomes [0,24,∣,24].[0, 24, |, 24].

  • Step 3: Normalize Row 2 to make the pivot 1: Row2 ← (1/24)·Row2:

    • Row2 becomes [0,1,∣,1].[0, 1, |, 1].

  • Step 4: Eliminate the entry above the pivot in column 2 by Row1 ← Row1 + 5·Row2:

    • Row1 becomes [1,0,∣,−5].[1, 0, |, -5].

  • Resulting RREF:

    • [1amp;0amp;∣amp;−5 0amp;1amp;∣amp;1]\begin{bmatrix}1 & 0 & | & -5 \ 0 & 1 & | & 1\end{bmatrix}

  • Read off the solution: x=−5, y=1.x = -5,\, y = 1.

Notes and study tips from the lecture

  • The matrix approach is especially valuable as the number of equations and variables grows (e.g., three equations and three variables).

  • For two equations and two variables, many students prefer substitution or elimination for speed, but practice with matrices builds fluency for larger systems.

  • The instructor emphasized collaboration for the heavier matrix sections (e.g., section 3.4) and suggested teamwork to divide up hard problems.

  • The Gauss–Jordan method is connected to the identity matrix: if the left side becomes the identity, the right side gives the solution directly (as in the example with x = 2, y = 5).

  • The identity matrix and the concept of the main diagonal (the entries from the upper-left to the lower-right) were discussed as part of recognizing when a matrix is in reduced row echelon form.

Practical takeaways for exams and assignments

  • Expect to identify whether a system has a unique solution, no solution, or infinitely many solutions by examining the equations (intersections vs. parallelism vs. dependence).

  • Be comfortable converting a system to an augmented matrix and performing row operations to reach RREF.

  • For larger systems, matrices are often the preferred method; practice with several examples to build speed and accuracy.

  • Use study groups to tackle the more time-consuming matrix problems; explaining steps to peers reinforces understanding.

Quick reference: common examples from the notes

  • Elimination leading to a single-variable equation: if you obtain −3c=51-3c = 51, then c=−17c = -17 and substituting back gives the other variable.

  • Substitution leading to inconsistency: y=32x+2  and  6x−4y=4⇒−8=4  (no  solution).y = \tfrac{3}{2}x + 2\;\text{and}\;6x - 4y = 4\Rightarrow -8 = 4\; (no\;solution).

  • Same-line (dependent) case: two equations representing the same line yield infinitely many solutions along that line, not a single point.

  • Matrix example to RREF: converting

    • [−1amp;5 3amp;9]∣[10 −6]\begin{bmatrix}-1 & 5 \ 3 & 9\end{bmatrix} \Big| \begin{bmatrix}10 \ -6\end{bmatrix}
      to RREF gives the solution, or shows that the system is inconsistent or dependent depending on the resulting form.

Final note from the lecture

  • Quizzes and office hours: quizzing serves as a practical study guide for exams; the instructor offered office hours to help with questions and stressed collaborative problem solving.

  • The material covered includes polynomials, factoring, Pythagorean theorem applications, and the matrix-based approach to systems, all of which are relevant for upcoming assessments.