## Grade 11 Mathematics Note

# System of Linear Equations.

Linear equations (ones that graph as straight lines) are simpler than non-linear equations, and the simplest linear system is one with two equations and two variables.

Now consider the following two-variable system of linear equations:

y = 3x – 2

y = –x – 6

Since the two equations above are in a system, we deal with them together at the same time. In particular, we can graph them together on the same axis system, like this:

A solution for a single equation is any point that lies on the line for that equation. A solution for a system of equations is any point that lies on each line in the system. For example, the red point at right is not a solution to the system, because it is not on either line:

The purple point at right is a solution to the system, because it lies on both of the lines:

In particular, this purple point marks the intersection of the two lines. Since this point is on both lines, it thus solves both equations, so it solves the entire system of equation. And this relationship is always true: For systems of equations, "solutions" are "intersections". You can confirm the solution by plugging it into the system of equations, and confirming that the solution works in each equation.

Example:

Equation; consistent and independent:

4x – 3y = - 6 …(i)

-4x + 2y = 16 …(ii)

Adding (i) and (ii), - y = 10 à y = -10

From (i) 4x = -36 à x = -9.

So, the system has a unique solution x = - 9, y = -10.

So, the system is consistent and independent.

Example;

Equation: consistent and dependent

-6x + 4y = 10 ..(i)

3x – 2y = -5 ..(ii)

Multiplying (ii) by -2

-6x + 4y = 10

= -6x + 4y = 10.

The two equations are one and the same. That is they coincide. Every solution of (i) is the solution of (ii) also. So, the system is consistent and dependent.

**Solutions of system of linear equations by applying matrix and Cramer’s rule:**

$3{\rm{x}} + \frac{4}{{\rm{y}}} = 10$

$ - 2{\rm{x}} + \frac{3}{{\rm{y}}} = - 1$

Let 1y = a.

Given system equations can be re-written as:

3x + 4a= 10 ..(i)

-2x + 3a = -1…(ii)

Cramers rule:

D = |34−23| = 9 + 8 = 17.

D_{1} = |104−13| = 30 + 4 = 34.

D_{2} = $\left| {\begin{array}{*{20}{c}}3&{10}\\{ - 2}&{ - 1}\end{array}} \right|$ = -3 + 20 = 17.

X = D1D = 347=2,

a = D2D = 17 = 1,

Then y = 1a = 11 = 1.

So, the solution is (2,1).

Matrix method:

A =[34−23], X =[xy], B = [10−1][10−1],

|A| = |34−23| = 9 + 8 = 17 ≠ 0

So, A^{-1} exists.

A_{11} = 3, A_{12} = 2, A_{21}= -4, A_{22} = 3.

Matrix of cofactors =(32−43),

Adj. A = (3−423),

A^{-1} = adj.A|A| = (3−423)17=(317−417217317)

Or, X = A^{-1}B.

Or, (xy) (317−417217317)(10−1) = 30+41720−317) = (21).

So, the solution is (2,1).