In the previous section, we saw adjoint and inverse of a matrix. In this section, we will see applications of determinants and matrices.
• First we will see consistent and inconsistent systems. It can be explained in 4 steps:
1. Consider a system of linear equations:
x + 2y = 4
2x + 3y = 5
(Recall that, in a linear equation, all variables will have a power of 1. If one or more variables have power greater that 1, it is a non-linear equation)
•
Solving this system, we get: x = -2 and y = 3
•
(-2, 3) is the only possible solution.
•
If we draw the graphs of the given equations, we will get two lines. Those two lines will intersect at one and only one point (-2,3)
•
The solution (-2,3) will satisfy both the equations.
2. Consider another system of linear equations:
6x - 2y = 16
3x - y = 8
•
If we draw the graphs of these equations, we will see that, both the equations represent the same line.
•
Any point which lies on one line will lie on the other line also.
•
So any solution which satisfy one equation will satisfy the other equation also.
♦ For example, (3,1) satisfies both equations.
♦ Another example is (0,8).
•
In this way, there will be an infinite number of solutions.
3. Consider yet another system of linear equations:
5x - y = 4
5x - y = -6
•
If we draw the graphs of these equations, we will see that, they represent parallel lines.
•
Parallel lines never meet. So this system has no solution.
4. We have seen three types of systems.
◼ Types I and II fall in the group: Consistent systems
•
We can write:
If the system of equations have one or more than one solutions, it is a consistent system.
◼ Type III falls in the group: Inconsistent systems.
•
We can write:
If the system of equations have no solution, it is a inconsistent system.
Determinants and matrices can be used to solve systems of linear equations. It can be explained in 5 steps:
1. Consider the system of equations
a1 x + b1 y + c1 z = d1
a2 x + b2 y + c2 z = d2
a3 x + b3 y + c3 z = d3
2. Let
$A = \left[\begin{array}{r}
a_1 &{ b_1 } &{ c_1 } \\
a_2 &{ b_2 } &{ c_2 } \\
a_3 &{ b_3 } &{ c_3 } \\
\end{array}\right],~X = \left[\begin{array}{r}
x \\
y \\
z \\
\end{array}\right]~~ \text{and}~~B = \left[\begin{array}{r}
d_1 \\
d_2 \\
d_3 \\
\end{array}\right]
$
3. Then the system of equations in (1) can be written as:
$\left[\begin{array}{r}
a_1 &{ b_1 } &{ c_1 } \\
a_2 &{ b_2 } &{ c_2 } \\
a_3 &{ b_3 } &{ c_3 } \\
\end{array}\right]~\left[\begin{array}{r}
x \\
y \\
z \\
\end{array}\right]~ = \left[\begin{array}{r} d_1 \\
d_2 \\
d_3 \\
\end{array}\right]
$
•
Consider the L.H.S
♦ A is 3 × 3 matrix. X is a 3 × 1 matrix.
♦ So multiplying A and X will give a 3 × 1 matrix.
•
In the R.H.S also, we have a 3 × 1 matrix. So after multiplication, we will be able to equate corresponding terms.
•
By equating corresponding terms, we will get the same system as in (1).
•
Thus it is clear that, the given system can be written as
AX = B
4. If we can find the matrix X, we will be able to write the values of x, y and z.
•
So our next task is to find X. For that, we adopt the following method:
◼ Remarks:
•
2 (magenta color): Here we premultiply the whole equation by A−1 .
•
3 (magenta color): Here we use the fact that, A−1A = I.
•
4 (magenta color): Here we use the fact that, IX = X.
5. We will be able to write the steps in (4), only if A−1 exists. That is., A must be a non-singular matrix.
•
If A is singular, we must calculate (adj A) B. Here two cases can arise.
I. (adj A) B ≠ O.
•
Then the system is inconsistent.
II. (adj A) B = O.
•
Then the system can be consistent or inconsistent.
♦ Consistent if there are infinite number of solutions.
♦ Inconsistent if there is no solution.
Let us see a solved example:
Solved example 20.23
Solve the system of equations:
x + 2y = 4
2x + 3y = 5
Solution:
1. The given system can be written in the form AX = B.
$A = \left[\begin{array}{r}
1 &{ 2 } \\
2 &{ 3 } \\
\end{array}\right],~X = \left[\begin{array}{r}
x \\
y \\
\end{array}\right]~~ \text{and}~~B = \left[\begin{array}{r}
4 \\
5 \\
\end{array}\right]
$
2. So X = A−1 B.
•
Check whether A−1 exists:
♦ |A| = (3 − 4) = −1
♦ |A| ≠ 0
♦ So A is a non-singular matrix. Therefore, A−1 exists.
3. Use the method in Solved example 20.21 to find A−1.
We get: $A^{-1} = \left[\begin{array}{r}
-3 &{ 2 } \\
2 &{ -1 } \\
\end{array}\right]$
4. Use matrix multiplication to find A−1B.
•
We get: X = A−1 B =
$\left[\begin{array}{r}
-3 &{ 2 } \\
2 &{ -1 } \\
\end{array}\right]~\left[\begin{array}{r}
4 \\
5 \\
\end{array}\right]~ = \left[\begin{array}{r} -2 \\
3 \\
\end{array}\right]
$
5. So the solution is: x = -2 and y = 3
In the next section, we will see a few more solved examples.
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