Sunday, April 28, 2024

20.14 - Applications of Determinants and Matrices

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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