20.16 - Miscellaneous Examples

In the previous section, we completed a discussion on determinants. We saw a solved example also. In this section, we will see some miscellaneous examples.

Solved example 20.27
If a, b, c are positive and unequal, show that the value of the determinant $\Delta ~=~\left |\begin{array}{r}                           
a    &{    b    }    &{    c    }    \\
b    &{    c    }    &{    a    }    \\
c    &{    a    }    &{    b    }    \\
\end{array}\right |
$ is negative.
Solution:
1. First we will simplify the given determinant:

◼ Remarks:
• 2 (magenta color): Apply R₁ → R₁ + R₂.
• 3 (magenta color): Apply R₁ → R₁ + R₃.
• 4 (magenta color): Apply C₃ → C₃ − C₁.
• 5 (magenta color): Apply C₂ → C₂ − C₁.
• 6 (magenta color): Expand along R₁.

2. Consider the above result. There are three terms:
(i) −(1/2)
(ii) (a+b+c)
(iii) (a−b)² + (a−c)² + (b−c)².

• Given that: a, b, c are +ve and unequal.
• So (ii) and (iii) cannot become -ve.
• Therefore, due to the presence of −(1/2), the result as a whole will become -ve.

Solved example 20.28
If a, b, c are in A.P, find the value of
$\Delta ~=~\left |\begin{array}{r}                           
2y+4    &{    5y+7    }    &{    8y+a    }    \\
3y+5    &{    6y+8    }    &{    9y+b    }    \\
4y+6    &{    7y+9    }    &{    10y+c    }    \\
\end{array}\right |
$.
Solution:

◼ Remarks:
• 2 (magenta color): Apply R₃ → R₃ − R₂.
• 3 (magenta color): Apply R₂ → R₂ − R₁.
• 4 (magenta color): Apply R₃ → R₃ − R₂.
• 5 (magenta color): Since, a, b, c are in A.P, we can put 2b = a+c.
• 6 (magenta color): All elements of R₃ are zeroes. So the value of the determinant is zero.

Solved example 20.29
Show that
$\Delta ~=~\left |\begin{array}{r}                         
(y+z)^2    &{    xy    }    &{    zx    }    \\
xy    &{    (x+z)^2    }    &{    yz    }    \\
xz    &{    yz    }    &{    (x+y)^2    }    \\
\end{array}\right | ~=~2xyz (x+y+z)^3
$.
Solution:

◼ Remarks:
• 2 (magenta color):
    ♦ Multiply R₁ by x
    ♦ Multiply R₂ by y
    ♦ Multiply R₃ by z
To balance these multiplications, the whole determinant should be multiplied by (1/xyz)
• 3 (magenta color):
Take out the common factors:
    ♦ x from C₁
    ♦ y from C₂
    ♦ z from C₃
• 4 (magenta color):
Apply two operations:
    ♦ C₂ → C₂ − C₁
    ♦ C₃ → C₃ − C₁
• 5 (magenta color):
    ♦ Apply the identity: a² − b² = (a+b)(a-b).
    ♦ This is applied to C₂ and C₃.
• 6 (magenta color): Take out (x+y+z) from C₁ and C₂.
• 7 (magenta color):
Apply R₁ → R₁ − R₂
• 8 (magenta color):
Apply R₁ → R₁ − R₃
• 9 (magenta color):
Apply C₂ → C₂ + (1/y)C₁
• 10 (magenta color):
Apply C₃ → C₃ + (1/z)C₁
• 11 (magenta color):
Expand along R₁.

Solved example 20.30
Use product
$\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ] \left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]$
to solve the system of equations
x - y + 2z = 1
2y − 3z = 1
3x − 2y + 4z = 2
Solution:
1. Use matrix multiplication to find the product.
• We get:
$\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ] \left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]~ = \left [\begin{array}{r}                         
1    &{    0    }    &{   0    }    \\
0    &{    1    }    &{   0    }    \\
0    &{    0    }    &{   1    }    \\
\end{array}\right ]
$

2. The product is an identity matrix. So it is clear that:
Inverse of $\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ]$ is $\left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]$

3. The given system can be written in the form AX = B.
$A = \left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ],~X = \left[\begin{array}{r}       
x        \\
y        \\
z        \\
\end{array}\right]~~ \text{and}~~B = \left[\begin{array}{r}                           
1        \\
1        \\
2        \\
\end{array}\right]
$

4. So X = A⁻¹ B.
• Check whether A⁻¹ exists:
We have already obtained the inverse. Therefore, A⁻¹ exists.

5. Use matrix multiplication to find A⁻¹B.
• We get: X = A⁻¹ B =
$\left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]~\left[\begin{array}{r}                           
1        \\
1        \\
2        \\
\end{array}\right]~ = \left[\begin{array}{r}                        0        \\
5        \\
3        \\
\end{array}\right]
$

6. So the solution is: x = 0, y = 5 and z = 3

Solved example 20.31
Prove that
$ \Delta ~=~ \left |\begin{array}{r}                         
a+bx    &{    c+dx    }    &{    p+qx    }    \\
ax+b    &{    cx+d    }    &{    px+q    }    \\
u    &{    v    }    &{   w    }    \\
\end{array}\right | ~=~ (1 - x^2) \left |\begin{array}{r}                         
a    &{    c    }    &{    p    }    \\
b    &{    d    }    &{    q    }    \\
u    &{    v    }    &{   w    }    \\
\end{array}\right |$
Solution:
1. First we will split the given matrix by applying property V.

◼ Remarks:
• 2 (magenta color):
We split R₁ so that, a, c and p are obtained in the first row. These are the elements that we want in the R₁ of the final result.

2. Now we simplify |A|:

◼ Remarks:
• 2 (magenta color): We split R₂ of |A|.
• 3 (magenta color): Consider the first determinant in (2). Every element in R₂ is proportional to the corresponding elements in R₁, by the same ratio 'x'. So this determinant becomes zero.

3. Next we simplify |B|:

◼ Remarks:
• 2 (magenta color): We split R₂ of |B|.
• 3 (magenta color): Consider the second determinant in (2). Every element in R₁ is proportional to the corresponding elements in R₂, by the same ratio 'x'. So this determinant becomes zero.
• 4 (magenta color): We take out the common factor 'x' from R₁ and R₂.
• 5 (magenta color): We want the elements a, c and p in R₁. So we interchange R₁ and R₂. The sign of the determinant will change when the two rows are interchanged.

4. Finally we add |A| and |B|. We get:

---

The link below gives a few more examples:

Miscellaneous Examples

---

In the next section, we will see some miscellaneous examples.

Previous

Contents

Copyright©2024 Higher secondary mathematics.blogspot.com

20.15 - Solved Examples on Matrix Method

In the previous section, we saw the matrix method for solving systems of linear equations. We saw a solved example also. In this section, we will see a few more solved examples.

Solved example 20.24
Solve the system of equations:
2x + 5y = 1
3x + 2y = 7
Solution:
1. The given system can be written in the form AX = B.
$A = \left[\begin{array}{r}                           
2        &{    5    }    \\
3        &{    2    }    \\
\end{array}\right],~X = \left[\begin{array}{r}       
x        \\
y        \\
\end{array}\right]~~ \text{and}~~B = \left[\begin{array}{r}                           
1        \\
7        \\
\end{array}\right]
$

2. So X = A⁻¹ B.
• Check whether A⁻¹ exists:
   ♦ |A| = (4 − 15) = −11
   ♦ |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: $A^{-1} = \left[\begin{array}{r}                           
-2/11        &{    5/11    }    \\
3/11        &{    -2/11    }    \\
\end{array}\right]$

4. Use matrix multiplication to find A⁻¹B.
• We get: X = A⁻¹ B =
$\left[\begin{array}{r}                           
-2/11        &{    5/11    }    \\
3/11        &{    -2/11    }    \\
\end{array}\right]~\left[\begin{array}{r}                           
1        \\
7        \\
\end{array}\right]~ = \left[\begin{array}{r}                        3        \\
-1        \\
\end{array}\right]
$

5. So the solution is: x = 3 and y = -1

Solved example 20.25
Solve the system of equations:
3x − 2y + 3z = 8
2x + y − z = 1
4x − 3y + 2z = 4
Solution:
1. The given system can be written in the form AX = B.
$A = \left[\begin{array}{r}                           
3        &{    -2    } &{    3    }    \\
2        &{    1    } &{    -1    }    \\
4        &{    -3    } &{   2    }    \\
\end{array}\right],~X = \left[\begin{array}{r}       
x        \\
y        \\
z        \\
\end{array}\right]~~ \text{and}~~B = \left[\begin{array}{r}                           
8        \\
1        \\
4        \\
\end{array}\right]
$

2. So X = A⁻¹ B.
• Check whether A⁻¹ exists:
   ♦ |A| = −17 (by expansion along any row or column)
   ♦ |A| ≠ 0
   ♦ So A is a non-singular matrix. Therefore, A⁻¹ exists.

3. Use the method in Solved example 20.20 to find A⁻¹.
We get: $A^{-1} = - \frac{1}{17} \left[\begin{array}{r}                           
-1        &{    -5    } &{    -1    }    \\
-8        &{    -6    } &{    9    }    \\
-10        &{    1    } &{   7    }    \\
\end{array}\right]$

4. Use matrix multiplication to find A⁻¹B.
• We get: X = A⁻¹ B =
$- \frac{1}{17} \left[\begin{array}{r}                           
-1        &{    -5    } &{    -1    }    \\
-8        &{    -6    } &{    9    }    \\
-10        &{    1    } &{   7    }    \\
\end{array}\right]~\left[\begin{array}{r}                           
8        \\
1        \\
4        \\
\end{array}\right]~ = \left[\begin{array}{r}                        1        \\
2        \\
3        \\
\end{array}\right]
$

5. So the solution is: x = 1, y = 2 and z = 3

Solved example 20.26
Sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of second number. Represent it algebraically and find the numbers by matrix method.
Solution:
1. Let the numbers be x, y and z.
2. Given: Sum is 6.
• So we can write: x + y + z = 6
3. Given: If we multiply third number by 3 and add second number to it, we get 11.
• So we can write: 3z + y = 11.
• This is same as 0x + y + 3z = 11
4. Given: By adding first and third numbers, we get double of second number.
• So we can write: x + z = 2y
• This is same as: x − 2y + z = 0
5. From (2), (3) and (4), we get three equations:
x + y + z = 6
0x + y + 3z = 11
x − 2y + z = 0
6. The given system can be written in the form AX = B.
$A = \left[\begin{array}{r}                           
1        &{    1    } &{    1    }    \\
0        &{    1    } &{    3    }    \\
1        &{    -2    } &{   1    }    \\
\end{array}\right],~X = \left[\begin{array}{r}       
x        \\
y        \\
z        \\
\end{array}\right]~~ \text{and}~~B = \left[\begin{array}{r}                           
6        \\
11        \\
0        \\
\end{array}\right]
$

7. So X = A⁻¹ B.
• Check whether A⁻¹ exists:
   ♦ |A| = 9 (by expansion along any row or column)
   ♦ |A| ≠ 0
   ♦ So A is a non-singular matrix. Therefore, A⁻¹ exists.

8. Use the method in Solved example 20.20 to find A⁻¹.
We get: $A^{-1} =  \frac{1}{9} \left[\begin{array}{r}                           
7        &{    -3    } &{    2    }    \\
3        &{    0    } &{    -3    }    \\
-1        &{    3    } &{   1    }    \\
\end{array}\right]$

9. Use matrix multiplication to find A⁻¹B.
• We get: X = A⁻¹ B =
$ \frac{1}{9} \left[\begin{array}{r}                           
7        &{    -3    } &{    2    }    \\
3        &{    0    } &{    -3    }    \\
-1        &{    3    } &{   1    }    \\
\end{array}\right]~\left[\begin{array}{r}                           
6        \\
11        \\
0        \\
\end{array}\right]~ = \left[\begin{array}{r}                        1        \\
2        \\
3        \\
\end{array}\right]
$

10. So the solution is: x = 1, y = 2 and z = 3

---

The link below gives a few more solved examples:

Exercise 20.6

---

In the next section, we will see some miscellaneous examples.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com

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.

Previous

Contents

Next

Copyright©2024 Higher secondary mathematics.blogspot.com