21.1 - Continuous Functions

In the previous section, we saw the two conditions which can be used to check continuity at any given point.

• In the solved examples that we saw in the previous section, we were checking the continuity at specified points like x = 0, x = 1 etc.,
• Now let us see some examples where the given functions are continuous at every point.

Solved example 21.5
Check the points where the constant function f(x) = k is continuous.
Solution:
1. Consider any arbitrary point c. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is: k
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = k
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point. So it can be any real number.
• Therefore we can write:
The given function is continuous at every real number.

Solved example 21.6
Prove that the identity function on real numbers given by f(x) = x is continuous at every real number.
Solution:
1. Consider any arbitrary point c. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point. So it can be any real number.
• Therefore we can write:
The given function is continuous at every real number.

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• In the above two solved examples 21.5 and 21.6, we see that, the given functions are continuous at every point. We say that, such functions are continuous functions. For such functions, there is no need to check for continuity at various individual points.
• We can write the definition for continuous functions:
A real function f is said to be a continuous function, if it is continuous at every point in the domain of f.

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Now we will see a special case. It can be written in steps:
1. In the solved examples 21.5 and 21.6, the domain is (−∞,∞). The functions are continuous between −∞ and ∞.
2. Let us see the case where domain is restricted. In fig.21.2 below, we see that, the domain is [a,b].

Fig.21.2

3. In the above fig., If the function f is to be a continuous function, then:
   ♦ f must be continuous at A.
   ♦ f must be continuous at B.
   ♦ f must be continuous at all the infinite points in between A and B.
4. If f is to be continuous at A, $\lim_{x\rightarrow a^{-}} f(x)$ should be equal to $\lim_{x\rightarrow a^{+}} f(x)$.
• But there is no way to find $\lim_{x\rightarrow a^{-}} f(x)$. This is because, we do not know how x approaches 'a' from the left.
• So there is no need to consider $\lim_{x\rightarrow a^{-}} f(x)$.
• To check the continuity at A, we need to check only one condition: $\lim_{x\rightarrow a^{+}} f(x) ~=~f(a)$ 
5. Similarly, if f is to be continuous at B, $\lim_{x\rightarrow b^{-}} f(x)$ should be equal to $\lim_{x\rightarrow b^{+}} f(x)$.
• But there is no way to find $\lim_{x\rightarrow b^{+}} f(x)$. This is because, we do not know how x approaches 'b' from the right.
• So there is no need to consider $\lim_{x\rightarrow b^{+}} f(x)$.
• To check the continuity at B, we need to check only one condition: $\lim_{x\rightarrow b^{-}} f(x) ~=~f(b)$
6. Now consider the case when the domain of f is a singleton. That is., the domain is a set which contains only one point.
• Then the graph will be a point. We cannot check left side limit or right side limit.
• In such a situation, we say that, f is a continuous function.

Let us see some solved examples:

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Solved example 21.7
Is the function defined by f(x) = |x|, a continuous function?
Solution:
• Fig.21.3 below shows the graph of the function f(x) = |x|.

Fig.21.3

Case 1: continuity of the left segment.
1. Consider any arbitrary point P on the left segment. Let us check whether the function is continuous at P.
2. Applying condition (i):
• Limiting value of f(x) at x = -p is: p
• It is clear that, $\lim_{x\rightarrow -p} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(-p) = |-p| = p
• We see that: $\lim_{x\rightarrow -p} f(x)~=~f(-p)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at P.
5. P was taken as an arbitrary point. So '-p' can be any real number less than zero.
• Therefore we can write:
The given function is continuous at every real number less than zero.

Case 2: continuity of the right segment.
1. Consider any arbitrary point Q on the right segment. Let us check whether the function is continuous at Q.
2. Applying condition (i):
• Limiting value of f(x) at x = q is: q
• It is clear that, $\lim_{x\rightarrow q} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(q) = |q| = q
• We see that: $\lim_{x\rightarrow q} f(x)~=~f(q)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at Q.
5. Q was taken as an arbitrary point. So q can be any real number greater than or equal to zero.
• Therefore we can write:
The given function is continuous at every real number greater than or equal to zero.

◼ Based on the results from the two cases. we can write:
The given function is continuous at all points.

Solved example 21.8
Discuss the continuity of the function defined by:
f(x) = x³ + x² − 1.
Solution:
1. Consider any arbitrary point c. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c³ + c² − 1
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c³ + c² − 1
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point. So it can be any real number.
• Therefore we can write:
The given function is continuous at every real number. Therefore, it is a continuous function.

Solved example 21.9
Is the function defined by $f(x) = \frac{1}{x},~x \ne 0$, a continuous function?
Solution:
• Fig.21.4 below shows the graph of the function $f(x) = \frac{1}{x}$.

Fig.21.4

Case 1: continuity of the left segment.
1. Consider any arbitrary point P on the left segment. Let us check whether the function is continuous at P. (Note that, it is impossible to consider any point whose x coordinate is zero. It is specially mentioned in the question)
2. Applying condition (i):
• Limiting value of f(x) at x = -p is: $-{\frac{1}{p}}$
• It is clear that, $\lim_{x\rightarrow -p} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(-p) = $\frac{1}{(-p)}~=~-{\frac{1}{p}}$
• We see that: $\lim_{x\rightarrow -p} f(x)~=~f(-p)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at P.
5. P was taken as an arbitrary point. So '-p' can be any real number less than zero.
• Therefore we can write:
The given function is continuous at every real number less than zero. However, the real number 'zero' can not be considered.

Case 2: continuity of the right segment.
1. Consider any arbitrary point Q on the right segment. Let us check whether the function is continuous at Q.
2. Applying condition (i):
• Limiting value of f(x) at x = q is: $\frac{1}{q}$
• It is clear that, $\lim_{x\rightarrow q} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(q) = $\frac{1}{q}$
• We see that: $\lim_{x\rightarrow q} f(x)~=~f(q)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at Q.
5. Q was taken as an arbitrary point. So 'q' can be any real number greater than zero.
• Therefore we can write:
The given function is continuous at every real number greater than zero. However, the real number 'zero' can not be considered.

◼ Based on the results from the two cases. we can write:
The given function is continuous at all real numbers other than zero.

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Based on the above solved example 21.9, we can discuss about the concept of infinity.
The graph is shown again in fig.21.5 below:

Fig.21.5

Case 1: The right segment.
This can be written in 5 steps:
1. Two points Q₁ and Q₂ are marked on the right segment.
   ♦ Q₂ is closer (horizontally) to zero than Q₁.
   ♦ So q₂ will be smaller than the q₁.
2. The input x values are given in the denominator.
• So smaller the denominator, larger will be the value of the function f(x).
• So f(q₂) will be larger than f(q₁).
• It is clear that, as x approaches zero from the right, f(x) will become larger and larger.
3. When x approaches zero from the right, it take values like 0.1, 0.001, 0.0001, etc.,     
• Let us find f(x) in such cases:
   ♦ When x = 0.1, f(x) = $\frac{1}{0.1} = 10$
   ♦ When x = 0.01, f(x) = $\frac{1}{0.01} = 100$
   ♦ When x = 0.001, f(x) = $\frac{1}{0.001} = 1000$
   ♦ When x = 0.0001, f(x) = $\frac{1}{0.0001} = 10000$
4. In the input x, we can give a million zeros after the decimal point. The resulting f(x) will be correspondingly large.
• We can make f(x) larger than the largest known number. All we need to do is to give the required number of zeros after the decimal point.
• "Larger than the largest known number" is denoted by the symbol +∞. It is read as plus infinity.
• +∞ is a concept. It is not a real number. We cannot do familiar calculations with +∞.
• For example:
   ♦ (+∞ + 4) gives +∞
   ♦ (+∞ ÷ 9) gives +∞
• We assume that:
+∞ is at the right end of the x-axis and at the top end of the y-axis.
5. So when $f(x) = \frac{1}{x}$, we can write: $\lim_{x\rightarrow 0^{+}} f(x)~=~+ \infty$

Case 2: The left segment.
This can be written in 5 steps:
1. Two points P₁ and P₂ are marked on the left segment.
   ♦ P₂ is closer (horizontally) to zero than P₁.
   ♦ So p₂ will be smaller (numerically) than the p₁.
2. The input x values are given in the denominator.
• So smaller the denominator, larger will be the value of the function f(x).
• So f(p₂) will be larger (numerically) than f(p₁).
• It is clear that, as x approaches zero, f(x) will become larger and larger numerically.
3. But on the left segment, the x values are −ve.
• If a "numerically larger value" is −ve, it is smaller in reality.
• So we can write:
As x approaches zero from the left, f(x) will become smaller and smaller.
4. When x approaches zero, it take values like −0.1, −0.001, −0.0001, etc.,     
• Let us find f(x) in such cases:
   ♦ When x = −0.1, f(x) = $\frac{1}{-0.1} = -10$
   ♦ When x = −0.01, f(x) = $\frac{1}{-0.01} = -100$
   ♦ When x = −0.001, f(x) = $\frac{1}{-0.001} = -1000$
   ♦ When x = −0.0001, f(x) = $\frac{1}{-0.0001} = -10000$
5. In the input x, we can give a million zeros after the decimal point. The resulting f(x) will be correspondingly small.
• We can make f(x) smaller than the smallest known number. All we need to do is to give the required number of zeros after the decimal point.
• "Smaller than the smallest known number" is denoted by the symbol −∞. It is read as minus infinity.
• −∞ is a concept. It is not a real number. We cannot do familiar calculations with −∞.
• For example:
   ♦ (−∞ + 4) gives −∞
   ♦ (−∞ ÷ 9) gives −∞
• We assume that:
−∞ is at the left end of the x-axis and at the bottom end of the y-axis.
6. So when $f(x) = \frac{1}{x}$, we can write: $\lim_{x\rightarrow 0^{−}} f(x)~=~− \infty$

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In the next section, we will see a few more solved examples.

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Chapter 21 - Continuity and Differentiability

In the previous section, we completed a discussion on determinants. In this section, we will see Continuity and Differentiability.

• We have already learned about limits and derivatives in a previous chapter. The reader is advised to revise those topics thoroughly, before attempting our present chapter.

• Consider fig.13.5 that we saw in that previous chapter. We can use that figure to learn the basics about continuity. It can be written in 4 steps:
1. In fig.13.5, we wanted the limit at x = 0.
• We saw that:
   ♦ The left side limit is 1.
   ♦ The right side limit is 2.
2. So, the limit at x = 0 does not exist.
3. In general, if at a point c, if the left side limit is not equal to the right side limit, we say that, the function is discontinuous at c. Also, c is a point of discontinuity.
4. Note that, in fig.13.5, at x = 0, we cannot draw the graph without lifting the pen from the plane of the paper. So there indeed, is a "discontinuity".

Let us see another example. It can be written in 5 steps:
1. Consider fig.13.7 that we saw in that previous chapter. We wanted the limit at x = 1.
• We saw that:
   ♦ The left side limit is 3.
   ♦ The right side limit is also 3.
2. So, the limit at x = 1 exists.
• But value of the function at x= 1, is not 3.
3. So in some cases,
   ♦ The left side and right side limits may be same at a point c.
   ♦ But value of the function at c, may be different from that limiting value.
4. The situation mentioned in the above step (3), is also a discontinuity. We say that, the function is discontinuous at c. And, c is a point of discontinuity.
5. Note that, in fig.13.7, at x = 1, we cannot draw the graph without lifting the pen from the plane of the paper. So there indeed, is a "discontinuity".

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• We have seen two examples which were related to two different situations. Those two  different situations gave us a basic idea about discontinuity. So now we can write "opposite situations" which will give us a basic idea about continuity. It can be written in 4 steps:
1. A function is said to be continuous at a point c, if two conditions are satisfied:
Condition (i):
   ♦ The left side limit at c
   ♦ is equal to
   ♦ The right side limit at c.
• Using symbols, we can write this condition as:
$\lim_{x\rightarrow c^{-}} f(x) ~=~ \lim_{x\rightarrow c^{+}} f(x)$

Condition (ii):
   ♦ Limiting value at c
   ♦ is equal to
   ♦ The value of the function at c.

2. It is possible to write the conditions in a simpler way.
• Consider the first condition that we wrote above. If this condition is satisfied, it means that, limit at c exists. So we can write:
◼  A function is said to be continuous at a point c, if two conditions are satisfied:
Condition (i):
Limit at c exists.
• Using symbols, we can write this condition as:
$\lim_{x\rightarrow c} f(x)$ exists.

Condition (ii):
   ♦ Limiting value at c
   ♦ is equal to
   ♦ The value of the function at c.
• Using symbols, we can write this condition as:
$\lim_{x\rightarrow c} f(x)~=~f(c)$.

3. We have seen numerous solved examples where we checked "whether limit at any given point c exists".
• All we need to do is: Find the limit at c.
• If the limiting value is in the form 0/0 or "a division by zero", we concluded that, the limit does not exist.
• Also note that, "finding the limit at c" is essential because, only then we will be able to apply the second condition.

4. When the two conditions are satisfied, we will be able to draw the graph in a single stroke, with out lifting the pen from the plane of the paper.

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Let us see some solved examples:

Solved example 21.1
Check the continuity of the function given by f(x) = 2x + 3 at x = 1.
Solution:
1. Applying condition (i):
• Limiting value of f(x) at x = 1 is:
2(1) + 3 = 5
• It is clear that, $\lim_{x\rightarrow 1} f(x)$ exists. So first condition is satisfied.
2. Applying condition (ii):
f(1) = 2(1) + 3 = 5
• We see that: $\lim_{x\rightarrow 1} f(x)~=~f(1)$
• So second condition is also satisfied.
3. Since both conditions are satisfied, the given function is continuous at x = 1.

Solved example 21.2
Check the continuity of the function given by f(x) = x² at x = 0.
Solution:
1. Applying condition (i):
• Limiting value of f(x) at x = 0 is:
(0)² = 0
• It is clear that, $\lim_{x\rightarrow 0} f(x)$ exists. So first condition is satisfied.
2. Applying condition (ii):
f(0) = (0)² = 0
• We see that: $\lim_{x\rightarrow 0} f(x)~=~f(0)$
• So second condition is also satisfied.
3. Since both conditions are satisfied, the given function is continuous at x = 0.

Solved example 21.3
Check the continuity of the function given by f(x) = |x| at x = 0.
Solution:
1. Applying condition (i):
This condition can be applied with greater clarity, if we draw the graph. It is shown in fig.21.1 below:

Fig.21.1

• When x approaches zero from the left, f(x) approaches zero. That is: $\lim_{x\rightarrow 0^{-}} f(x) = 0$
• When x approaches zero from the right, f(x) approaches zero. That is: $\lim_{x\rightarrow 0^{+}} f(x) = 0$
• Both left side and right side limits are the same. That means, $\lim_{x\rightarrow 0} f(x)$ exists. So first condition is satisfied.
2. Applying condition (ii):
• From the graph, we have: f(0) = 0
• We see that: $\lim_{x\rightarrow 0} f(x)~=~f(0)$
• So second condition is also satisfied.
3. Since both conditions are satisfied, the given function is continuous at x = 0.

Solved example 21.4
Show that the function f given by
$f(x) = \begin{cases} x^3+3,  & \text{if}~x\ne0 \\[1.5ex] 1, & \text{if}~x=0 \end{cases}$
is not continuous at x = 0.
Solution:
1. Applying condition (i):
• Limiting value of f(x) at x = 0 is:
(0)³ + 3 = 3
• It is clear that, $\lim_{x\rightarrow 0} f(x)$ exists. So first condition is satisfied.
2. Applying condition (ii):
f(0) = 0
• We see that: $\lim_{x\rightarrow 0} f(x)~\ne~f(0)$
• So second condition is not satisfied.
3. Since both conditions are not satisfied, the given function is not continuous at x = 0.

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In the next section, we will see continuous functions.

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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:

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The link below gives a few more examples:

Miscellaneous Examples

---

In the next section, we will see Continuity and Differentiability.

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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.

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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

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In the next section, we will see a few more solved examples.

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20.13 - Solved Examples

In the previous section, we saw adjoint and inverse of a matrix. In this section, we will see some solved examples.

Solved example 20.20
If $A~=~\left [\begin{array}{r}   
1       &{    3    } &{    3    }    \\
1       &{    4    } &{    3    }    \\
1       &{    3    } &{    4    }    \\
\end{array}\right ]$, then verify that A(adj A) = |A| I. Also find A⁻¹.
Solution:
Part (i):
1. First we will find the Minors and Cofactors of A.

2. So the Cofactor matrix of A is: $\left [\begin{array}{r}   
7       &{    -1    } &{    -1    }    \\
-3       &{    1    } &{    0    }    \\
-3       &{    0    } &{    1    }    \\
\end{array}\right ]$

3. Transpose of the Cofactor matrix will give the adjoint matrix. So we get:

adj A = $\left [\begin{array}{r}   
7       &{    -3    } &{    -3    }    \\
-1       &{    1    } &{    0    }    \\
-1       &{    0    } &{    1    }    \\
\end{array}\right ]$

4. Use matrix multiplication to find A(adj A). We get:

$\left [\begin{array}{r}   
1       &{    3    } &{    3    }    \\
1       &{    4    } &{    3    }    \\
1       &{    3    } &{    4    }    \\
\end{array}\right ] \left [\begin{array}{r}   
7       &{    -3    } &{    -3    }    \\
-1       &{    1    } &{    0    }    \\
-1       &{    0    } &{    1    }    \\
\end{array}\right ] ~=~\left [\begin{array}{r}   
1       &{    0    } &{    0    }    \\
0       &{    1    } &{    0    }    \\
0       &{    0    } &{    1    }    \\
\end{array}\right ]$

5. Let us calculate |A|:

|A| = 1(16 − 9) − 3(4 − 3) + 3(3 − 4)
= 7 − 3 − 3 = 1

6. Now we can calculate |A|I:

$(1) \left [\begin{array}{r}   
1       &{    0    } &{    0    }    \\
0       &{    1    } &{    0    }    \\
0       &{    0    } &{    1    }    \\
\end{array}\right ] ~=~\left [\begin{array}{r}   
1       &{    0    } &{    0    }    \\
0       &{    1    } &{    0    }    \\
0       &{    0    } &{    1    }    \\
\end{array}\right ]$

7. From (4) and (6), we see that: A(adj A) = |A| I

Part (ii):
We have already calculated |A| and (adj A) in part (i). So we can calculate A⁻¹ as follows:

$A^{-1} = \frac{1}{|A|}(\text{adj A}) = \frac{1}{1} \left [\begin{array}{r}   
7       &{    -3    } &{    -3    }    \\
-1       &{    1    } &{    0    }    \\
-1       &{    0    } &{    1    }    \\
\end{array}\right ]= \left [\begin{array}{r}   
7       &{    -3    } &{    -3    }    \\
-1       &{    1    } &{    0    }    \\
-1       &{    0    } &{    1    }    \\
\end{array}\right ]$

Solved example 20.21
If $A~=~\left [\begin{array}{r}   
2        &{    3    }    \\
1        &{    -4    }    \\
\end{array}\right ]$ and $B~=~\left [\begin{array}{r}   
1        &{    -2    }    \\
-1        &{    3    }    \\
\end{array}\right ]$, then verify that (AB)⁻¹ = B⁻¹A⁻¹.
Solution:
1. First we will test whether A and B are invertible.
(i) |A| = −8 − 3 = −11
|A| ≠ 0. So A is invertible.
(ii) |B| = 3-2 = 1
|B| ≠ 0. So B is invertible.

2. Next we will find A⁻¹.
• adj A =  $\left [\begin{array}{r}   
-4        &{    -3    }    \\
-1        &{    2    }    \\
\end{array}\right ]$
(Recall the shortcut method explained in fig.20.5 of the previous section)

• So we get:
$A^{-1} = \frac{1}{|A|}(\text{adj A}) = -{\frac{1}{11}} \left [\begin{array}{r}   
-4        &{    -3    }    \\
-1        &{    2    }    \\
\end{array}\right ]$   

3. Next we will find B⁻¹.
• adj B =  $\left [\begin{array}{r}   
3        &{    2    }    \\
1        &{    1    }    \\
\end{array}\right ]$
(Recall the shortcut method explained in fig.20.5 of the previous section)
• So we get:
$B^{-1} = \frac{1}{|B|}(\text{adj B}) = \frac{1}{1} \left [\begin{array}{r}   
3        &{    2    }    \\
1        &{    1    }    \\
\end{array}\right ]= \left [\begin{array}{r}   
3        &{    2    }    \\
1        &{    1    }    \\
\end{array}\right ]$

4. Use matrix multiplication to find B⁻¹A⁻¹. We get:
$B^{-1} A^{-1} = \left [\begin{array}{r}   
3        &{    2    }    \\
1        &{    1    }    \\
\end{array}\right ]~\times~ -{\frac{1}{11}} \left [\begin{array}{r}   
-4        &{    -3    }    \\
-1        &{    2    }    \\
\end{array}\right ] = {\frac{1}{11}} \left [\begin{array}{r}   
14        &{    5    }    \\
5        &{    1    }    \\
\end{array}\right ]$

5. Use matrix multiplication to find AB. We get:
$AB = \left [\begin{array}{r}   
2        &{    3    }    \\
1        &{    -4    }    \\
\end{array}\right ]~\left [\begin{array}{r}   
1        &{    -2    }    \\
-1        &{    3    }    \\
\end{array}\right ] = \left [\begin{array}{r}   
-1        &{    5    }    \\
5        &{    -14    }    \\
\end{array}\right ]$

6. Next we will find |(AB)|.
|(AB)| = 14 − 25 = -11

7. Now we can calculate (AB)⁻¹.
• adj (AB) =  $\left [\begin{array}{r}   
-14        &{    -5    }    \\
-5        &{    -1    }    \\
\end{array}\right ]$
(Recall the shortcut method explained in fig.20.5 of the previous section)

• So we get:
$(AB)^{-1} = \frac{1}{|(AB)|}(\text{adj (AB)}) = \frac{1}{(-11)} \left [\begin{array}{r}   
-14        &{    -5    }    \\
-5        &{    -1    }    \\
\end{array}\right ] = \frac{1}{11} \left [\begin{array}{r}   
14        &{    5    }    \\
5        &{    1    }    \\
\end{array}\right ]$

8. Comparing the results in (4) and (7), we can write:
(AB)⁻¹ = B⁻¹A⁻¹.

Solved example 20.22
Show that the matrix   $A~=~\left [\begin{array}{r}   
2        &{    3    }    \\
1        &{    2    }    \\
\end{array}\right ]$ satisfies the equation A² − 4A + I = O, where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A⁻¹.
Solution:
Part (i): To prove that, A² − 4A + I = O
1. Use matrix multiplication to find A². We get:
$A^2 = AA = \left [\begin{array}{r}   
2        &{    3    }    \\
1        &{    2    }    \\
\end{array}\right ]~\left [\begin{array}{r}   
2        &{    3    }    \\
1        &{    2    }    \\
\end{array}\right ] = \left [\begin{array}{r}   
7        &{    12    }    \\
4        &{    7    }    \\
\end{array}\right ]$

2. Next we will find 4A.
$4A = 4 \left [\begin{array}{r}   
2        &{    3    }    \\
1        &{    2    }    \\
\end{array}\right ]= \left [\begin{array}{r}   
8        &{    12    }    \\
4        &{    8    }    \\
\end{array}\right ]$

3. Substituting the above matrices in the given equation, we get:

• We see that, L.H.S = R.H.S.
• So matrix A satisfies the given equation.

Part (ii): To find A⁻¹.
1. First we need to test whether A is invertible.
• We have: |A| = (4-3) = 1.
• Since |A| ≠ 0, A is invertible.

2. To find A⁻¹, we rearrange the equation that was proved in part (i).

◼ Remarks:
• 3 (magenta color): In this line, we post multiply the whole equation by A⁻¹.
• 4 (magenta color): In this line, we apply the fact that, AA⁻¹ = I.

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The link below gives a few more solved examples:

Exercise 20.5

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In the next section, we will see the applications of determinants and matrices.

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20.12 - Adjoint And Inverse of A Matrix

In the previous section, we saw Minors and Cofactors. In this section, we will see adjoint and inverse of a matrix.

Adjoint of a Matrix

Some basics can be written in 6 steps:
1. Consider a square matrix A = [a_ij]_n.
2. We know how to write the Cofactor A_ij of any element a_ij of the matrix A.
3. So we can write a new matrix which contains the Cofactors.
• For example, if the original matrix is
$A ~=~\left [\begin{array}{r}                           
a_{11}      &{    a_{12}    } &{    a_{13}    }    \\
a_{21}      &{    a_{22}    } &{    a_{23}    }    \\
a_{31}      &{    a_{32}    } &{    a_{33}    }    \\
\end{array}\right ]$,
then the corresponding matrix of Cofactors will be:
$\left [\begin{array}{r}                           
A_{11}      &{    A_{12}    } &{    A_{13}    }    \\
A_{21}      &{    A_{22}    } &{    A_{23}    }    \\
A_{31}      &{    A_{32}    } &{    A_{33}    }    \\
\end{array}\right ]$

4. Now, the transpose of the matrix of Cofactors will be:
$\left [\begin{array}{r}                           
A_{11}      &{    A_{21}    } &{    A_{31}    }    \\
A_{12}      &{    A_{22}    } &{    A_{32}    }    \\
A_{13}      &{    A_{23}    } &{    A_{33}    }    \\
\end{array}\right ]$
• This matrix obtained by transposing, is called adjoint matrix.
5. Adjoint of matrix A is denoted by adj A.
6. For the example matrix A mentioned in (3),
adj A = $\left [\begin{array}{r}                           
A_{11}      &{    A_{21}    } &{    A_{31}    }    \\
A_{12}      &{    A_{22}    } &{    A_{32}    }    \\
A_{13}      &{    A_{23}    } &{    A_{33}    }    \\
\end{array}\right ]$

Solved example 20.19
Find adj A for $A~=~\left [\begin{array}{r}   
2       &{    5    }    \\
-3       &{    7    }    \\
\end{array}\right ]$
Solution:
1. First we will find the Minors and Cofactors:

2. So the matrix of Cofactors is:
$\left [\begin{array}{r}   
7       &{    3    }    \\
-5       &{    2    }    \\
\end{array}\right ]$

3. The transpose of the above matrix, is the required adjoint matrix. So we get:

$\text{adj A}~=~\left [\begin{array}{r}   
7       &{    -5    }    \\
3       &{    2    }    \\
\end{array}\right ]$

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For any square matrix of order 2, there is a direct method to find the adjoint. It can be written in 4 steps:
1. Fig.20.5(a) below shows the original matrix A.

Fig.20.5

2. Consider the yellow diagonal, which is drawn from left-top to right-bottom. We need to interchange the elements in this diagonal.   
3. Consider the red diagonal, which is drawn from right-top to left-bottom. We should not interchange the elements in this diagonal. But we must change the signs of those elements.
4. The resulting new matrix is the adj A. It is shown in fig.20.5(b) above.

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Now we will see four theorems related to adjoint matrices.

Theorem I
• If A is any given square matrix of order n, then
A(adj A) = (adj A)A = |A|I
    ♦ Where I is the identity matrix of the same order n.

Verification can be written in steps:
1. Let $A ~=~\left [\begin{array}{r}                           
a_{11}      &{    a_{12}    } &{    a_{13}    }    \\
a_{21}      &{    a_{22}    } &{    a_{23}    }    \\
a_{31}      &{    a_{32}    } &{    a_{33}    }    \\
\end{array}\right ]$.
• Then adj A will be: $\left [\begin{array}{r}       
A_{11}      &{    A_{21}    } &{    A_{31}    }    \\
A_{12}      &{    A_{22}    } &{    A_{32}    }    \\
A_{13}      &{    A_{23}    } &{    A_{33}    }    \\
\end{array}\right ]$
2. We know how to write the product A(adj A).
• Let us write the first element (intersection of R₁ and C₁):
a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃
• We see that:
    ♦ First term is the product of a₁₁ and it’s Cofactor.
    ♦ Second term is the product of a₁₂ and it’s Cofactor.
    ♦ Third term is the product of a₁₃ and it’s Cofactor.            
    ♦ Also, we are taking the sum of the three terms.
3. So we can write:
The first element in A(adj a) is: The determinant |A|
4. All diagonal elements of A(adj a), will be similar to the form written in (2).
• So all diagonal elements in A(adj a), will become |A|
5. Let us write the second element (intersection of R₁ and C₂) of A(adj a):
a₁₁ A₂₁ + a₁₂ A₂₂ + a₁₃ A₂₃
• We see that:
    ♦ First term is the product of a₁₁ and a different Cofactor.
    ♦ Second term is the product of a₁₂ and a different Cofactor.
    ♦ Third term is the product of a₁₃ and a different Cofactor.
    ♦ Also, we are taking the sum of the three terms.
6. So we can write:
The second element in A(adj a) is: zero
7. All non-diagonal elements of A(adj a), will be similar to the form written in (5).
• So all non-diagonal elements in A(adj a), will become zero.
8. Thus we get:
$A(\text{adj A})~=~\left[\begin{array}{r}   
|A|    &{    0    }    &{    0    }    \\
0    &{    |A|    }    &{    0    }    \\
0    &{    0    }    &{    |A|    }    \\
\end{array}\right]
~=~|A|\left[\begin{array}{r}                           
1    &{    0    }    &{    0    }    \\
0    &{    1    }    &{    0    }    \\
0    &{    0    }    &{    1    }    \\
\end{array}\right]~=~|A|I$
9. Similarly, we can show that (adj A)A = |A|I
10. Based on (8) and (9), we can write:
A(adj A) = (adj A) A = |A|I

Theorem II
If A and B are non-singular matrices of the same order, then AB and BA are also non-singular matrices of the same order.
(We will see the proof of this theorem in higher classes)
    ♦ A square matrix A is said to be singular if |A| = 0
    ♦ A square matrix A is said to be non-singular if |A| ≠ 0

Theorem III
If A and B are square matrices of the same order, then |AB| = |A| |B|.
(We will see the proof of this theorem in higher classes)

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Now we can write about an interesting result. It can be written in 4 steps:

1. Let A be a square matrix of order 3.
2. Let us try to simplify (adj A)A: 

◼ Remarks:
• 1 (magenta color): Here we use theorem I
• 2 (magenta color): In the R.H.S, the unit matrix is multiplied by a constant value. So all non-diagonal elements of the resulting matrix will be zeroes.
• 3 (magenta color): Here we take determinants on both sides.
• 5 (magenta color): Here we apply theorem III in the L.H.S.

3. We see that:
• We started the calculations with a square matrix A of order 3.
• In the final result, the determinant of A has a power of 2.
4. So we can write the general form:
If A is a square matrix of order n, then
|(adj a)| = |A|^n-1.
(We will see the actual proof in higher classes)

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Theorem IV
A square matrix A is invertible if and only if A is a non-singular matrix.
• Proof can be written in 3 steps:
1. Let A be a invertible matrix of order n. Also, let I be an identity matrix of the same order n.
• Then, from the discussion that we had on matrices in the previous chapter, we can write: AB = BA = I.
    ♦ Where B is a square matrix of order n.
2. Consider the equation AB = I
• Writing determinants of matrices on both sides, we get:
|AB| = |I|
• This is same as |AB| = 1
3. Applying theorem III on the L.H.S, we get: |A| |B| = 1.
• |A| and |B| are numbers. If their product is 1, it means that, none of them can be zero.
• So we get: |A| ≠ 0.
• That means, A is a non-singular matrix.

Converse of Theorem IV
If A is a non-singular square matrix, then it is invertible.
• Proof can be written in 3 steps:
1. Consider theorem I:
A(adj A) = (adj A) A = |A|I
2. |A| is a number. So we can use it to divide the whole equation. We get:
$A \left(\frac{1}{|A|}(\text{adj A}) \right)~=~ \left(\frac{1}{|A|}(\text{adj A}) \right)A~=~I$
3. We can rearrange this equation into a familiar form, if we put 'B' in the place of $\frac{1}{|A|}(\text{adj A})$.
• We get: AB = BA = I
4. The above result implies that, matrix B is the inverse of matrix A.
• So we can write:
A is invertible and the inverse is $\frac{1}{|A|}(\text{adj A})$

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

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