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

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

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