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