TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e)

Students must practice these TS Intermediate Maths 1A Solutions Chapter 3 Matrices Ex 3(e) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e)

I.
Question 1.
Find the adjoint and inverse of the following matrices. (March 2002)
i) \(\left[\begin{array}{rr}
2 & -3 \\
4 & 6
\end{array}\right]\)
Answer:
If A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) then adj A = \(\left[\begin{array}{rr}
\mathrm{d} & -\mathrm{b} \\
-\mathrm{c} & \mathrm{a}
\end{array}\right]\)
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 1

ii) \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 2

iii) Find the adjoint and inverse of the matrix \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\).
Answer:
Find cofactors of elements in the matrix as
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 3

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e)

iv) \(\left|\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right|\)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 4

Question 2.
If A = \(\left[\begin{array}{cc}
\mathrm{a}+\mathrm{i b} & \mathrm{c}+\mathrm{i d} \\
-\mathrm{c}+\mathrm{i d} & \mathrm{a}-\mathrm{i b}
\end{array}\right]\), a2 + b2 + c2 + d2 = 1
Answer:
det A = (a + ib) (a – ib) – (c + id) (- c + id)
= (a2 – i2 b2) – (- c2 + i2d2)
= a2 + b2 + c2 + d2 (∵ i2 = -1)
= 1
Adj A = \(\left[\begin{array}{cc}
a-i b & -c-i d \\
c-i d & a+i b
\end{array}\right]\)
A-1 = \(\frac{{Adj} A}{{det} A}=\left[\begin{array}{cc}
a-i b & -c-i d \\
c-i d & a+i b
\end{array}\right]\)

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\), then find (A’)-1. (Board Model Paper)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 5

Question 4.
If A = \(\left[\begin{array}{rrr}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right]\), then show that the adjoint of A = 3A, find A-1
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 6

Question 5.
If abc ≠ 0; find the inverse of \(\left[\begin{array}{lll}
\mathrm{a} & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\) (May 2006)
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 7

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e)

II.
Question 1.
If A = \(\left[\begin{array}{lll}
\mathrm{b}+\mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{b}-\mathrm{a} \\
\mathrm{c}-\mathrm{b} & \mathrm{c}+\mathrm{a} & \mathrm{a}-\mathrm{b} \\
\mathrm{b}-\mathrm{c} & \mathrm{a}-\mathrm{c} & \mathrm{a}+\mathrm{b}
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathrm{b}+\mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{b}-\mathrm{a} \\
\mathrm{c}-\mathrm{b} & \mathrm{c}+\mathrm{a} & \mathrm{a}-\mathrm{b} \\
\mathrm{b}-\mathrm{c} & \mathrm{a}-\mathrm{c} & \mathrm{a}+\mathrm{b}
\end{array}\right]\), then show that ABA-1 is a diagonal matrix.
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 8

Question 2.
If 3A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\), then show that A-I = A’.
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 9
∴ A.A’ = I and by definition A’ = A-1
similarly A’.A = I

Question 3.
If A = \(\left[\begin{array}{rrr}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\), then show that A-1 = A3
Answer:
TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(e) 10
So, the multiplicative inverse of A exists and it is A3.
∴ A-1 = A3

Question 4.
If AB = I or BA = I, then prove that A is invertible and B = A-1.
Answer:
Given AB = I
⇒ |AB| = |I|
⇒ |A| |B| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix.
Also BA = I
⇒ |B| |A| = |I|
⇒ |A| |B| = 1
⇒ |A| *0
∴ A is a non-singular matrix.
⇒ A is invertible
⇒ A-1 exists AB = I
⇒ A-1 AB = A-1I
⇒ (A-1 A) B = A-1I
⇒ IB = A-1I
⇒ B = A-1.

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