TS Inter 1st Year Maths 1A Matrices Important Questions Short Answer Type

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Matrices Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Matrices Important Questions Short Answer Type

Question 1.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), , then show that for all the positive integers n, Aⁿ = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) [May 98, 91]
Answer:

∴ S(k + 1) is true.
∴ By the principle of mathematical induction, S(n) is true for all n ∈ N.
∴ Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), ∀ n ∈ N.

Question 2.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 1 \\
0 & 1 & -1 \\
3 & -1 & 1
\end{array}\right]\), then find A³ – 3A² – A – 3I, where I is unit matrix of order 3. [Mar. 19 (TS); Mar. 11, 98; May 98]
Answer:

Question 3.
If I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and E = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), then show that (aI + bE)³ = a³I + 3a²bE, where I is unit matrix of order 2. [Mar. 16 (TS), 15(AP), 10; May 05]
Answer:

Question 4.
If θ – Φ = \(\frac{\pi}{2}\), then show that \(\left[\begin{array}{cc}
\cos ^2 \theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^2 \theta
\end{array}\right]\left[\begin{array}{cc}
\cos ^2 \phi & \cos \phi \sin \phi \\
\cos \phi \sin \phi & \sin ^2 \phi
\end{array}\right]\) = 0 [May 15 (TS); May 11, 09, 96; Mar. 04]
Answer:

Question 5.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), then show that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\) for any integer n ≥ 1, by using mathematical induction. [May 08, 02]
Answer:

∴ S(k + 1) is true.
∴ By using the principle of mathematical Induction, S(n) is true for all n ∈ N.
∴ Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\) ∀ n ∈ N.

Question 6.
For any n × n matrix A, prove that A can be uniquely expressed as a sum of a symmetric matrix and a skew symmetric matrix. (Mar. ‘03)
Answer:
Let A be a square matrix.

∴ A can be expressed as a sum of a symmetric matrix and a skew symmetric matrix.

Question 7.
Show that \(\left|\begin{array}{ccc}
\mathbf{1} & \mathbf{a} & \mathbf{a}^2 \\
\mathbf{1} & \mathbf{b} & \mathbf{b}^2 \\
\mathbf{1} & \mathbf{c} & \mathbf{c}^2
\end{array}\right|\) = (a – b) (b – c) (c – a). [Mar. 17 (TS). 05]
Answer:

= (a – b) (b – c) (c – a) [0(c² – c) – 1(0 – 1) + (a + b) (0 – 0)]
= (a – b) (b – c) (c – a) (1) = (a – b) (b – c) (c – a) = RHS

Question 8.
Show that \(\left|\begin{array}{lll}
b c & b+c & 1 \\
c a & c+a & 1 \\
a b & a+b & 1
\end{array}\right|\) = (a – b) (b – c) (c – a). [Board Paper]
Answer:

= (a – b) (c – a) [b (1 – 0) – 1 (c – 0) + 0 (ac + bc – ab)]
= (a – b) (c – a) [b – c] = (a – b) (b – c) (c – a) = R.H.S.

Question 9.
Show that \(\left|\begin{array}{ccc}
b+c & c+a & a+b \\
a+b & b+c & c+a \\
a & b & c
\end{array}\right|\) = a³ + b³ + c³ – 3abc. [May 13, 07; Mar. 08]
Answer:

= (a + b + c) [c² – bc – ac + ab + a² – 2ab + b²]
= (a + b + c) [a² + b² + c² – ab – bc – ca]
= a³ + b³ + c³ – 3abc

Question 10.
If \(\left|\begin{array}{ccc}
a & a^2 & 1+a^3 \\
b & b^2 & 1+b^3 \\
c & c^2 & 1+c^3
\end{array}\right|\) = 0 and \(\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right|\) ≠ 0 then show that abc = – 1. [Mar. 14, 04; May. 98, 95]
Answer:

Question 11.
Show that \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\) = 0. [May. 08]
Answer:

Question 12.
Let A and B be invertiable matrices then show that (AB)^-1 = B^-1 A^-1. [May. 03]
Answer:
A is invertible matrix then A^-1 exists and AA^-1 = A^-1 A = I
B is an invertible matrix then B^-1 exists and BB^-1 = B^-1B = I
Now (AB) (B^-1 A^-1) = A(BB^-1)A^-1 = A(T)A^-1 = AA^-1 = I
∴ (AB) (B^-1 A^-1) = I ………………… (1)
(B^-1 A^-1) (AB) = B^-1 (A^-1 A) B = B^-1 (I) B = B^-1 B = 1
∴ (B^-1 A^-1) (AB) = I ………………….. (2)
From (1) & (2)
(AB) (B^-1 A^-1) = (B^-1 A^-1) (AB) = I
AB is invertiable and (AB)^-1 = B^-1 A^-1.

Question 13.
Find the adjoint and the inverse of the matrix A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\) [May 14; Mar. 08]
Answer:
Given A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\)
Cofactor of 1 is A₁₁ = +(16 – 9) = 7
Cofactor of 3 is A₁₂ = – (4 – 3) = – 1
Cofactor of 3 is A₁₃ = (3 – 4) = – 1
Cofactor of 1 is B₁₁ = – (12 – 9) = -3
Cofactor of 4 is B₁₂ = (4 – 3) = 1
Cofactor of 3 is B₁₃ = – (3 – 3) = 0
Cofactor of 1 is C₁₁ = (9 – 12) = – 3
Cofactor of 3 is C₁₂ = – (3 – 3) = 0
Cot actor of 4 is C₁₃ = (4 – 3) = 1
∴ Cofactor matrix,

Now, det A = 1(16 – 9) – 3(4 – 3) + 3 (3 – 4)
= 1(7) – 3(1) + 3(- 1) = 7 – 3 – 3 = 1
Hence A is invertiable.
A^-1

Question 14.
Show that A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\) is non-singular and find A^-1. [Mar. 17 (TS). 12, 98; May 89]
Answer:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
3 & 2 & 3 \\
1 & 1 & 2
\end{array}\right]\)
det A = 1 (4 – 3) – 2 (6 – 3) + 1 (3 – 2)
= 1 – 6 + 1 = – 4 ≠ 0
∴ A is a non-singular matrix.
Cofactor of 1 is A₁ = + (4 – 3) = 1
Cofactor of 2 is B₁ = – (6 – 3) = – 3
Cofactor of 1 is C₁ =+(3 – 2) = 1
Cofactor of 3 is A₂ = – (4 – 1) = – 3
Cofactor of 2 is B₂ = + (2 – 1) = 1
Cofactor of 3 is C₂ = – (1 – 2) = + 1
Cofactor of 1 is A₃ = + (6 – 2) = 4
Cofactor of 1 is B₃ = – (3 – 3) = 0
Cofactor of 2 is C₃ = + (2 – 6) = – 4
∴ Cofactor matrix of A is B

Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\). [Mar. 05; May 98]
Answer:

Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\). [Mar. 08, 89]
Answer:
\(\left[\begin{array}{rrr}
-2 & 3 & 1 \\
1 & -2 & 0 \\
2 & -2 & -1
\end{array}\right],\left[\begin{array}{rrr}
-2 & 3 & 1 \\
1 & -2 & 0 \\
2 & -2 & -1
\end{array}\right]\)

Question 15.
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 is 3A’. Find A^-1. [Mar. 19 (AP), May 08]
Answer:
Given A = \(\left[\begin{array}{rrr}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right]\)

Cofactor of – 1 is A₁ = + (1 – 4) = – 3
Cofactor of – 2 is B₁ = – (2 + 4) = – 6
Cofactor of – 2 is C₁ = + (- 4 – 2) = – 6
Cofactor of 2 is A₂ = – (- 2 – 4) = 6
Cofactor of 1 is B₂ = + (- 1 + 4) = 3
Cofactor of – 2 is C₂ = – (2 + 4) = – 6
Cofactor of 2 is A₃ = + (4 + 2) = 6
Cofactor of – 2 is B₃ = – (2 + 4) = – 6
Cofactor of 1 is C₃ = + (- 1 + 4) = 3
∴ Cofactor matrix of

∴ Adj A = 3A’
det A = – 1 (1 – 4) + 2 (2 + 4) – 2 (- 4 – 2)
= – 1 (- 3) + 2(6) – 2 (- 6) = + 3 + 12 + 12 = 27 ≠ 0
∴ A is invertiable.
A^-1 = \(\frac{{adj} A}{{det} A}=\frac{1}{27}\left[\begin{array}{ccc}
-3 & 6 & 6 \\
-6 & 3 & -6 \\
-6 & -6 & 3
\end{array}\right]\)

Question 16.
If abc ≠ 0, find the inverse of \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\). [Mar. 06; Oct. 96]
Answer:
Let A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\)
Cofactor of a is A₁ + (bc – 0) = bc
Cofactor of 0 is B₁ = – (0 – 0) = 0
Cofactor of 0 is C₁ = + (0 – 0) = 0
Cofactor of 0 is A₂ = – (0 – 0) = 0
Cofactor of b is B₂ = + (ac – 0) = ac
Cofactor of 0 is C₂ = – (0 – 0) = 0
Cofactor of 0 is A₃ = + (0 – 0) = 0
Cofactor of 0 is B₃ = – (0 – 0) = 0
Cofactor of c is C₃ = + (ab – 0) = ab
∴ Cofactor matrix of

Question 17.
If 3A = \(\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\), then show that A^-1 = A’. [Mar. 14, 09; May. 12]
Answer:

Question 18.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\), then find (A’)^-1. [Board Paper]
Answer:
Given A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\)
Cofactor of 1 is A₁ = + (- 1 – 8) = – 9
Cofactor of 0 is A₂ = – (- 2 – 6) = 8
Cofactor of – 2 is A₃ = (- 8 + 3) = – 5
Cofactor of – 2 is B₁ = – (0 + 8) = – 8
Cot actor of – 1 is B₂ = + (1 + 6) = 7
Cofactor of 2 is B₃ = – (4 – 0) = – 4
Cofactor of 3 is C₁ = + (0 – 2) = – 2
Cofactor of 4 is C₂ = – (2 – 4) = 2
Cofactor of 1 is C₃ = (- 1 + 0) = – 1