AP Inter 1st Year Maths 1A Solutions Chapter 3 మాత్రికలు Ex 3(f)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Chapter 3 మాత్రికలు Exercise 3(f) will help students to clear their doubts quickly.

AP Inter 1st Year Maths 1A Solutions Chapter 3 మాత్రికలు Exercise 3(f)

I. కింది ఇచ్చిన ప్రతి మాత్రికకూ కోటి కనుక్కోండి.

Question 1.
$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right|$$
= 0 – 0
= 0
| 1 | = 1 ≠ 0
∴ ρ(A) = 1

Question 2.
$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|$$
= 1 – 0
= 1 ≠ 0
∴ ρ(A) = 2

Question 3.
$$\left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right|$$
= 0 – 0
= 0
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 4.
$$\left[\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right|$$
= 0 – 1
= -1 ≠ 0
∴ ρ(A) = 2

Question 5.
$$\left[\begin{array}{ccc} 1 & 0 & -4 \\ 2 & -1 & 3 \end{array}\right]$$
Solution:
$$\left|\begin{array}{cc} 1 & -4 \\ 2 & 3 \end{array}\right|$$
= 3 + 8
= 11 ≠ 0
∴ ρ(A) = 2

Question 6.
$$\left[\begin{array}{lll} 1 & 2 & 6 \\ 2 & 4 & 3 \end{array}\right]$$
Solution:
$$\left|\begin{array}{ll} 2 & 6 \\ 4 & 3 \end{array}\right|$$
= 6 – 24
= -18 ≠ 0
∴ ρ(A) = 2

II.

Question 1.
$$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$$
Solution:
Det A = $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$$
= 1(1 – 0) – 0(0 – 0) + 0(0 – 0)
= 1 – 0 + 0
= 1 ≠ 0
∴ ρ(A) = 3

Question 2.
$$\left[\begin{array}{ccc} 1 & 4 & -1 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ccc} 1 & 4 & -1 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end{array}\right|$$
= 1(6 – 0) – 2(8 + 1) + 0(0 + 3)
= 6 – 18
= -12 ≠ 0
∴ ρ(A) = 3

Question 3.
$$\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]$$ [(T.S) Mar. ’15]
Solution:
Det A = $$\left|\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right|$$
= 1(6 – 4) – 2(4 – 3) + 0(8 – 9)
= 2 – 2 + 0
= 0
∴ ρ(A) ≠ 3, ρ(A) < 3
ఉపమాత్రిక నిర్ధారకం $$\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|$$
= 3 – 4
= -1 ≠ 0
∴ ρ(A) = 2

Question 4.
$$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$ [Mar. ’08]
Solution:
A = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$
det A = 0, ρ(A) ≠ 3.
ప్రతి 2 × 2 ఉపమాత్రిక det సున్న
∴ ρ(A) ≠ 2
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 5.
$$\left[\begin{array}{cccc} 1 & 2 & 0 & -1 \\ 3 & 4 & 1 & 2 \\ -2 & 3 & 2 & 5 \end{array}\right]$$
Solution:
ఉపమాత్రిక B నిర్ధారకం = $$\left|\begin{array}{ccc} 1 & 2 & 0 \\ 3 & 4 & 1 \\ -2 & 3 & 2 \end{array}\right|$$
= 1(8 – 3) – 2(6 + 2)
= 5 – 16
= -11 ≠ 0
మాత్రిక కోటి = 3

Question 6.
$$\left[\begin{array}{cccc} 0 & 1 & 1 & -2 \\ 4 & 0 & 2 & 5 \\ 2 & 1 & 3 & 1 \end{array}\right]$$
Solution:
ఉపమాత్రిక A నిర్ధారకం = $$\left[\begin{array}{lll} 0 & 1 & 1 \\ 4 & 0 & 2 \\ 2 & 1 & 3 \end{array}\right]$$
= -1(12 – 4) + 1(4 – 0)
= -8 + 4
= -4 ≠ 0
∴ ρ(A) = 3