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

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

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

I.

Question 1.
ఈ క్రిందివాటిని ఒకే మాత్రికగా వ్రాయండి.
(i) [2 1 3] + [0 0 0]
(ii) $$\left[\begin{array}{c} 0 \\ 1 \\ -1 \end{array}\right]+\left[\begin{array}{c} -1 \\ 1 \\ 0 \end{array}\right]$$
(iii) $$\left[\begin{array}{ccc} 3 & 9 & 0 \\ 1 & 8 & -2 \end{array}\right]+\left[\begin{array}{ccc} 4 & 0 & 2 \\ 7 & 1 & 4 \end{array}\right]$$
(iv) $$\left[\begin{array}{cc} -1 & 2 \\ 1 & -2 \\ 3 & -1 \end{array}\right]+\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ -2 & 1 \end{array}\right]$$
Solution:

Question 2.
A = $$\left[\begin{array}{cc} -1 & 3 \\ 4 & 2 \end{array}\right]$$, B = $$\left[\begin{array}{cc} 2 & 1 \\ 3 & -5 \end{array}\right]$$, X = $$\left[\begin{array}{ll} x_1 & x_2 \\ x_3 & x_4 \end{array}\right]$$, A + B = X అయితే x1, x2, x3, x4 ల విలువలు కనుక్కోండి.
Solution:
A + B = X కనుక

∴ x1 = 1, x2 = 4, x3 = 7, x4 = -3

Question 3.
A = $$\left[\begin{array}{ccc} -1 & -2 & 3 \\ 1 & 2 & 4 \\ 2 & -1 & 3 \end{array}\right]$$, B = $$\left[\begin{array}{ccc} 1 & -2 & 5 \\ 0 & -2 & 2 \\ 1 & 2 & -3 \end{array}\right]$$, C = $$\left[\begin{array}{ccc} -2 & 1 & 2 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \end{array}\right]$$ అయితే A + B + C ని కనుక్కోండి.
Solution:

Question 4.
A = $$\left[\begin{array}{ccc} 3 & 2 & -1 \\ 2 & -2 & 0 \\ 1 & 3 & 1 \end{array}\right]$$, B = $$\left[\begin{array}{ccc} 3 & 2 & -1 \\ 2 & -2 & 0 \\ 1 & 3 & 1 \end{array}\right]$$, X = A + B అయితే, మాత్రిక X ను కనుక్కోండి.
Solution:

Question 5.
$$\left[\begin{array}{cc} x-3 & 2 y-8 \\ z+2 & 6 \end{array}\right]=\left[\begin{array}{cc} 5 & 2 \\ -2 & a-4 \end{array}\right]$$ అయితే x, y, z, a విలువలను కనుక్కోండి. [May ’06]
Solution:
$$\left[\begin{array}{cc} x-3 & 2 y-8 \\ z+2 & 6 \end{array}\right]=\left[\begin{array}{cc} 5 & 2 \\ -2 & a-4 \end{array}\right]$$
∴ x – 3 = 5
⇒ x = 3 + 5 = 8
2y – 8 = 2
⇒ 2y = 8 + 2 = 10
⇒ y = 5
z + 2 = -2
⇒ z = -2 – 2 = -4
a – 4 = 6
⇒ a = 4 + 6 = 10

II.

Question 1.
$$\left[\begin{array}{ccc} x-1 & 2 & 5-y \\ 0 & z-1 & 7 \\ 1 & 0 & a-5 \end{array}\right]=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 7 \\ 1 & 0 & 0 \end{array}\right]$$ అయితే x, y, z, a ల విలువలు కనుక్కోండి.
Solution:
$$\left[\begin{array}{ccc} x-1 & 2 & 5-y \\ 0 & z-1 & 7 \\ 1 & 0 & a-5 \end{array}\right]=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 0 & 4 & 7 \\ 1 & 0 & 0 \end{array}\right]$$
∴ x – 1 = 1
⇒ x = 1 + 1 = 2
5 – y = 3
⇒ y = 5 – 3 = 2
z – 1 = 4
⇒ z = 4 + 1 = 5
a – 5 = 0
⇒ a = 5

Question 2.
A = $$\left[\begin{array}{ccc} 1 & 3 & -5 \\ 2 & -1 & 5 \\ 2 & 0 & 1 \end{array}\right]$$ అయితే జాడ A కనుక్కోండి. [May ’13]
Solution:
జాడ A = ప్రధాన వికర్ణ మూలకాల మొత్తం
= 1 – 1 + 1
= 1

Question 3.
A = $$\left[\begin{array}{ccc} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 4 & 5 & -6 \end{array}\right]$$, B = $$\left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$$ అయితే B – A, 4A – 5B లను కనుక్కోండి.
Solution:

Question 4.
A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right]$$, B = $$\left[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right]$$ అయితే 3B – 2A ను కనుక్కోండి.
Solution:
A = $$\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right]$$, B = $$\left[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right]$$