Home TG Inter Study Material TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(a)

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

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

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

Question 1.
Write the following as a single matrix.
(i) [2 1 3] + [0 0 0]
Answer:
[2 1 3] + [0 0 0] = [2 + 0 1 + 0 3 + 0]
= [2 1 3]

(ii) 011+110
Answer:
011+110 = 011+11+0 = 121

(iii) [319802]+[470124]

Answer:
TS-Inter-1st-Year-Maths-1A-Solutions-Chapter-3-Matrices-Ex-3a-1

(iv) 113221+012101
Answer:
TS-Inter-1st-Year-Maths-1A-Solutions-Chapter-3-Matrices-Ex-3a-2

Question 2.
If A = [1432], B = [2315], X = [x1x3x2x4] and A + B = X then find the values of x1, x2, x3 and x4.

Answer:
A + B = X
TS-Inter-1st-Year-Maths-1A-Solutions-Chapter-3-Matrices-Ex-3a-3
⇒ x1 = 1, x2 = 4, x3 = 7, x4 = – 3.

Question 3.
If A = 112221343 B = 101222523 and C = 212110221 then find A + B + C.

Answer:
A + B + C =
TS-Inter-1st-Year-Maths-1A-Solutions-Chapter-3-Matrices-Ex-3a-4

Question 4.
If A = 321223101, B = 324111032 and X = A + B then find X.

Answer:
X = A + B
TS-Inter-1st-Year-Maths-1A-Solutions-Chapter-3-Matrices-Ex-3a-5

Question 5.
If [x3z+22y86] = [522a4] then find the values of x, y, z and a. [May 2006, Mar. 14]

Answer:
Given [x3z+22y86] = [522a4]
We have x – 3 = 5, 2y – 8 = 2, z + 2 = – 2, a – 4 = 6
⇒ x = 8, y = 5, z = – 4, a = 10

II.
Question 1.
If x1012z105y7a5 = 101240370 then find the values x, y, z and a.
Answer:
Given x1012z105y7a5 = 101240370
we have x – 1 = 1, 5 – y = 3, z – 1 = 4,
a – 5 = 0
⇒ x = 2, y = 2, z = 5, a = 5

Question 2.
Find the trace of 121310551

Answer:
Trace of the given matrix
= 1 – 1 + 1 = sum of the diagonal elements
= 1

Question 3.
If A = 024135246 and B = 100210301 find A – B and 4A – 5B.

Answer:
TS-Inter-1st-Year-Maths-1A-Solutions-Chapter-3-Matrices-Ex-3a-6

Question 4.
If A = [132231] and B = [312213] find 3B – 2A.

Answer:
TS-Inter-1st-Year-Maths-1A-Solutions-Chapter-3-Matrices-Ex-3a-7

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