Question 36 (Choice 2) Evaluate the product AB, where A =[■8(1&−1&0@2&3&4@0&1&2)] and B = [■8(2&2&−4@−4&2&−4@2&−1&5)] Hence the solve system of linear equations x – y = 3 2x + 3y + 4z = 17 y + 2z = 7
Finding the product
[■8(1&−1&0@2&3&4@0&1&2)] [■8(2&2&−4@−4&2&−4@2&−1&5)]
=[■8(1(2)+(⤶7−1)(−4)+0(2)&1(2)+(−1)(2)+0(−1)&1(−4)+(−1)(−4)+0(5)@2(2)+3(−4)+4(2)&2(2)+3(2)+4(−1)&2(−4)+3(−4)+4(5)@0(2)+1(−4)+2(2)&0(2)+1(2)+2(−1)&0(−4)+1(−4)+2(5))]
= [■8(6@0@0)" " ■8(0@6@0)" " ■8(0@0@6)]
= 6[■8(1@0@0)" " ■8(0@1@0)" " ■8(0@0@1)]
Thus,
AB = 6I
A × 𝑩/𝟔 = I
We know that
AA-1 = I
So 𝟏/𝟔 [■8(𝟐&𝟐&−𝟒@−𝟒&𝟐&−𝟒@𝟐&−𝟏&𝟓)] is inverse of [■8(1&−1&0@2&3&4@0&1&2)]
Given equations are
x – y = 3
2x + 3y + 4z = 17
y + 2z = 7
Writing the equation as AX = D
[■8(1&−1&2@0&2&−3@3&−2&4)] [■8(𝑥@𝑦@𝑧)] = [■8(3@17@7)]
Here A = [■8(1&−1&0@2&3&4@0&1&2)], X = [■8(𝑥@𝑦@𝑧)] & D = [■8(3@17@7)]
Now,
AX = D
X = A-1 D
Putting A-1 = 𝟏/𝟔 𝑩=𝟏/𝟔 [■8(𝟐&𝟐&−𝟒@−𝟒&𝟐&−𝟒@𝟐&−𝟏&𝟓)]
So, our equation becomes
[■8(𝑥@𝑦@𝑧)] = 𝟏/𝟔 [■8(𝟐&𝟐&−𝟒@−𝟒&𝟐&−𝟒@𝟐&−𝟏&𝟓)] [■8(3@17@7)]
[■8(𝑥@𝑦@𝑧)] = 𝟏/𝟔 [■8(2(3)+2(17)−4(7)@−4(3)+2(17)−4(7)@2(3)−1(17)+5(7))]
[■8(𝑥@𝑦@𝑧)] = 𝟏/𝟔 [■8(12@−6@24)]
[■8(𝑥@𝑦@𝑧)] = [■8(2@−1@4)]
Hence x = 2 , y = −1 & z = 4

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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