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If A = [3 2 1 4 -1 2 7 3 -3], then find A -1 and hence solve the following system of equations : 3x + 4y + 7z = 14, 2x − y + 3z = 4, x + 2y − 3z = 0

OR

If A = [2 1 1 1 0 1 0 2 -1] find the inverse of A using elementary row transformations and hence solve the following matrix equation XA = [1  0  1 ].

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/


Find A-1 and hence solve equations: 3x + 4y + 7z = 14 | Solve matrix

Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3 Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4 Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5 Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 6 Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 7 Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 8 Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 9 Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 10 Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Part 11

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Transcript

Question 25 If A = [■8(3&2&[email protected]&−1&[email protected]&3&−3)], then find A-1 and hence solve the following system of equations : 3x + 4y + 7z = 14, 2x − y + 3z = 4, x + 2y − 3z = 0 Given 3x + 4y + 7z = 14, 2x − y + 3z = 4, x + 2y − 3z = 0 Write equation as AX = B [■8(3&4&[email protected]&−1&[email protected]&2&−3)][■8(𝑥@𝑦@𝑧)] = [■8([email protected]@0)] i.e. A’X = B Hence A =[■8(3&2&[email protected]&−1&[email protected]&3&−3)], A’= [■8(3&4&[email protected]&−1&[email protected]&2&−3)] , X = [■8(𝑥@𝑦@𝑧)] & B =[■8([email protected]@0)] Now, A’X = B X = (A’)-1 B X = (A-1)’ B Finding A-1 Calculating |A| |A|= |■8(3&2&[email protected]&−1&[email protected]&3&−3)| = 3(−1(-3) – 3(2)) − 2 (4(–3) − 7(2)) + 1 (4(3) − 7(–1)) = 3(–3) – 2(–26) + 1(19) = –9 + 52 + 19 = 62 So, |A|≠ 0 ∴ The system of equation is consistent & has a unique solutions Now, A-1 = 1/(|A|) adj (A) adj A = [■8(A11&A12&[email protected]&A22&[email protected]&A32&A33)]^′ = [■8(A11&A21&[email protected]&A22&[email protected]&A23&A33)] A = [■8(3&2&[email protected]&−1&[email protected]&3&−3)] 𝐴11 = (–1)1+1 [–1(–3) – 3(2) ] = 3 – 6 = –3 𝐴12 = (–1)1+2 [4(–3) – 7(2)]= −[−12−14] = − (–26) = 26 𝐴13 = (–1)1+3 [4(3) – 7(–1)] = 12 + 7 = 19 𝐴21 = (–1)2+1 [2(–3) – 3(1)] = –(–6 – 3) = 9 𝐴22 = (–1)2+2 [3(–3) – 7(1)] = −9 − 7 = −16 𝐴23 = (–1)2+3 [3(3) – 7(2)] = – [9 – 14] = 5 𝐴31 = (–1)3+1 [2(2) – (–1)(1)] = 4 + 1 = 5 𝐴32 = (–1)3+2 [3(2) – 4(1)] = – [6 – 4] = –2 𝐴33 = (–1)3+3 [3(–1) – 4(2)] = –3 – 8 = –11 Thus adj A = [■8(−3&9&[email protected]&−16&−[email protected]&5&−11)] & |A| = 62 Now, A-1 = 1/(|A|) adj A A-1 = 1/62 [■8(−3&9&[email protected]&−16&−[email protected]&5&−11)] Solving X = (A-1)’ B [■8(𝑥@𝑦@𝑧)] = 1/62 [■8(−3&9&[email protected]&−16&−[email protected]&5&−11)]^′ [■8([email protected]@0)] [■8(𝑥@𝑦@𝑧)] = 1/62 [■8(−3&26&[email protected]&−16&[email protected]&−2&−11)][■8([email protected]@0)] " " [■8(𝑥@𝑦@𝑧)]" =" 1/62 [█(−3(14)+26(4)+19(0)@9(14)+(−16)(4)+5(0)@5(14)+(−2)(4)+(−11)(0))] " " [■8(𝑥@𝑦@𝑧)]" =" 1/62 [■8([email protected]@62)] " " [■8(𝑥@𝑦@𝑧)]" =" [■8([email protected]@1)] "∴ x = 1, y = 1 and z = 1 " Question 25 If A = [■8(2&1&[email protected]&0&[email protected]&2&−1)] find the inverse of A using elementary row transformations and hence solve the following matrix equation XA = [1 0 1 ]. Given A = [■8(2&1&[email protected]&0&[email protected]&2&−1)] We know that A = IA [■8(2&1&[email protected]&0&[email protected]&2&−1)] = [■8(1&0&[email protected]&1&[email protected]&0&1)] A R1 ↔ R2 [■8(𝟏&0&[email protected]&1&[email protected]&2&−1)] = [■8(0&1&[email protected]&0&[email protected]&0&1)] A R2 → R2 – 2R1 [■8(1&0&[email protected]𝟐−𝟐(𝟏)&1−2(0)&1−2(1)@0&2&−1)] = [■8(0&1&[email protected]−2(0)&0−2(1)&0−2(0)@0&0&1)] A [■8(1&0&[email protected]𝟎&1&−[email protected]&2&−1)] = [■8(0&1&[email protected]&−2&[email protected]&0&1)] A R3 → R3 – 2R2 [■8(1&0&[email protected]&1&−[email protected]−2(0)&𝟐−𝟐(𝟏)&−1−2(−1))] = [■8(0&1&[email protected]&−2&[email protected]−2(1)&0−2(−2)&1−2(0))] A [■8(1&0&[email protected]&1&−[email protected]&𝟎&1)] = [■8(0&1&[email protected]&−2&[email protected]−2&4&1)] A R1 → R1 – R3 [■8(1−0&0−0&𝟏−𝟏@0&1&−[email protected]&0&1)] = [■8(0−(−2)&1−4&0−[email protected]&−2&[email protected]−2&4&1)] A [■8(1&0&𝟎@0&1&−[email protected]&0&1)] = [■8(2&−3&−[email protected]&−2&[email protected]−2&4&1)] A R2 → R2 + R3 [■8(1&0&[email protected]+0&1+0&−𝟏+𝟏@0&0&1)] = [■8(2&−3&−[email protected]+(−2)&−2+4&[email protected]−2&4&1)] A [■8(1&0&[email protected]&1&𝟎@0&0&1)] = [■8(2&−3&−[email protected]−1&2&[email protected]−2&4&1)] A 𝐼 = [■8(2&−3&−[email protected]−1&2&[email protected]−2&4&1)] A R1 → R1 – R3 [■8(1−0&0−0&𝟏−𝟏@0&1&−[email protected]&0&1)] = [■8(0−(−2)&1−4&0−[email protected]&−2&[email protected]−2&4&1)] A [■8(1&0&𝟎@0&1&−[email protected]&0&1)] = [■8(2&−3&−[email protected]&−2&[email protected]−2&4&1)] A R2 → R2 + R3 [■8(1&0&[email protected]+0&1+0&−𝟏+𝟏@0&0&1)] = [■8(2&−3&−[email protected]+(−2)&−2+4&[email protected]−2&4&1)] A [■8(1&0&[email protected]&1&𝟎@0&0&1)] = [■8(2&−3&−[email protected]−1&2&[email protected]−2&4&1)] A 𝐼 = [■8(2&−3&−[email protected]−1&2&[email protected]−2&4&1)] A This is similar to I = A-1 A Hence A-1 = [■8(2&−3&−[email protected]−1&2&[email protected]−2&4&1)] Now, We need to solve XA = [1 0 1 ] X = [1 0 1 ]A–1 Putting value of A–1 X = [1 0 1 ][■8(2&−3&−[email protected]−1&2&[email protected]−2&4&1)] X = [1 0 1 ]_(1 × 3) [■8(2&−3&−[email protected]−1&2&[email protected]−2&4&1)]_(3 × 3) X = [■8(1(2)+0(−1)+1(−2)&1(−3)+0(2)+1(4)&1(−1)+0(1)+1(1))]_(1 × 3) X = [■8(2−2&−3+4&−1+1)] X = [■8(0&1&0)]

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.