# Example 33

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 33 Use product 1−1202−33−24 −20192−361−2 to solve the system of equations x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 Consider the product 1−1202−33−24 −20192−361−2 = 1 −2+ −1 9+2(6)1 0+ −1 2+2(1)1 1+ −1 −3+2(−2)0 −2+2 9+(−3)(6)0 0+2 2+(−3)(1)0 1+0 −3+(−3)(−2)3 −2+ −2 9+(6)3 0+ −2 2+4(1)3 1+ −2 −3+4(−2) = −2−9+120−2+21+3−40+18−180+4−30−6+6−6−18+240−4+43+6−8 = 100 010 001 We know that AA-1 = I So −20102−361−2 is inverse of 1−1202−33−24 i.e. 𝟏−𝟏𝟐𝟎𝟐−𝟑𝟑−𝟐𝟒−𝟏= −𝟐𝟎𝟏𝟗𝟐−𝟑𝟔𝟏−𝟐 Given equations are x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 Step 1 Write the equation as AX = B 1−1202−33−24 𝑥𝑦𝑧 = 112 Here A 1−1202−33−24, X = 𝑥𝑦𝑧 & B = 112 Now, AX = B X = A-1 B Here A = 1−1202−33−24 So, A-1 = 1−1202−33−24−1= −20192−361−2 X = A-1 B Putting Value 𝑥𝑦𝑧 = −20192−361−2 112 = −20192−361−2 112 𝑥𝑦𝑧 = −2 1+0 1+1(2)9 1+2 1−3(2)6 1+1 1−2(2) = −2+0+29+2−66+1−4 𝑥𝑦𝑧 = 053 Hence x = 0 , y = 5 & z = 3

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Example 33 You are here

Example 34 Important

Chapter 4 Class 12 Determinants

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .