Examples

Chapter 4 Class 12 Determinants
Serial order wise

Transcript

Example 16 Solve the system of equations 2x + 5y = 1 3x + 2y = 7 Step 1 Write equation as AX = B [โ 8(2&5@3&2)] [โ 8(๐ฅ@๐ฆ)] = [โ 8(1@7)] A = [โ 8(2&5@3&2)] , X = [โ 8(๐ฅ@๐ฆ)] , B = [โ 8(1@7)] Step 2 Calculate |A|, A = [โ 8(2&5@3&2)] |A| = |โ 8(2&5@3&2)| = 2 ร 2 โ 5 ร 3 = 4 โ 15 = โ 11 Since |๐จ| โ  0, System is consistent and the system of equations has a unique solution AX = B X = A-1 B Step 3 Calculating X = A-1 B Here, A-1 = 1/(|A|) adj (A) adj A = [โ 8(2&5@3&2)] = [โ 8(2&โ5@โ3&2)] Now, A-1 = 1/(|A|) adj A Putting values = ๐/(โ๐๐) [โ 8(๐&โ๐@โ๐&๐)] & B = [โ 8(1@7)] Now X = A-1 B [โ 8(๐ฅ@๐ฆ)] = (โ1)/11 [โ 8(2&โ5@โ3&2)] [โ 8(1@7)] [โ 8(๐ฅ@๐ฆ)] = (โ1)/11 [โ 8(2(1)+(โคถ7โ5)7@โ3(1)+2(7))] [โ 8(๐ฅ@๐ฆ)] = (โ1)/11 [โ 8(2โ35@โ3+14)] = (โ1)/11 [โ 8(โ33@11)] = [โ 8(โ33 ร (โ1)/11@11 ร (โ1)/11)] [โ 8(๐ฅ@๐ฆ)] = [โ 8(3@โ1)] Hence x = 3 & y = โ 1