ii Add the products along the three full diagonals that extend from upper left

Ii add the products along the three full diagonals

This preview shows page 126 - 131 out of 151 pages.

(ii) Add the products along the three full diagonals that extend from upper left to lower right. (iii) Subtract the products along the three full diagonals that extend from lower left to upper right. Let’s work the previous example, now using “ Sarrus scheme ”: E.g. Find - 2 3 - 1 4 - 7 1 1 2 - 5 Solution: det A = - 2 3 - 1 4 - 7 1 1 2 - 5 = - 2 3 - 1 4 - 7 1 1 2 - 5 - 2 3 4 - 7 1 2 = ( - 2)( - 7)( - 5) + (3)(1)(1) + ( - 1)(4)(2) - (1)( - 7)( - 1) - (2)(1)( - 2) - ( - 5)(4)(3) = - 70 + 3 - 8 - 7 + 4 + 60 = - 18 . NB: Sarrus rule does not work for n × n matrices where n 4. Properties of determinants: 1. The sign of a determinant changes if any two columns or rows are inter- changed. 123
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E.g. From an example, we saw that - 2 3 - 1 4 - 7 1 1 2 - 5 = - 18 . If we interchange row 1 and row 3, the new determinant; 1 2 - 5 4 - 7 1 - 2 3 - 1 = 18 . Interchanging of columns has the same effect. 2. The determinant of a matrix with a zero column or zero row is zero. E.g. 1 0 - 5 4 0 1 - 2 0 - 1 = 0 and 1 2 0 0 = 0. 3. If any two rows or columns of a matrix are equal, its determinant is zero. E.g. 1 2 - 5 8 - 5 - 2 8 - 5 - 2 = 0 since R 2 = R 3 . 1 2 4 1 - 3 - 5 5 - 3 - 4 0 - 5 - 4 2 1 - 5 2 = 0 since C 1 = C 4 . 4. If any row or column of a matrix is multiplied by a non-zero constant k , it’s determinant is multiplied by k . E.g. If A = - 2 3 - 1 4 - 7 1 1 2 - 5 , then if we multiply column 3 by -5, i.e. we have new matrix B = - 2 3 5 4 - 7 - 5 1 2 25 , 124
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then det B = - 5 det A = - 5( - 18) = 90. Suppose we multiply the whole matrix A by 2, then each row will be multiplied by two. So det 2 A = 2 × 2 × 2 det A = 8 × ( - 18) = - 144 . This is because each row or column is a multiplied by 2, hence the multiple of each row multiplies the det A , in this case three 2’s since we have 3 rows or columns. In general If A is an n × n matrix, then det( kA ) = k n det A. 5. If any row or column of a matrix is multiplied by a non-zero constant k and the result is added to another row or column, it’s determinant remains the same. We can show this by performing row operation. E.g. A = - 2 3 - 1 4 - 7 1 1 2 - 5 using row operation. det A = - 2 3 - 1 4 - 7 1 1 2 - 5 = 0 7 - 11 0 - 15 21 1 - 5 - 2 R 1 + 2 R 3 R 2 - 4 R 3 Expanding along C 1 = 1 × 7 - 11 - 15 21 = 7 × 21 - ( - 15)( - 11) = 147 - 165 = - 18 . Which still yields the same result. NB: Care must be taken when performing row operation as to which row we multiply. Performing row operation to find the determinant is quite helpful when 125
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we deal with higher order square matrices, e.g. 4 × 4, 5 × 5, etc. E.g. Consider a 4 × 4 matrix D = 1 3 5 7 3 1 3 5 5 3 1 3 7 5 3 1 1 3 5 7 3 1 3 5 5 3 1 3 7 5 3 1 = 1 3 5 7 0 - 8 - 12 - 16 0 - 12 - 24 - 32 0 - 16 - 32 - 48 R 2 - 3 R 1 R 3 - 5 R 1 R 4 - 7 R 1 = 1 × - 8 - 12 - 16 - 12 - 24 - 32 - 16 - 32 - 48 = ( - 4)( - 4)( - 16) 2 3 4 3 6 8 1 2 3 = - 256 0 - 1 - 2 0 0 - 1 1 2 3 R 1 - 2 R 3 R 2 - 3 R 3 = - 256 × 1 × - 1 - 2 0 - 1 = - 256 × (1 - 0) = - 256 . 126
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a b + c a 2 b c + a b 2 c a + b c 2 = a a + b + c a 2 b a + b + c b 2 c a + b + c c 2 C 2 + C 1 = ( a + b + c ) a 1 a 2 b 1 b 2 c 1 c 2 = ( a + b + c ) a 1 a 2 b - a 0 b 2 - a 2 c - a 0 c 2 - a 2 R 2 - R 1 R 3 - 1 R 1 Expand along C 2 = - ( a + b + c ) b - a ( b - a )( b + a ) c - a ( c - a )( c + a ) = - ( a + b + c )( b - a )( c - a ) 1 b + a 1 c + a = - ( a + b + c )( b - a )( c - a )( c + a - b - a ) = - ( a + b + c )( b - a )( c - a )( c - b ) 6. det( AB ) = det A det B 7. det( A + B ) 6 = det A + det B.
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