CALCULUS – II
Determinants
by
Dr. Eman Saad
&
Dr. Shorouk Ossama
References
 Robert Wrede and Murrary R. Spiegel,
Theory and Problems of Advanced Calculas,
2nd Edition, 2002.
• The determinant of (2x2) matrix:
Where:
Then the determinant of A is:
-
•The determinant of (3x3) matrix:
+
+
Where:
Then the determinant of A is:
Is known as the Minor of a11, (M11). the minor
of an element is obtained by deleting the row
and column through the element and writing
down the determinant of the elements of the
remaining 2 x 2 sub matrix with a + or – sign
attached.
The minors of a12 and a13 are:
Where the signs attached should be noted.
The signs associated with the factors alternate,
starting with a + at the top left as we move
across or down from the top left-hand corner
as shown:
The sign associated with Cij is + if i+j is even
and - if i+j is odd, can be expressed as (-1)i+j
So the number (-1)i+j is denoted by Cofactor
a13, Row 1 & column 3, so (-1)i+j = (-1)4 = (even) +
a21, Row 2 & colunm 1, so (-1)i+j = (-1)3 = (odd) -
Example: Finding Minors and Cofactors for
C11 = (-1)1+1 M11 = M11 = 16
so
C32 = (-1)2+3 M32 = - M32 = -26
Example: Find the determinant of the matrix
+
+
= 3 (-4) – [(-2) (-2) - 5 (3)] = -1
Example: Find the determinant of the matrix
2 0
+
0
-
1
+
= -2 ( 1 + 2) = -6
Properties of Determinant
1. det AT = det A,
where AT is the transpose of A. The
determinant of a square matrix and its
transpose are equal since.
Example: Evaluate
= -3
Since the determinant has two zero in the first
column, it is advantageous to use Rule 1. the
determinant of the transpose of A is given by:
2. If every element of any single row or
column of the matrix A is multiplied by a
scalar K, then the determinant of this matrix
is k det A.
= a11 ka22a33 - a11 ka23a32 - a12 ka21a33 + a12 ka23a31 + a13 ka21a32 a13 ka22a31 = k det A
Example: Evaluate
Δ = 11 x
= 11 x ( -13 - 72 + 1) = -924
3. If B obtained from A by interchanging two
rows (or columns) then det B = - det A
For example, rows 1 and 3 are interchanged, so that:
= a31 a22a13 – a31 a23a12 – a32 a21a13 + a32 a23a11 + a33 a21a12 – a33
a22a11 = - det A
The same is true whichever or columns pairs are exchanged.
Example: Evaluate
-1
Interchange rows 1 and 2 using rule 3:
2
0 -1
+
- + 4
-1 2 0 -1
Expanding by row 1, remembering the sign role for cofactors:
2
-1 4
4 =2
-1 0
4. Expansion by any row or column
By grouping the terms differently, we can write, for example:
det A = a31 (a12 a23 – a13 a22) - a32 (a11 a23 – a13 a21) + a33 (a11 a22
– a12 a21)
= a31 C31 + a32 C32 + a33 C33
In terms of cofactors. Here the elements a31, a32 ,a33 constitute
the third row, and we call this the expansion of det A by the
third row.
It can be shown that the expansion can be written down
similarly using any row or column. Thus:
det A = a12 C12 + a22 C22 + a32 C32
is an expansion by the second column.
Example: Evaluate det A, where
2
1
0 -1
Since column 3 contains three zeros, expand by this
column. The cofactor of the element in rows 3,
column 3, is the 3 x 3 determinant obtained from A
by deleting the third row and third column in A. it is
associated with a + sign.
= 2 (4 + 10) = 28
• The Adjoint Matrix:
We can now rewrite the formula for the inverse:
Example: Find the inverse of:
Evaluate first det A = 1 x [-2-1] -2 x [0+1] – 1 x [0-1] = -4
The cofactors are:
C11= -3
C12= -1 C13= -1
C21= 5
C22= -1 C23= 3
C31= -1
C32= 1 C33= 1
Exercise:
If
Find: det A det B det(AB)
adjA det A-1
det AT
Thanks