Consider row reducing the standard 2x2 matrix. Suppose that a is nonzero.
1/a R1 -> R1
R2 - cR1 -> R2
Now notice that we cannot make the lower right corner a 1 if
d - cb/a = 0
ad - bc = 0
Definition of the Determinant
We call ad - bc the determinant of the 2 by 2
it tells us when it is possible to row reduce the matrix and find a solution
to the linear system.
The determinant of the matrix
3(2) - 1(5) = 6 - 5 = 1
Determinants of Three by Three Matrices
We define the determinant of a triangular matrix
det = abc
Notice that if we multiply a row by a constant k then the new determinant
is k times the old one. We list the effect of all three row operations
The effect of the the three basic row operations on
the determinant are as follows
- Multiplication of a row by a constant multiplies the determinant
by that constant.
- Switching two rows changes the sign of the determinant.
- Replacing one row by that row + a multiply of another row has no
effect on the determinant.
To find the determinant of a matrix we use the operations to make the matrix
triangular and then work backwards.
Find the determinant of
We use row operations until the matrix is triangular.
1/2 R1 <-> R1
(Multiplies the determinant by
R2 - 2R1 -> R2
on the determinant)
Note that we do not need to zero out the upper
middle number. We only need to zero out the bottom left numbers.
R3 + 2R2 -> R3
(No effect on the determinant)
Note that we do not need to make the middle number
The determinant of this matrix is 48. Since this matrix has 1/2 the determinant
of the original matrix, the determinant of the original matrix has
determinant = 48(2) = 96.
We call the square matrix I with all 1's down the diagonal and zeros
everywhere else the identity matrix. It has the unique property
that if A is a square matrix with the same dimensions then
AI = IA = A
If A is a square matrix then the inverse A-1
of A is the
unique matrix such that
The inverse of a matrix exists if and only if the
determinant is nonzero.
To find the inverse of a matrix, we write a new extended matrix with the
identity on the right. Then we completely row reduce, the resulting
matrix on the right will be the inverse matrix.
First note that the determinant of this matrix is
-2 + 1 = -1
inverse exists. Now we set the augmented matrix as
R1 <-> R2
R2 - 2R1 -> R2
R1 + R2 -> R1
Notice that the left hand part is now the identity. The right hand
side is the inverse. Hence
Solving Equations Using Matrices
Suppose we have the system
2x - y = 3
x - y = 4
Then we can write this in matrix form
Ax = b
We can multiply both sides by A-1:
A-1A x = A-1b
x = A-1b
Hence our solution is
x = -1
and y = 5
The Easy Way
A graphing calculator can be used to work all of the above problems.