Method 2 (an example of dinosaur mathematics - should be extinct)

The inverse of a 3×3 matrix is given by:

`A^-1=("adj"A)/(detA)`

"adj A" is short for "the adjoint of A". We use cofactors (that we met earlier) to determine the adjoint of a matrix.

Cofactors

Recall: The cofactor of an element in a matrix is the value obtained by evaluating the determinant formed by the elements not in that particular row or column.

Example 2a

Consider the matrix:

`((5,6,1),(0,3,-3),(4,-7,2))`

The cofactor of 6 is

`|(0,-3),(4,2)|=0+12=12`

The cofactor of -3 is

`|(5,6),(4,-7)|=-35-24=-59`

We find the adjoint matrix by replacing each element in the matrix with its cofactor and applying a + or - sign as follows:

`((+,-,+),(-,+,-),(+,-,+))`

and then finding the transpose of the resulting matrix. The transpose means the 1st column becomes the 1st row; 2nd column becomes 2nd row, etc.

Example 2b

Find the inverse of the following by using the adjoint matrix method:

`A=((5,6,1),(0,3,-3),(4,-7,2))`

Solution

Step 1:

Replace elements with cofactors and apply + and -

`((+(-15),-(12),+(-12)),(-(19),+(6),-(-59)),(+(-21),-(-15),+(15)))`

`=((-15,-12,-12),(-19,6,59),(-21,15,15))`

 

Step 2

Transpose the matrix:

`"adj"A = ((-15,-19,-21),(-12,6,15),(-12,59,15))`

Before we can find the inverse of matrix A, we need det A:

`|(5,6,1),(0,3,-3),(4,-7,2)|` `=5(-15)+4(-21)` `=-159`

Now we have what we need to apply the formula

`A^-1=("adj"A)/detA`

So

`A^-1=("adj"A)/detA`

`=1/-159((-15,-19,-21),(-12,6,15),(-12,59,15))``

`A^-1=((0.094,0.119,0.132),(0.075,-0.038,-0.094),(0.075,-0.371,-0.094))`

Example 2c

Find the inverse of

`((-2,6,1),(0,3,-3),(4,-7,3))`

using Method 2.

Solution

`text(C of) A` `=((+(-12),-(12),+(-12)),(-(25),+(-10),-(-10)),(+(-21),-(6),+(-6)))`

`=((-12,-12,-12),(-25,-10,10),(-21,-6,-6))`

Interchange rows and columns:

`"adj"A=((-12,-25,-21),(-12,-10,-6),(-12,10,-6))`

`"det"A`

`=|(-2,6,1),(0,3,-3),(4,-7,3)|`

`=2(9-21)+4(-21)`

`=-60`

So

`A^-1=("adj"A)/(detA)`

`=1/-60((-12,-25,-21),(-12,-10,-6),(-12,10,-6))`

`=( (1/5,5/12,7/20),(1/5,1/6,1/10),(1/5,-1/6,1/10))`

`=((0.2,0.417,0.35),(0.2,0.167,0.1),(0.2,-0.167,0.1))`


Now let's see how to do all this more appropriately using a computer...

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