00:01
We're going to find the characteristic polynomial of the matrix a equal negative 2, negative 1, 0, 0, 4, 3 and negative 2, 5, 0 and we know that characteristic polynomial, let's call it pa of lambda where lambda will be its independent variable is defined as the determinant of a minus lambda times the identity matrix which in this case must be the 3 by 3 identity matrix.
00:36
And that will be the determinant of the matrix negative 2 minus lambda minus 1, 0 0, 4 minus lambda, 3 and negative 2, 5 minus lambda.
00:56
And we can develop that determinant for example by using the first row and so the first factor will be negative 2 minus lambda which is in position 1, 1 so the indices of the position of this factor is equal to 2 and that's an even number so the expression, the factor negative 2 minus lambda preserves its sign.
01:20
That is we don't change the sign of that factor and that's multiplied by the determinant of a sub matrix in this case 2 by 2 which is obtained by getting rid of the first row and first column which is the position of the factor negative 2 minus lambda.
01:41
The sub matrix we get is 4 minus lambda, 3, 5 and negative lambda...