00:01
Problem number 32, we say that let the characteristic binomial b, landa, is equal to the determinant of a minus landa times a unity.
01:08
And 4 .7 .6 we have b lambda is equal to lambda 1 minus lambda multiplying by lambda 2 minus lambda until we reach lambda n minus which is equal to lambda 2 squared minus lambda 1 plus lambda 2 times lambda plus lambda 1 plus lambda 1 lambda 2 multiplied by lambda 3 minus lambda until we reach lambda n minus lambda which is equal to negative lambda 3 plus lambda squared multiplying by landa 1 plus landa 2 plus landa 3 minus landa 1, landa 2 plus landa 1, landa 3 plus landa 2 until we reach landa nanda nanda minus lambda which is equal to negative 1 power n lambda plus negative 1 power n plus negative 1 power n minus 1 lambda n minus 1 plus londa 2 plus until we reach lambda n plus until we reach landa 1, land 2, land 3 till run the end.
04:30
By equating the right hand side of the equation and the right side of the equation to the right hand side of the equation 5 .7 .5, we will have negative 1 .5, 4n landa power n plus p1 landa for n plus p1 landa for n minus 1 plus p n is equal to negative 1 for landa landa plus negative 1 for n minus n this is the right -hand side lambda 1 plus lambda n for lambda n minus 1 plus lambda plus lambda 1 plus lambda 1 lambda 2 until we reach lambda land the same the same right hand side at this equation from equating the two equation we get by equating the coefficients in both sides with the same power of lambda of both sides of terms with the same power of lambda.
07:14
We get that b1 is equal to lambda 1 plus lambda 2 plus lambda n power negative 1 n minus 1 b2 is equal to lambda 1 lambda 2 plus lambda 1 lambda 1 round the 3 plus lambda n minus 1 1 1 1 negative 1 n minus 2 until we reach b n is equal to lambda 1, lambda, n, n, 1, multiplying by, london 2, number 2, multiplying by, under 3 until we reach, lambda n.
08:42
We can say that this these coefficients, we can say that this is the beta serum...