Solving Determinants: Tips for Midterm Exam

[salman213]

Hi, I have a midterm tomorrow for my Lin Alg course and I was doing some review probs and I can't seem to understand this one..

http://img119.imageshack.us/img119/239/39786409wb4.jpg

Can someone help me and explain how to do this one!

I know I can just find the whole vector

x1,x2,x3,x4 by just multiplying by A^-1 but on a midterm I don't thinK I will want to waste time and find the inverse of such a matrix if there is another way to approach this problem.

 

[HallsofIvy]

IF you knew A^-1, then x₃ would just be third row of A^-1 time that vector- in other words, since the given vector has a 1 in the second place, zeros elsewhere, it would be the number in the third row, second column of A^-1. Can you think of a way of finding that number only? (Especially since you are given the determinant!)

 

[salman213]

thanks a lot

Cof (a) (2,3)
------------ = would be that specific element
det a

right?
!

 

[radou]

I'm not sure what you mean by "cof(2,3)", but, since you're given a Cramer system AX = B (A is a regular matrix), the solution is X = (x1, x2, x3, x4), where xi = Dj / D, where Dj is the determinant of the matrix which is created by substituting the j-th column of the matrix A with B, and D = det A.

 

[salman213]

Yea somone asked that question today in morning cause exam is in afternoon and he said to do it that way ur saying.. so i guess i get what u said thanks!

but i treid what i was saying and it gives same answer since basically Cofactor of A of the 2,3 term divided by determinant of A is exactly the 3,2 term in the A^-1 and therefore thas waht we want. It gives same answer so I guess you can do it etiehr way but I am sure he would want us to use cramers rule to get like ur saying

det A with 3rd column replaced with 0,1,0,0 over determinant of A

thanks!