# Thread: Linear Algebra question regards Matrix and (....)

1. ## Linear Algebra question regards Matrix and (....)

The first,

Evaluate det(A) and use your answer to give the values of t which is a real number for which A is non singular

[(3-t) 0 7 5]
[(t-3) t 1 1] = A, (4x4 matric)
[0 -t 4t 2]
[0 0 2t+4 t+5]

I'm not sure about how to handle this question. Is it possible to handle this question be reduce this matrix to row-echelon form? so that I can find the determinant afterward.

The second question, I'm not sure if what I did is right or not for the first part, and for the second part I have no idea how to do it.

Consider the following three points in R^3

P=(1,-1,2) Q=(1,2,-3) R=(-1,4,1)

a) calculate QP (hat), QR (hat) and QP(hat) x QR(hat)

QP(hat) =(1, -1, 2) - (1,2,-3) = (0,-3,5)
QR(hat) =(-1,4,1) - (1,2,-3)= (-2,2,4)
QP(hat) x QR(hat) = (0,-3,5) x (-2,2,4) = 0x-2 + (-3x2) + (5x4)=20-6= 14 (is it)?

b) Use the cross product to find the area of the triangle with vertices P,Q and R

What cross-product I supposed to use? and how do I use it...

Best Regards
Junks

2. ## Re: Linear Algebra question regards Matrix and (....)

Hey junkwisch.

Hint (for Q1): Do you know the inductive formula for a determinant? A determinant of a 2x2 matrix [a b; c d] is ad - bc and you can use the inductive form to get det(A) for all matrices of higher order (i.e. rows/columns).

Take a look at the inductive formula here (if you don't have appropriate notes):

Determinant -- from Wolfram MathWorld

3. ## Re: Linear Algebra question regards Matrix and (....)

It seems very peculiar that you would be given problems like this if you have no idea how to even start! Are you sure you have been paying attention in class?

There are several ways to find the determinant of a 4 by 4 matrix. One is to "expand" by columns (or rows). Expanding by the first column (because it already has two "0"s) gives
$\displaystyle (3- t)\left|\begin{array}{ccc}t & 1 & 1 \\ -t & 4t & 2 \\ 0 & 2t+ 4 & t+ 5\end{array}\right|$$\displaystyle - (t- 3)\left|\begin{array}{ccc} 0 & 7 & 5 \\ -t & 4t & 2 \\ 0 & 2t+ 4 & t+ 5\end{array}\right|$
Of course, you then expand the 3 by 3 matrices in the same way.

But, yes, the simplest way to find a determinant is to "row reduce" it to a triangular matrix. You can use the row operations "swap two rows" or "add a multiple of one row to another row" without change. The only thing to be careful with is that if you "multiply a row by a number" that mulltiplies the determinant by the same number so to get back to the original determinant, you have to divide by it.

As for the second problem, first, "QP(hat) x QR(hat) = (0,-3,5) x (-2,2,4) = 0x-2 + (-3x2) + (5x4)=20-6= 14" is wrong because the cross product of two vectors is a vector, not a number! Whoever gave you this problem expects you to know the theorem "The length of the cross product of two vectors is equal to the area of the parallelogram having the two vectors as sides." Of course, the triangle having those the vectors as sides is half the parallelogram formed by any two of the vectors.

So the area of the triangle is one half the length of the cross product of any two of the three vectors forming the triangle. As to "what cross product", did it not occur that they meant the one they had just told you to calculate?