Find the determinant?

Question

Find the determinant of the nxn matrix with 2's on the diagonal, 1's above the diagonal, and 0's below the diagonal.

Answer 1

Hi,

|.2..1.|
|.0..2.|

This gives (2)(2) - (1)(0) = 4 - 0 = 4

I hope this helps!! :-)

Answer 2

properly, one thank you to teach that's to teach the column vectors are actually not linearly autonomous. If we strengthen them out: | a² a²+2a+a million a²+4a+4 a²+6a+9 | | b² b²+2b+a million b²+4b+4 b²+6b+9 | | c² c²+2c+a million b²+4c+4 b²+6c+9 | | d² d²+2nd+a million d²+4d+4 d²+6d+9 | If we are able to make some non-0 linear mix of {a², a²+2a+a million, a²+4a+4, a²+6a+9} to be identically 0 for any integer a, then we've stumbled on a linear mixture that proves the column vectors are actually not linearly autonomous, and subsequently, the determinant could equivalent 0 for all a, b, c, and d. assume there exists p, q, r, s, such that: pa² + q(a²+2a+a million) + r(a²+4a+4) + s(a²+6a+9) = 0 for all a. Then: (p + q + r + s)a² + (2q + 4r + 6s)a + (q + 4r + 9s) = 0 If we are able to discover values of p, q, r, and s such that the above is authentic, then we've our linear mixture. p + q + r + s = 0 2q + 4r + 6s = 0 q + 4r + 9s = 0 This has non-0 thoughts, as a results of fact that's an equation of much less equations than unknowns. The rank of the coefficient matrix is at maximum 3, and so the nullity must be a minimum of a million. subsequently, there exist non-0 p, q, r, s such that: pa² + q(a²+2a+a million) + r(a²+4a+4) + s(a²+6a+9) = 0 for all a, and subsequently, p, q, r, s grant coefficients for a linear mix of the column vectors of the matrix which grant us the 0 vector. subsequently the matrix has 0 determinant.

Answer 3

2^n. The 1's above the diagonal are multiplied by the zero's.

Answer 4

The matrix is in lower triangular form and the
determinant of such a matrix is the product
of its diagonal elements. So the answer is 2^n.

Answer 5

Easy.....2^n.

Answer 6

It's simple. 2^n.