Question

Evaluate the determinant for the following matrix:

Answer 1

The determinant is 0

Step-by-step explanation

For an n×n matrix A, the determinant of A, det(A), can be obtained by expanding along the kth row:

`\det(A) = a_(k1) C_(k1) + a_(k2) C_(k2) + \ldots + a_(kn) C_(kn)`

where
`a_(k1)` is the entry of A in the kth row, 1st column,
`a_(k2)` is the entry of A in the kth row, 2nd column, etc., and
`C_(kn)` is the kn cofactor of A, defined as
`C_(kn) = (-1)^(k+n) M_(kn)`.

`M_(kn)` is the kn minor, obtained by getting the determinant of the matrix which is the matrix A with row k and column n deleted.

Applying this here, we can expand along the 1st row. For convenience, let G be the matrix given by

`image`

where the first row has been bolded.

The determinant of G is therefore

`image`

Note that g₁₁ is the matrix element of G that is in the 1st row, 1st column,

g₁₂ is the matrix element of G that is in the 1st row, 2nd column, etc. Then we have

`image`

M₁₁ is the determinant of the matrix that is matrix G with row 1 and column 1 removed. The bold entires are the row and the column we delete.

The determinant of a 2×2 matrix is

`image`

so it follows that

Applying the same for M₁₂ and M₁₃, we have

and

so therefore

`image`

Answer 2

a11 1 * [ (2*5) - (5*2)] = 0
a12 = -4 *(5*5) - 2*1) = 23*-4 = -92
a13 = 4 * (5*5) - (1*2) = 23 * 4 = 92
a21 = 0
a22 = 2 * (5-4) = 2
a23 = -2 * (5-4) - -2
a31 = 0
a32 = -5*(2-20) = 90
a33 = 5 * (2-10) = -90

Adding these gives zero

Answer is B:- 0