Question

Evaluate the determinant for the following matrix:

Answer 1

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

Answer 2

ANSWER

The determinant is 0

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

$G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ 5 & 2 & 2 \\ 1 & 5 & 5 \end{bmatrix}$

where the first row has been bolded.

The determinant of G is therefore

$\begin{aligned} \text{det}(G) &= g_{11}C_{11} + g_{12}C_{12} + g_{13}C_{13} \end{aligned}$

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

$\begin{aligned} \text{det}(G) &= g_{11}(-1)^{1+1}M_{11} + g_{12}(-1)^{1+2}M_{12} + g_{13}(-1)^{1+3}M_{13} \\ &= g_{11} M_{11} - g_{12}M_{12} + g_{13}M_{13} \end{aligned}$

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.

$\begin{aligned} G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ \bf 5 & 2 & 2 \\ \bf 1 & 5 & 5 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 2&2 \\ 5&5 \end{bmatrix} \right) \end{aligned}$

The determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

so it follows that

$\begin{aligned} G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ \bf 5 & 2 & 2 \\ \bf 1 & 5 & 5 \end{bmatrix} \implies M_{11} &= \det\left(\begin{bmatrix} 2&2 \\ 5&5 \end{bmatrix} \right) \\ &= (2)(5) - (2)(5) \\ &= 0 \end{aligned}$

Applying the same for M₁₂ and M₁₃, we have

$\begin{aligned} G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ 5 & \bf 2 & 2 \\ 1 & \bf 5 & 5 \end{bmatrix} \implies M_{12} &= \det\left(\begin{bmatrix} 5&2 \\ 1&5 \end{bmatrix} \right) \\ &= (5)(5) - (2)(1) \\ &= 23 \end{aligned}$

and

$\begin{aligned} G=\begin{bmatrix} \bf 1 & \bf 4 & \bf 4\\ 5 & 2 & \bf 2 \\ 1 & 5 & \bf 5 \end{bmatrix} \implies M_{13} &= \det\left(\begin{bmatrix} 5&2 \\ 1&5 \end{bmatrix} \right) \\ &= (5)(5) - (2)(1) \\ &= 23 \end{aligned}$

so therefore

$\begin{aligned} \text{det}(G) &= g_{11} M_{11} - g_{12}M_{12} + g_{13}M_{13} \\ &= (1)(0) - (4)(23) + (4)(23) \\ &= 0 -4(23) + 4(23) \\ &= 0 \end{aligned}$