Respuesta :
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
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
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:
[tex] \det(A) = a_{k1} C_{k1} + a_{k2} C_{k2} + \ldots + a_{kn} C_{kn} [/tex]
where [tex] a_{k1} [/tex] is the entry of A in the kth row, 1st column, [tex] a_{k2} [/tex] is the entry of A in the kth row, 2nd column, etc., and [tex] C_{kn} [/tex] is the kn cofactor of A, defined as [tex] C_{kn} = (-1)^{k+n} M_{kn} [/tex].
[tex] M_{kn} [/tex] 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
[tex] G=\begin{bmatrix}
\bf 1 & \bf 4 & \bf 4\\
5 & 2 & 2 \\
1 & 5 & 5
\end{bmatrix} [/tex]
where the first row has been bolded.
The determinant of G is therefore
[tex] \begin{aligned}
\text{det}(G) &= g_{11}C_{11} + g_{12}C_{12} + g_{13}C_{13}
\end{aligned} [/tex]
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
[tex] \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} [/tex]
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.
[tex] \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} [/tex]
The determinant of a 2×2 matrix is
[tex] \det\left(
\begin{bmatrix}
a & b \\
c& d
\end{bmatrix}
\right) = ad-bc [/tex]
so it follows that
[tex] \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} [/tex]
Applying the same for M₁₂ and M₁₃, we have
[tex] \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} [/tex]
and
[tex] \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} [/tex]
so therefore
[tex] \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} [/tex]
