Question

Let $A = \begin{pmatrix} 1 & 0 & 3\\ 0 & 4 & 5 \\ 1 & 2 & 6 \end{pmatrix}$

What is the contactor of the entry A₂₃ = 5

1. 2
2. -2
3. 10
4. -10

Answer 1

The co-factor of 5 is obtained by eliminating the row and column of 5. By doing this, we obtain the matrix below

$\begin{pmatrix} 1 & 0 \\ 1&2 \end{pmatrix}$

the determinant of the matrix above is $1(2) -0=2$

Next, we check the sign associated with matrix. The matrix sign used to write the co-factor of each elements is given below;

$\begin{pmatrix} + & - & +\\ -&+&-\\+&-&+ \end{pmatrix}$

Therefore the sign associated with 5 is -1. Hence the co-factor of 5 is $-1(2) = -2$