what would the value of determinant of a matrix be if a specific entry changed?

The determinant, is after all a multilinear map on the column / row vectors.

Therefore, $$ \det\begin{pmatrix} 1&3&4 \\ 5&2&(a+2) \\6&-2&3 \end{pmatrix} = \det\begin{pmatrix} 1&3&4 \\ 5&2&a \\6&-2&3 \end{pmatrix} +\det\begin{pmatrix} 1&3&0 \\ 5&2&2 \\6&-2&0 \end{pmatrix} $$

• The determinant of the second matrix is?

• If I changed $2$ to $3$ or $4$ or $b$, the answer would be?

Expand the $|A|$ along column 3.

\begin{align} |A|&= \begin{vmatrix} 1&3&4\\ 5&2&a\\ 6&-2&3 \end{vmatrix} \\ &= 4 \begin{vmatrix} 5&2\\ 6&-2 \end{vmatrix}-a \begin{vmatrix} 1&3\\ 6&-2 \end{vmatrix}+3 \begin{vmatrix} 1&3\\ 5&2 \end{vmatrix} \\ &=4(-22)-a(-20)+3(-13) \\ &=-127+20a \end{align}

The property of the matrix (in general, for any size $n\times n$): $$\begin{vmatrix} a+b&c+d \\ e&f \end{vmatrix} = \begin{vmatrix} a&c \\ e&f \end{vmatrix}+\begin{vmatrix} b&d \\ e&f \end{vmatrix};\\ \begin{vmatrix} a+b&c \\ d+e&f \end{vmatrix} = \begin{vmatrix} a&c \\ d&f \end{vmatrix}+\begin{vmatrix} b&c \\ e&f \end{vmatrix};$$ Hence: $$\begin{vmatrix} 1&3&4 \\ 5&2&(a+2) \\6&-2&3 \end{vmatrix} =\begin{vmatrix} 1&3&c+(4-c) \\ 5&2&a+2 \ \ \ \ \ \ \ \ \ \ \\6&-2&d+(3-d) \end{vmatrix}=\\ \begin{vmatrix} 1&3&c \\ 5&2&a \\6&-2&d \end{vmatrix}+\begin{vmatrix} 1&3&4-c \\ 5&2&2 \\6&-2&3-d \end{vmatrix} $$ For easy calculation of determinant you can create as many zeros as possible (especially in rows/columns).