Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 13/Solution 1

By the formula of determinant, we choose the first row to start,

${\begin{vmatrix}1&0&0&1\\10&0&3&10\\7&1&5&9\\-1&0&0&1\\\end{vmatrix}}=1{\begin{vmatrix}0&3&10\\1&5&9\\0&0&1\\\end{vmatrix}}-1{\begin{vmatrix}10&0&3\\7&1&5\\-1&0&0\\\end{vmatrix}}$

Then, the first term on the right hand side of the equation can be further computed by applying the formula of determinant on the first column:

$1{\begin{vmatrix}0&3&10\\1&5&9\\0&0&1\\\end{vmatrix}}=-1{\begin{vmatrix}3&10\\0&1\\\end{vmatrix}}=-3.$

For the second term, we apply the formula on the third row to obtain

$-1{\begin{vmatrix}10&0&3\\7&1&5\\-1&0&0\\\end{vmatrix}}=(-1)(-1){\begin{vmatrix}0&3\\1&5\\\end{vmatrix}}=-3$

By adding them up, we get the determinant of the given matrix as $\color {blue}-6.$