## Algebra and Trigonometry 10th Edition

$15$
The general form of a matrix of order $2 \times 2$ is: $det \ A=\begin{bmatrix} p & q \\ r & s\end{bmatrix}=ps-qr$ Since Row-3 has the most zeroes, we choose $R_3$. Now, $det \ A =1 \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}-(-1) \begin{bmatrix} 4 & -1 \\ 2 & 2 \end{bmatrix}+0=(2+3)+(8+2)=15$