I'm having a problem finding the determinant of the following matrix using elementary row operations. I know the determinant is -15 but confused on how to do it using the elementary row operations.
Here is the matrix
$$\begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix}$$
Thank you
Best Answer
Note that the determinant of a lower (or upper) triangular matrix is the product of its diagonal elements. Using this fact, we want to create a triangular matrix out of your matrix \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} So, I will start with the last row and subtract it from the second row to get \begin{bmatrix} 2 & 3 & 10 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix} Now, I want to get rid of the $2$ in the first row. I thus multiply the last row by $2$ and subtract it from the first row to obtain: \begin{bmatrix} 0 & 1 & 16 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix} Finally, I subtract the second row from the first one to obtain \begin{bmatrix} 0 & 0 & 15 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix} We now have $$\det \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} = \det \begin{bmatrix} 0 & 0 & 15 \\ 0 & 1 & 1 \\ 1 & 1 & -3 \end{bmatrix}$$
Now, I will transform the RHS matrix to an upper diagonal matrix. I can exchange the first and the last rows. Exchanging any two rows changes the sign of the determinant, and therefore $$\det \begin{bmatrix} 2 & 3 & 10 \\ 1 & 2 & -2 \\ 1 & 1 & -3 \end{bmatrix} = -\det \begin{bmatrix} 1 & 1 & -3 \\ 0 & 1 & 1 \\ 0 & 0 & 15 \end{bmatrix}$$
The matrix on the RHS is now an upper triangular matrix and its determinant is the product of its diagonal elements, which is $15$. With the minus sign, the $\det$ of our initial matrix is thus $-15$.