I know that I can find the determinant by working my way from 4×4 to 3×3 to 2×2 determinant. However that is long.
Which other method can I use?
\begin{bmatrix}
2 & 1 & 0 & 6 \\
2 & 5 & 3 & 0 \\
1 & 6 & 0 &-2 \\
0 &-2 & 5 & -1
\end{bmatrix}
determinantlinear algebramatrices
I know that I can find the determinant by working my way from 4×4 to 3×3 to 2×2 determinant. However that is long.
Which other method can I use?
\begin{bmatrix}
2 & 1 & 0 & 6 \\
2 & 5 & 3 & 0 \\
1 & 6 & 0 &-2 \\
0 &-2 & 5 & -1
\end{bmatrix}
Best Answer
Substract twice the third row both from the second and the first rows to get:
$$\begin{pmatrix}0&-11&0&10\\0&-7&3&4\\1&6&0&-2\\0&-2&5&-1\end{pmatrix}$$
Develop by the first column and get:
$$\det\begin{pmatrix}-11&0&10\\-7&3&4\\-2&5&-1\end{pmatrix}=33+0-350+220-0+60=-37$$