Answer:
Yes, it is invertible
Step-by-step explanation:
We need to find in the matrix determinant is different from zero, since iif it is, that the matrix is invertible.
Let's use co-factor expansion to find the determinant of this 4x4 matrix, using the column that has more zeroes in it as the co-factor, so we reduce the number of determinant calculations for the obtained sub-matrices.We pick the first column for that since it has three zeros!
Then the determinant of this matrix becomes:
[tex]4\,*Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] +0+0+0[/tex]
And the determinant of these 3x3 matrix is very simple because most of the cross multiplications render zero:
[tex]Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] =1 \,(3\,*\,1-0)+4\,(0-0)+6\,(0-0)=3[/tex]
Therefore, the Det of the initial matrix is : 4 * 3 = 12
and then the matrix is invertible
