So the question would be:For which values of t is the matrix A invertible?

[junsugal]

Homework Statement

Consider the matrix A = 1 1 2 1
1 2 3 4
2 4 7 2t-6
2 2 6-t t
Find the values of t for which the matrix A is invertible

Homework Equations

The Attempt at a Solution

For A to be invertible, the determinant should not be 0.

So what I did is find the determinant of the 4x4 matrix.
Since this 4x4 matrix doesn't have any zero.
I find 4 3x3 determinants which leads to 3 2x2 determinants. And a total of 12 2x2 determinants.
After I sum all of them, my equation lead to 4t^2-7t+50 is not equals to zero.
And I can't solve for t.
This is weird because when I use online calculator for matrix, for t is 0,1,3, or any random number I put in, there is an inverse matrix.

 

[vela]

The determinant of that matrix is 2t²-17t+26. I would suggest row-reducing the matrix a bit to make finding the determinant less tedious.

 

[Deveno]

you needn't even row-reduce that much. clearing out the first column reduces the problem to finding a 3x3 determinant, and one can use the rule of sarrus:

\begin{vmatrix}a&b&c\\d&e&f\\g&h&k\end{vmatrix} = aek + bfg + cdh - ceg - afh - bdk