# Thread: Invertible matrix

1. ## Invertible matrix

I have to determine all values of h for which A is invertible and I really don't know what should be my first step( If anyone could guide me through this that would be awesome. Here's the matrix:

1 1 0 1
1 0 0 1
0 1 2
h + 1
0 1 1
h

2. 1 1 0 1
1 0 0 1
0 1 2 h + 1
0 1 1 h

Sorry if this is confusing the fourth column is 1,1,h+1, h
the first column is 1,1,0,0
second- 1,0,1,1
third - 0,0,2,1

3. You mean then that the matrix is
$\begin{bmatrix}1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}$

I see two ways to do that. One is to use the fact that a matrix is invertible if and only if its determinant is non-zero. The other is to row-reduce this to triangular form and use the fact that a matrix is invertible if and only if, reduced to triangular form, it has no zeros on its main diagonal. Since a simple way of determining the determinant of a matrix is to reduce to triangular form, those are essentially the same.

4. thank you ! now i have to figure out how to reduce it.

5. Originally Posted by anna123456
thank you ! now i have to figure out how to reduce it.
Quiet easy,
$R_2->(-1)R_1+R_2$
$R_3->(-1)R_1+R_3$
then
$R_3->(-1)R_2+R_3$
$R_4->(-1)R_2+R_4$ , $R_2,_3,_4...$ are row operations
And so on...until you get to reduced row echelon form.
Hope it helps

6. Thank you ! Ok here's what I got:
$
\begin{bmatrix}2/h & -3/h+1 & 1/h & -2/h \\ 1 & -1 & 0 & 0 \\ 2/h & -3/h & 1/h & -2/h \\ 1 & -2 & 1 & -1\end{bmatrix}
$

That's the inverse(A^-1)
Now I'm not really sure what to do with this. How do I figure out what values of h make this matrix invertible?

7. Originally Posted by anna123456
Thank you ! Ok here's what I got:
$
\begin{bmatrix}2/h & -3/h+1 & 1/h & -2/h \\ 1 & -1 & 0 & 0 \\ 2/h & -3/h & 1/h & -2/h \\ 1 & -2 & 1 & -1\end{bmatrix}
$

That's the inverse(A^-1)
Now I'm not really sure what to do with this. How do I figure out what values of h make this matrix invertible?
I think all you had to do is get the
$\begin{bmatrix}1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}$
Matrix to rref(reduced row echelon form) and find it's det.
detA(in rref)= $a_1,_1*a_2,_2*a_3,_3*a_4,_4$
and find for which h detA doesn't equal to 0.

8. We haven't learnt this formula yet so I can't use it(

9. Originally Posted by HallsofIvy
You mean then that the matrix is
$\begin{bmatrix}1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}$

I see two ways to do that. One is to use the fact that a matrix is invertible if and only if its determinant is non-zero. The other is to row-reduce this to triangular form and use the fact that a matrix is invertible if and only if, reduced to triangular form, it has no zeros on its main diagonal. Since a simple way of determining the determinant of a matrix is to reduce to triangular form, those are essentially the same.
Is there any way to find the determinant without using any formulas? when I reduce my matrix to echelon form it looks like
$
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & h \\ 0 & 0 & 1 & 0\end{bmatrix}
$

so is my answer all values of h except for 0 make this matrix invertible?

10. I'm really stuck ( Because it seems to easy to just say that h=/= 0.

11. Ok I think this looks a bit less complicated but the only answer i can come up with is h=/= 0...
$
\begin{bmatrix}1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}
$

---> Switch /R1 and R2

$
\begin{bmatrix}1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}
$

--->R2-R3
$
\begin{bmatrix}1 & 0 & 0 & 1 \\ 1 & 1 & -2 & -h \\ 1 & 0 & 2 & h+ 1 \\ 0 & 1 & 1 & h\end{bmatrix}
$

---> R3-R1
$
\begin{bmatrix}1 & 0 & 0 & 1 \\ 1 & 1 & -2 & -h \\ 0 & 0 & 2 & h \\ 0 & 1 & 1 & h\end{bmatrix}
$

----> R2+R3
$
\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & h \\ 0 & 1 & 1 & h\end{bmatrix}
$

--->R4-R2
$
\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & h \\ 0 & 0 & 1 & h\end{bmatrix}
$

--->R3- R4
$
\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & h\end{bmatrix}
$

---> R4-R3
$
\begin{bmatrix}1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & h\end{bmatrix}
$

--->R1- R4/h
$
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & h\end{bmatrix}
$

12. Originally Posted by anna123456
Is there any way to find the determinant without using any formulas? when I reduce my matrix to echelon form it looks like
$
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & h \\ 0 & 0 & 1 & 0\end{bmatrix}
$

so is my answer all values of h except for 0 make this matrix invertible?
One more "row operation" would be to swap the third and fourth rows.

That will give you $
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & h\end{bmatrix}
$

Now you also need to note that
1) If you "add a multiple of one row to another" the determinant of a matrix
remains unchanged.

2) if you "multiply one row by a number", the determinant of a matrix is multiplied by that number.

3) if you "swap two rows", the determinant of a matrix is multiplied by -1.

Since you have not "multiplied one row by a number", the determinant of your original matrix must be $\pm$ the determinant of this matrix: that is, $\pm h$.

The determinant of your original matrix is non-zero if and only if h is non-zero.