# Matrix problem

• Feb 24th 2010, 06:09 PM
anna123456
Matrix problem
Sorry for the long post but I'm really stuck with this problem. I have to find all of the h values for which this matrix is invertible. I tried to find the inverse so that explains extra 4 columns. I've asked this question before and i was suggested to use a determinant but the problem is we haven't learned any formulas for the determinant yet except for the 2*2 matrix. I reduced it to echelon form but i don't know what to do next.
$
\begin{bmatrix}1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & h+ 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & h& 0 & 0 & 0 & 1\end{bmatrix}
$

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

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

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

--->R1 - R2
$
\begin{bmatrix}1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 1 & -2 & -h & 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & h& 0 & 0 & 0 & 1\end{bmatrix}
$

---> R3-R4
$
\begin{bmatrix}1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & -3 & -2h & 1 & 0 & -1 & -1 \\ 0 & 1 & 1 & h& 0 & 0 & 0 & 1\end{bmatrix}
$

--->R3/2 +R4
$
\begin{bmatrix}1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & -3/2 & -h & 1/2 & 0 & -1/2 & -1/2 \\ 0 & 1 & -1/2 & 0& 1/2 & 1/2 & -1/2 & 1/2\end{bmatrix}
$

--->2R3 +6R4
$
\begin{bmatrix}1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & -2h & 4 & -6 & 2 & -4 \\ 0 & 0 & 3 & 0& 3 & -6 & 3 & -3\end{bmatrix}
$

--->R4/3& (R3/2h+R1
$
\begin{bmatrix}1 & 0 & 0 & 0 & 2/h & -3/h+1 & 1/h & -2/h \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & - 2/h & 3/h & -1/h & -2/h \\ 0 & 0 & 1 & 0&1 & -2 & 1 & -1\end{bmatrix}
$
• Feb 24th 2010, 07:48 PM
tonio
Quote:

Originally Posted by anna123456
Sorry for the long post but I'm really stuck with this problem. I have to find all of the h values for which this matrix is invertible. I tried to find the inverse so that explains extra 4 columns. I've asked this question before and i was suggested to use a determinant but the problem is we haven't learned any formulas for the determinant yet except for the 2*2 matrix. I reduced it to echelon form but i don't know what to do next.
$
\begin{bmatrix}1 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & h+ 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & h& 0 & 0 & 0 & 1\end{bmatrix}
$

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

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

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

--->R1 - R2
$
\begin{bmatrix}1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 1 & -2 & -h & 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & h& 0 & 0 & 0 & 1\end{bmatrix}
$

---> R3-R4
$
\begin{bmatrix}1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & -3 & -2h & 1 & 0 & -1 & -1 \\ 0 & 1 & 1 & h& 0 & 0 & 0 & 1\end{bmatrix}
$

--->R3/2 +R4
$
\begin{bmatrix}1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & -3/2 & -h & 1/2 & 0 & -1/2 & -1/2 \\ 0 & 1 & -1/2 & 0& 1/2 & 1/2 & -1/2 & 1/2\end{bmatrix}
$

--->2R3 +6R4
$
\begin{bmatrix}1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & -2h & 4 & -6 & 2 & -4 \\ 0 & 0 & 3 & 0& 3 & -6 & 3 & -3\end{bmatrix}
$

--->R4/3& (R3/2h+R1
$
\begin{bmatrix}1 & 0 & 0 & 0 & 2/h & -3/h+1 & 1/h & -2/h \\ 0 & 1 & 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & - 2/h & 3/h & -1/h & -2/h \\ 0 & 0 & 1 & 0&1 & -2 & 1 & -1\end{bmatrix}
$

A pity you can't use determinant because it'd be pretty fast. Anway, your matrix is $A=\begin{pmatrix}1&1&0&1\\1&0&0&1\\1&0&2&h+1\\0&1& 1&h\end{pmatrix}$ .

Now reduce this matrix (without adding anything!) to echelon form and check what the condition on h must be so that you won't get a row of zeroes...

Tonio
• Feb 25th 2010, 09:16 AM
anna123456
I know that would be a lot easier but i did reduce it to echelon form and i got
$
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{bmatrix}
$

This reduced echelon form doesn't help me at all((
• Feb 25th 2010, 07:26 PM
tonio
Quote:

Originally Posted by anna123456
I know that would be a lot easier but i did reduce it to echelon form and i got
$
\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{bmatrix}
$

This reduced echelon form doesn't help me at all((

Apparently you don't know what "to reduce to echelon form means" , whether you use Gauss or Gauss-Jordan methods: you only have to make zeros every column below the upper one and "go down" from entry 1-1 to entry 2-2 and etc., until you reach the last row. Then you check under what conditions one row (many times the last one) becomes all zeroes...
Also, you could NOT have reached the above form for your matrix UNLESS you made some sharp assumption on h.

Tonio