# Thread: Negative Determinent of a matrix

1. ## Negative Determinent of a matrix

First post, so bear with me lol.

Solve the following system of equations using matrix inversion method:

10x + 14y = 818
8x + 11y = 646

Of course, first order of business is to calculate the determinent, which works out to -2 (10 x 11 - 14 x 8 = -2...unless i calculated it wrong.)

My question is where do I go from here that I have a negative determinent, assuming i calculated it right?

Cheers

Ibrox

2. Hello,
Originally Posted by ibrox
First post, so bear with me lol.

Solve the following system of equations using matrix inversion method:

10x + 14y = 818
8x + 11y = 646

Of course, first order of business is to calculate the determinent, which works out to -2 (10 x 11 - 14 x 8 = -2...unless i calculated it wrong.)

My question is where do I go from here that I have a negative determinent, assuming i calculated it right?

Cheers

Ibrox
Welcome to the forums

You have $\begin{pmatrix} 10 & 14 \\ 8 & 11 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 818 \\ 646 \end{pmatrix}$
You calculated the determinant of the matrix $A=\begin{pmatrix} 10 & 14 \\ 8 & 11 \end{pmatrix}$

Now compute the inverse of A (which exists since the determinant is not 0) : The inverse of a matrix $B=\begin{pmatrix} a&b \\ c&d \end{pmatrix}$ is $\frac{1}{\text{det}(B)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

Apply this formula here, to get $A^{-1}$

You had $A \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 818 \\ 646 \end{pmatrix}$
So multiply to the left by $A^{-1}$ :

$A^{-1}A \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}= A^{-1} \begin{pmatrix} 818 \\ 646 \end{pmatrix}$

3. Originally Posted by ibrox
First post, so bear with me lol.

Solve the following system of equations using matrix inversion method:

10x + 14y = 818
8x + 11y = 646

Of course, first order of business is to calculate the determinent, which works out to -2 (10 x 11 - 14 x 8 = -2...unless i calculated it wrong.)

My question is where do I go from here that I have a negative determinent, assuming i calculated it right?

Cheers

Ibrox
The determinant is used in finding the inverse of the coefficient matrix .....

6.3 - The Inverse of a Square Matrix

4. Okay, I get how to calculate a matrix, its just you proceed as normal...thanks guys.

5. Sorry, one more thing:

The part b to this question is "Suppose now that both right hand-side values are increased by an amount "delta (i.e. triangle..dont have symbol lol). What value of delta causes x to = 35 and what is the value of y at this point?

Here's what i did:

x = [23] + [-5.5 7] [delta]
y = [42] [ 4 -5] [ 0 ]

Which of course gives:

23 - 5.5delta = 35
42 + 4delta

Then:
-5.5 delta = 12
delta = -2.18

Am I correct? If not, where did I go wrong and where do I go from here?

Thanks again.

Ibrox

6. Hello, ibrox!

I interpreted the set-up differently . . .

Suppose now that both right hand-side values are increased by an amount $\Delta$.
What value of $\Delta$ makes $x = 35$
and what is the value of $y$ at this point?

I think the new equations are: . $\begin{array}{ccc}10x + 14y &=& 818 + \Delta \\ 8x +11y &=& 646 + \Delta\end{array}$

We have: . $\begin{pmatrix}10 & 14 \\ 8 & 11 \end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} \;=\; \begin{pmatrix}818+\Delta \\ 646+\Delta \end{pmatrix}$ .[1]

The inverse matrix is still: . $A^{-1} \:=\:\begin{pmatrix}\text{-}\frac{11}{2} & 7 \\ 4 & \text{-}5\end{pmatrix}$

Multiply both sides of [1] by $A^{-1}\!:\quad \begin{pmatrix}\text{-}\frac{11}{2} & 7 \\ 4 & \text{-}5 \end{pmatrix}\begin{pmatrix}10&14\\8&11\end{pmatri x}\begin{pmatrix}x\\y\end{pmatrix} \;=\; \begin{pmatrix}\text{-}\frac{11}{2} & 7 \\ 4 & \text{-}5 \end{pmatrix}\begin{pmatrix}818+\Delta \\ 646+\Delta\end{pmatrix}$

And we get:. $\begin{pmatrix}x\\y\end{pmatrix} \;=\;\begin{pmatrix}(\text{-}4499-\frac{11}{2}\Delta) + (4522+7\Delta) \\ (3272 + 4\Delta) + (\text{-}3230 - 5\Delta) \end{pmatrix} \;=\;\begin{pmatrix}23 + \frac{3}{2}\Delta \\ 42 - \Delta \end{pmatrix}$

If $x = 35$, we have: . $23+\tfrac{3}{2}\Delta \:=\:35 \quad\Rightarrow\quad \boxed{\Delta \,=\,8}$

Then: . $y \:=\:42 - 8 \quad\Rightarrow\quad\boxed{ y \,=\,34}$