# For what values of $x$ does $A^{-1}$ exist?

Let $A = \begin{bmatrix}4&-1&2\\5&x&7\\x&-1&3\end{bmatrix}$

I'm trying to find the values of $x$ such that $A^{-1}$ exists.

What I have tried:

$4(3x + 7) - (-1)(15)(7x) + (2)(-5)(x^2) = 0$

$-10x^2 + 117x + 28 = 0$

but I don't think that's right!

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You have the right idea, but you have typos. – Brian Fitzpatrick Feb 11 '14 at 4:10
What's written after "What I have tried" ? The matrix's determinant or what? If so, in the second and third summands of the first line there must be a minus sign and not multiplication between the second and third factor in each... – DonAntonio Feb 11 '14 at 4:11
@BrianFitzpatrick wait what typos? – Cheryl Feb 11 '14 at 4:12
the computed determinant is wrong – janmarqz Feb 11 '14 at 4:13
@Cheryl you should have $(15-7x)$ rather than $(15)(7x)$ and $(-5-x^2)$ rather than $(-5)(x^2)$ – Omnomnomnom Feb 11 '14 at 4:14

Note that \begin{align*} \det A&=4\cdot\left(3x+7\right)-(-1)\cdot\left(15-7x\right)+2\cdot\left(-5-x^2\right)\\ &=-2x^2+5x+33 \\ &=-2(x+3)\left(x-\frac{11}{2}\right) \end{align*} Hence $A$ is invertible if and only if $x\neq -3$ and $\displaystyle x\neq\frac{11}{2}$.

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i have shown him using software as well – dato datuashvili Feb 11 '14 at 4:46

syms x
>> A=[4 -1 2;5 x 7;x -1 3]

A =

[ 4, -1, 2]
[ 5,  x, 7]
[ x, -1, 3]

>> p=det(A)

p =

- 2*x^2 + 5*x + 33

r=solve(p)

r =

-3
11/2

that means that $x$ must not be $-3$ or $11/2$

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