Matrices trouble, finding k so matrix is rank 2

mr_coffee
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Hello everyone, I have a problem...
I am suppose to Find the value of k for which the matrix:

A =
-4 9 14
2 7 16
-7 -2 k

has rank 2.
k = ?

I row reduced until i couldn't do it anymore and i got the following:
-4 9 14
0 23 46
0 0 92k + 2116

now I'm lost on how I'm suppose to find a value for k that will make this mantrix rank 2. ANy help would be great
 
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Just put 92k + 2116 = 0 since the rank of the matrix is determined by the number of non-zero rows when in row echelon form.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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