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I'm about to use this diagram to make computing 3x3 determinants clear and easy for my students. I tried to add this picture to my lecture but it's blurred due to the bad quality of the picture. So i want to create exactly the same figure using latex.  I've searched and read a few topics about highlighting elements and diagonals in matrix but it does not work in this case. I want to add plus signs and minus signs as in the picture, and highlight elements in the matrice with elements outside the matrice.  I appreciate all your help and your advice. Thank you in advance ! Is there any way to create multiple insert statements in a ms-access query? Resize (enlarge) an image without resampling (in favor of quality loss)? How to resize an image or selection without resampling the pixels? I have some pixel art that I want to enlarge without losing the hard, non-anti-aliased, pixel edges. In other words, 1px should become 4px of the same color— fake big pixels. Is there any way to achieve this procedurally?  Example (of what I don't want):  before resize:    after resize:    Open to other tools/methods beyond photoshop, thanks in advance. Change style properties of several layers simultaneously? How can I install a .bin file? Traveling to the US with new passport, visa on the old one I have a valid B1/B2 US visa expiring about a year from now. The passport it's in expired a few years ago, and I recently got a new one.   The passport number does not match, but the national ID number, mentioned in both, does (it used to be the same as your national ID, but not anymore)  Can I travel to the US carrying both passports? Or is it asking for trouble? (I read a few years ago that it was possible, but that might have changed)  Should I just get a new visa to avoid potential complications? How to analyze lightly varying senses of adjective *very* Use of very as an adjective is (in my experience) most frequently attested in phrases like     ...the very person I was looking for.   To use adjective very with the indefinite article sounds quite bizarre in this case:     ?...a very person I was looking for.   But consider this sentence, where indefinite a is used:     Visualizing a very transformation of the master-slave relation, Thomas presses further: the slave is brother to the master, even doubly a brother.   OED treats the two uses as separate senses of the word, with the following definitions:           (first use):&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;With limitation (usually expressed by the or a possessive) to particular instances      (second use): Really or truly entitled to the name or designation   Here are two respective examples quoted in OED:     What would you say to me now, and I were your verie, verie Rosalind? (1616)   Thence we went into Queen Mary's room, and saw that beautiful portrait—that very queen and very woman. (1857)   Clearly, the difference in meaning between the two senses is quite light, and it seems to me like a kind of difference which is recurrent enough to have its own term. What is the proper term? Are there other English words having similarly related pairs of senses? And lastly, is OED jumping the gun in treating these as two senses? Could the difference in meaning instead be due to the choice of determiner? Generating a DEM and DSM from correct LiDAR point classification I have downloaded classified LAS data from    I am trying to generate accurate DEM and DSM for further analysis. I created a LAS dataset with a projected CRS of  NAD_1983_UTM_Zone_19N in meters and the z CRS in meters as well. My question is on which classifications do I choose for the DEM and DSM. Here is the filter options I have in my dataset:     According to unh LiDAR data report class 2 is ground.     What I have done so far:  DEM I chose the class 2 points and used the las dataset to raster.   DSM In the predefined settings I chose first return and used the las dataset to raster tool.   Is this an accurate way of generating these two rasters? Do I not have to take into account the the unassigned class 1, noise class 7, reserved 11, reserved 17, reserved 18?  Additionally (I can ask this as a separate question if it gets requested to), when using the las dataset to raster tool the sampling value is defaulted to 10. I would like to change it to 2 or 3 (would be meters) to match the LiDAR resolution. I also plan on using the generated DEM and DSM for a least cost path analysis and I want them to be higher resolution. Will making the sampling_value 2 or 3 throw off the results or should I leave it at 10? I'm trying to solve this problem from Hatcher 3.3.24.  Let $M$ be a closed connected 3-manifold, and write $H_1(M, \mathbb{Z})$ as $\mathbb{Z^r} \oplus T$ where $T$ is torsion. Show that $H_2(M, \mathbb{Z})$ is $\mathbb{Z^r}$ if $M$ is orientable or $\mathbb{Z^{r-1}} \oplus \mathbb{Z_2}$ if $M$ is not orientable. In particular $r \geq 1$ if $M$ is not orientable. Using exercise 6 construct examples showing there are no other restrictions on the homology groups of closed 3-manifolds. (In the nonorientable case consider the manifold $N$ obtained from $S^2 \times I$ by identifying $S^2 \times \{0\}$ with $S^2 \times \{1\}$ via a reflection of $S^2$.)  I think I've solved the first part, here's my proof:   First note the following theorem: If $M$ is a closed connected and orientable n-manifold then $H_n(M, \mathbb{Z}) = \mathbb{Z}$ and $Tors(H_{n-1}(M, \mathbb{Z})) = 0$. If nonorientable, then $H_n(M, \mathbb{Z}) = 0$, $Tors(H_{n-1}(M, \mathbb{Z})) = \mathbb{Z_2}$.  Now note by Poincare duality and UCT we have that $H_2(M) \cong H^1(M) \cong \mathbb{Z^r} \oplus T_0 \cong \mathbb{Z^r}$.   In the nonorientable case consider the exact sequence:  $0 \longrightarrow H_2(M) \longrightarrow H_1(M) \longrightarrow H_0(M) \longrightarrow 0$  But we know $H_1(M) = \mathbb{Z^r} \oplus Z_2$ from the earlier theorem and $H_0(M) = \mathbb{Z}$.   Therefore $H_2(M) \cong ker(\mathbb{Z^r} \oplus Z_2 \rightarrow \mathbb{Z}) \cong \mathbb{Z^{r-1}} \oplus \mathbb{Z_2}$.   For the rest of the question I'm quite lost. A Tweedie distribution can model skewed data with a point mass at zero when the parameter $p$ (exponent in the mean-variance relationship) is between 1 and 2.   Similarly a zero-inflated (whether otherwise continuous or discrete) model may have a large number of zeros.  I'm having trouble understanding why it is that when I do prediction or calculate fitted values with these kinds of models, all of the predicted values are non-zero.  Can these models actually predict exact zeros?  For example  library(tweedie) library(statmod) # generate data y &lt;- rtweedie( 100, xi=1.3, mu=1, phi=1)  # xi=p x &lt;- y+rnorm( length(y), 0, 0.2) # estimate p out &lt;- tweedie.profile( y~1, p.vec=seq(1.1, 1.9, length=9)) # fit glm fit &lt;- glm( y ~ x, family=tweedie(var.power=out$p.max, link.power=0)) # predict pred &lt;- predict.glm(fit, newdata=data.frame(x=x), type="response")   pred now does not contain any zeros.  I thought the usefulness of models such as the Tweedie distribution comes from its ability to predict exact zeros and the continuous part.  I know that in my example the variable x is not very predictive. Interpreting coefficient in a linear regression model with categorical variables I will give my examples with R calls. First a simple example of a linear regression with a dependent variable 'lifespan', and two continuous explanatory variables.  data.frame(height=runif(4000,160,200))-&gt;human.life human.life$weight=runif(4000,50,120) human.life$lifespan=sample(45:90,4000,replace=TRUE) summary(lm(lifespan~1+height+weight,data=human.life))  Call: lm(formula = lifespan ~ 1 + height + weight, data = human.life)  Residuals: Min       1Q   Median       3Q      Max  -23.0257 -11.9124  -0.0565  11.3755  23.8591   Coefficients:              Estimate Std. Error t value Pr(&gt;|t|)     (Intercept) 63.635709   3.486426  18.252   &lt;2e-16 *** height       0.007485   0.018665   0.401   0.6884     weight       0.024544   0.010428   2.354   0.0186 *   --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1   Residual standard error: 13.41 on 3997 degrees of freedom Multiple R-squared: 0.001425,   Adjusted R-squared: 0.0009257  F-statistic: 2.853 on 2 and 3997 DF,  p-value: 0.05781   In order to find the estimate of 'lifespan' when the value of 'weight' is 1, I add (Intercept)+height=63.64319  Now what if I have a similar data frame, but one where one of the explanatory variables is categorical?  data.frame(animal=rep(c("dog","fox","pig","wolf"),1000))-&gt;animal.life animal.life$weight=runif(4000,8,50) animal.life$lifespan=sample(1:10,replace=TRUE) summary(lm(lifespan~1+animal+weight,data=animal.life))  Call: lm(formula = lifespan ~ 1 + animal + weight, data = animal.life)  Residuals: Min      1Q  Median      3Q     Max  -4.7677 -2.7796 -0.1025  3.1972  4.3691   Coefficients:             Estimate Std. Error t value Pr(&gt;|t|)     (Intercept) 5.565556   0.145851  38.159  &lt; 2e-16 *** animalfox   0.806634   0.131198   6.148  8.6e-10 *** animalpig   0.010635   0.131259   0.081   0.9354     animalwolf  0.806650   0.131198   6.148  8.6e-10 *** weight      0.007946   0.003815   2.083   0.0373 *   --- Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1   Residual standard error: 2.933 on 3995 degrees of freedom Multiple R-squared: 0.01933,    Adjusted R-squared: 0.01835  F-statistic: 19.69 on 4 and 3995 DF,  p-value: 4.625e-16   In this case, to find the estimate of 'lifespan' when the value of 'weight' is 1, should I add each of the coefficients for 'animal' to the intercept: (Intercept)+animalfox+animalpig+animalwolf? Or what is the proper way to do this?  Thanks Sverre Securely erasing all data from a hard drive I've seen the identity $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ used here recently.  I checked for proofs here   I couldn't have figured out the combinatorial proof by myself, and the inductive proof assumes you already know the answer...  So my question is : do you know how to prove directly through computation that $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ ? I understand that you have to have visas to enter some countries, but I was wondering whether you have to pay for each visa separately or you can get a visa to go everywhere. You see videos and lists on what you should pack to travel the world, but no lists on the documentation needed. How to disable SMS notification during phone call? I call somebody, or somebody calls me, we are talking. Now, during the call I got SMS and I cannot hear anything because the SMS notification is playing.  So how to disable this notification when call is active? Thank you in advance.  Samsung Galaxy Ace 2, Android 2.3.  Note: I am interested in setting this once and for good. Not something I have to remember to switch off each time I make/get a call. Character vs Charm - Pronunciation Is there a rule to understand how the group "Cha" has to be pronounced?  "Character" sounds with a hard first syllable, while "Charm" sound softer, but I don't find how to tell which sound to use before earing someone saying the word. It could be because of the double consonant "rm" vs "ra"? Or is it just a matter of knowing the rule for every single word?  EDIT: More specifically, let's talk about UK English pronunciation. Just like if we have any quadratic equation which has complex roots, then we are not able to factorize it easily. So we apply quadratic formula and get the roots.  Similarly if we have a cubic equation which has two complex roots (which we know conjugate of each other) and one fractional root, then we are not able to find its first root by hit &amp; trial.  So my question is like quadratic formula, is there exist any thing like cubic formula which help in solving cubic equations?  For example, I have an equation $$2x^3+9x^2+9x-7=0\tag{1}$$ and I have to find its solution which I am not able to find because it has no integral solution. Its solutions are $\dfrac {1}{2}$, $\dfrac{-5\pm \sqrt{3}i}{2} $, I know these solutions because this equation is generated by myself.  So how can I solve equations like these?  Also while typing this question, I thought about the derivation of quadratic formula, which is derived by completing the square method.  So I tried to apply ‘completing the cube’ method on the general equation $ax^3+bx^2+cx+d=0$ but it didn't help.  So please help me in finding a cubic formula or to solve the equations like given in example by an alternative method. I'm trying to show that a finite sum of eigenspaces (with distinct eigenvalues) is a direct sum.   I have $ \alpha : V \to V $. The eigenspaces are $ V_{\lambda_i} = \ker(\alpha - \lambda_i id_V )$ for $ 1 \leq i \leq n $. My attempt at a proof:  $ A + B $ is a direct sum iff $ A \cap B = \{0\} $. If $ v \neq 0 \in V_{\lambda_i} \cap V_{\lambda_j} $ for some $i,j, i \neq j $, then $ \alpha(v) = \lambda_i v $ and $ \alpha(v) = \lambda_j v $. So $(\lambda_i - \lambda_j)v = 0 $, and so $ \lambda_i = \lambda_j $. This is a contradiction, so any pair of the eigenspaces have trivial intersection. Therefore $ \cap_{i=1}^n V_{\lambda_i} = \{0\} $, and so we have a direct sum.  Is this ok?  Thanks Can you please explain this to me?          Could you please explain this to me?   I am unable to figure out which to use which situation. I did google, and some posts say they are both the same, even if the second one is more formal. Other posts say could is the past tense of can, but in the above example, I don't think could is used as past tense of can.  What is correct way to use these words? Where does $xe^x$ solution come from when the characteristic polynomial is square?