L 13 # 0 L 7 5 67 8 8 1 9 3 Gl- 7 [L...

$$ \begin{array}{|c|cccccc|}\hline X & {3} & {7} & {9} & {12} & {14} \\ \hline P(X) & {\frac{4}{13}} & {\frac{1}{13}} & {\frac{3}{13}} & {\frac{1}{13}} & {\frac{2}{13}} \\ \hline\end{array} $$

To find these two matrices together. Here. First thing we need to do is check the dimensions will work. Okay? First matrix we've got is two rows three columns. That's a two by three. The second matrix is a three row one column matrix. These inner two dimensions have to match which they do. That means is a resulting matrix will be dimension the outer dimensions. Okay, So we're gonna have to buy one. Matrix is our answer just two spots here. Okay, we'll fill in the first spot. This is gonna be row one. Column one. Okay, so everything in row one thinks everything in column one negative one times six plus zero times negative four plus seven. Thanks one. Okay, this should give us negative six plus zero plus seven. That should give us a one in that first little spot right there. Okay, The second guy this is gonna be row two column one. Okay, so we're gonna take everything in row two times that same column right there. So three times six less negative five that was negative for plus two times one spent 18 plus 20 plus two should give us 40. Okay? So one in the first row, 40 in the second round, and that is the result of this modification.

This question gives us a sequence and asked us to determine a formula. What we know that the numerator for each term is one, and the denominators essentially add the numbers after each other times, too. So, as we said, the numerator is one and the denominator times two minus one and this works. If you plug in 1 to 3, you'll got 1 1/3 1 fifth onwards.

So on this problem, we're given this matrix and were asked to use matrix capability of a graphing utility to find the determinant. So I went to Desmond's dot com and went to math tools matrix calculator and got this matrix calculator. So I have a four x four matrix. So I go new matrix And I got four rows and four columns And the first entry is a zero and then a minus three in an eight and that too. And then eight one a minus one and six And then I -4 in a six. Mhm and zero and a nine. And then uh minus seven, two, zeros zero, zero and 14. They had dinner now to find the determiner of this, I use the D E T button right here go determine it a matrix A And there it is 7000 441 7441.

This question asked us to write a formula for the given sequence. What we know is that we alternate between positive and negative values. Therefore, each term is multiplied by negative one to the power and minus one. And then we know the numerator. Zahra Old perfect squares therefore were multiplying remote point by n squared. And then lastly, the denominators are one more than end. Therefore, the denominator is simply n plus one or one plus.