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Ben and Jerry Jerry start with the same number of trading cards. After Ben gives 12 of his cards to Jerry then has twice as many cards as Ben does. How many cards did Ben have at the start?
The easy way to answer this is to plug in numbers. Because you are given "12" and you are given "twice as many", the answer has to be a variable of 12 or 6. If you start plugging in numbers, you will find that 36 is the answer... Ben and Jerry start with 36 cards. Ben gives 12 of his cards to Jerry. Ben now has 24 and Jerry now has 48. 24 is one half of 48. For the actual math, you have to use variables. Please stick with it, this gets a little ugly if the child you are working with is as young as mine is. Ben at the start is equal to B0 Jerry at the start is equal to J0 At the start B0=J0 Now, Ben gives 12 of his cards to Jerry. The new value for Ben is B1 and B1 = B0-12. The new value for Jerry is J1 and J1 = J0+12. Also, we know that the new Jerry is twice the amount of the new Ben...J1 =B1 x 2. So, the below are known: 1: B0=J0 2: B1 = B0-12 3: J1 = J0+12 4: J1 = B1 x 2 So, using the last known, start to substitute the other values to get everything equal to the same variable. In this case, I'm going to solve it for B0. J1 = B1 x 2 J1 = (B0-12)x2 - Substitute B1 for B0-12, see line 2 J0+12=(B0-12)x2 - Substitute J1 for J0+12, see line 3 B0+12=(B0-12)x2 - Substitute J0 for B0, see line 1 1/2B0+6 = B0-12 - Divide the whole equation by 2 +6=B0-12-1/2B0 - Subtract 1/2B0 from both sides +18=B0-1/2B0 - Add 12 to both sides +18=1/2B0 - Simplify equation 36=B0 - Multiply both sides by 2 There is probably an easier way to do the above, but that's how I worked it out on paper.
How do you find a determinant of a square matrix in maths?
For a 2x2 matrix, with elements a, b, c and , the determinant is ad - bc.However, for larger matrices it is more complicated. It would have been neater to illustrate this if I could use subscripts but the browser that we are required to use is pretty basic and therefore rubbish!Suppose the elements of an n*n matrix are x(i,j) where i is the row and j is the column. Consider the product x(1, j1)*x(2,j2)*...*x(n,jn) where all the all the js are different. [This is the product of n elements of the matrix such that there is one element from each row and one from each column.] There are n! = n*(n-1)*...*2*1 such terms.Now swap pairs of these term so that the js are in ascending order. For each swap, change the sign of the term so a term requiring an odd number of swaps will have a negative sign and one requiring an even number (including 0 swaps) will be positive.Add these terms together.The following is an illustration for a 3x3 matrix. The 3! = 3*2*1 = 6 terms are:t1 = x(1,1)*x(2,2)*x(3,3)t2 = x(1,1)*x(2,3)*x(3,2) = - x(1,1)*x(3,2)*x(2,3) swap second and thirdt3 = x(1,2)*x(2,1)*x(3,3) = - x(2,1)*x(1,2)*x(3,3) swap first and secondt4 = x(1,2)*x(2,3)*x(3,1) = x(3,1)*x(1,2)*x(2,3) swap first and third, swap second and new thirdt5 = x(1,3)*x(2,1)*x(3,2) = x(2,1)*x(3,2)*x(1,3) swap first and second, swap new second and thirdt6 = x(1,3)*x(2,2)*x(3,1) = - x(3,1)*x(2,2)*x(1,3) swap first and third.The determinant is t1+t2+t3+t4+t5+t6= x(1,1)*x(2,2)*x(3,3) - x(1,1)*x(3,2)*x(2,3) - x(2,1)*x(1,2)*x(3,3) + x(3,1)*x(1,2)*x(2,3) + x(2,1)*x(3,2)*x(1,3) - x(3,1)*x(2,2)*x(1,3)Group theory ensures that the order in which you do the swaps and the parity (odd or even number) is determined by the order of the js and so fixed for each of these terms. Half the terms will be positive and half negative.
