Matrix InversesDate: 04/07/98 at 21:26:12 From: Jonathan Hornbuckle Subject: Linear algebra, matrix inverses Can you find a 2 x 2 matrix inverse to itself where every entry is non-zero and an integer? Can you find a 2 x 2 matrix with all distinct and non-zero entries which has no inverse? Thanks! Date: 04/13/98 at 09:30:25 From: Doctor Lester Subject: Re: Linear algebra, matrix inverses For your first question, we want to find a matrix A such that AA = I, the identity matrix. Let A be the matrix: [a b] [c d] Now we'll try to find values for a, b, c and d that satisfy your requirements. We need: [a b][a b] = [1 0] [c d][c d] [0 1] Multiplying out the lefthand side we get: [a^2+bc ab+bd] = [1 0] [ac+dc bc+d^2] [0 1] Comparing each element in the lefthand matrix with the corresponding element in the righthand matrix, we get the four equations: i) a^2 + bc = 1 ii) ab + db = 0 iii) ac + dc = 0 iv) d^2 + bc = 1 We now need to find out what restrictions these put on our possible choices for a, b, c and d. Subtracting (iv) from (i), you find a simple relation between a and d. Try this to see exactly how they are related. It tells you that once a value for a has been chosen, there are only two possible choices for d (set in terms of a). Let's call these choices 1 and 2. Let's make an arbitrary choice for the value of a, and call it k. Obviously, k cannot be 0, or we have introduced a zero element (which you don't want). It turns out that k cannot be +1 or -1 either (but we don't know that yet!). Find out now what following choice 1 or 2 implies. It turns out that only one of these can lead to a matrix satisfying your conditions. For choice 1 (or 2 depending on how you've labelled them!), equation (ii) gives us kb = 0 (try it). This requires that either k or b is 0, again introducing zero elements. So follow choice 2 instead, for d. For this choice of d in terms of k, now (ii) and (iii) are automatically satisfied, whatever our choices of b and c. (Please check this!) So can we have ANY non-zero values for b and c? Well, (i) and (iv) both give us an equation listing bc in terms of k. Please find out what this is. You will find that, as long as k is not +1 or -1, then bc = <an integer>, so you can always choose b and c to fit this equation. You should have found that, to build an example matrix of this form, you must set a to any arbitrary number. We have seen that this cannot be 0, -1 or +1. Now you need to set d equal to -a and then choose b and c as you wish, such that bc = 1 - a^2. But make sure you understand the steps above that lead us to this recipe. Once you've made such a matrix, try working out the inverse just to check. Obviously, you could have just guessed a solution, but that wouldn't have shown you why that solution worked. I have tried to leave much of the working for you to try yourself, so please write back if this has confused things for you! For your second question, to find a matrix with non-zero integer entries that is not invertible, you just need to know that the determinant of a matrix is zero if and only if the matrix has no inverse. Do you know how to evaluate determinants? The determinant of a matrix is just a special number associated with a matrix. For 2x2 matrices of the form: [a b] [c d] the determinant is just the value ad - bc. So we just have to choose elements so that ad - bc is zero. For example you can easily make a matrix such that ad = 12 and bc = 12 but where a and d factorise 12 in a different way from b and c (because 3x4 = 12 = 2x6). Then ad - bc = 12 - 12 = 0. -Doctor Lester, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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