How to find the square of a matrix

Matrices 2. Solving Square Systems of Linear Equations; Inverse Matrices Solving square systems of linear equations; inverse matrices. Linear algebra is essentially about solving systems of linear equations, an important application of mathematics to real-world problems in engineering, business, and science, especially the social sciences. Here we willjust stick tothe mostimportant case, wherethe system is square, i.e., there are as many variables as there are equations. In low dimensions such systemslook asfollows(wegive a2 × 2 system and a 3 × 3 system): a11x1+ a12x2= b1a11x1+ a12x2+ a13x3= b1(7) a21x1+ a22x2= b2a21x1+ a22x2+ a23x3= b2a31x1+ a32x2+ a33x3= b3In these systems, the aijand biare given, and we want to solve for the xi. As a simple mathematical example, consider the linear change of coordinates given by the equations x1= a11y1+ a12y2+ a13y3x2= a21y1+ a22y2+ a23y3x3= a31y1+ a32y2+ a33y3If we know the y-coordinates of a point, then these equations tell us its x-coordinates immediately. But if instead we are given the x-coordinates, to find the y-coordinates we must solve a system of equationslike(7) above, with the yias the unknowns. Using matrix multiplication, we can abbreviatethe system onthe rightin(7) by   x1b1(8) Ax = b,x =  x2 ,b =  b2 ,x3b3where Aisthe square matrix of coefficients(aij). (The 2×2 system and the n×nsystem would be written analogously; all of them are abbreviated by the same equation Ax = b, notice.) Youhavehad experience with solving small systemslike(7) by elimination: multiplying the equations by constants and subtracting them from each other, the purpose being to eliminate all the variablesbut one. When eliminationisdone systematically,itis an efficient method. Here however we want to talk about another method more compatible with handheld calculators and MatLab, and which leads more rapidly to certain key ideas and results in linear algebra. Inversematrices.Referring tothesystem(8),supposewecan find asquarematrix M, the same size as A, such that (9) MA= I(the identitymatrix).1
            2MATRIXINVERSESWe canthen solve(8) by matrix multiplication, using the successive steps, Ax = b M(Ax)= Mb (10) x = Mb; where the step M(Ax)= x isjustified by M(Ax)=(MA)x,by associative law; = Ix,by(9); = x,because Iis the identity matrix . Moreover,the solutionis unique, since(10) gives an explicitformulaforit. The same procedure solves the problem of determining the inverse to the linear change of coordinates x = Ay, as the next example illustrates. 12 −32 Example 2.1 Let A= and M= . Verify that Msatisfies(9) 23 2 −1 above, and use it to solve the first system below for xiand the second for the yiin terms of the xi: x1+2x2= −1 x1= y1+2y22x1+3x2=4 x2=2y1+3y2      1 2 −3 2 1 0 Solution. We have = , by matrix multiplication. To 2 3 2 −1 0 1 x1−32 −1 11 solve the first system, we have by (10), = = , so the x22 −14 −6 solution is x1=11,x2= −6. By reasoning similartothatused aboveingoingfrom Ax = b to x = Mb, the solution to x = Ay is y= Mx, so that we get y1= −3x1+2x2y2=2x1−x2as the expression for the yiin terms of the xi. Our problem now is: how do we get the matrix M? In practice, you mostly press a key on the calculator, or type a Matlab command. But we need to be able to work abstractly with thematrix —i.e.,with symbols,notjustnumbers,andforthissometheoreticalideas are important. The first is that Mdoesn’t always exist. Mexists ⇔|A|6=0.The implication ⇒ follows immediately from the law M-5 in section M.1 (det(AB)= det(A)det(B)), since MA= I⇒|M||A|= |I|=1 ⇒|A|6=0.