Instruction
1
The basis of the matrix algebra are operations on matrices and find their main characteristics. To find the adjoint matrix you need to perform the transpose and formed on the basis of its result a new matrix of the corresponding algebraic additions.
2
The transpose of a square matrix is a record of its elements in a different order. The first column is changed to the first row, the second on the second etc. in General it looks like this (see picture).
3
The second stage of finding the adjoint matrix – search for algebraic additions. These numerical characteristics of the matrix elements are obtained by computing the minors. Those, in turn, are determinants of the initial matrix order, smaller by 1, and obtained by striking out the appropriate rows and columns. For example, M11 = (a22•a33 – a23•a32). The cofactor is different from the minor coefficient equal to (-1) degree of amount of item numbers: A11 = (-1)^(1+1)• (a22•a33 – a23•a32).
4
Consider the example: find the adjoint matrix to a is given. For convenience, take the third right. This will allow faster understanding of the algorithm without heavy calculations, because the calculation of determinants of matrices of the third order enough of all four elements.
5
Swipe the transpose of a given matrix. Here you want to swap the first row on the first column, the second to the second and third – the third.
6
Write down expressions for the search of algebraic additions in total there will be 9 by the number of elements of the matrix. Be careful with the sign, it is better to refrain from calculations in mind, and to paint everything in detail.
7
A11 = (-1)2•(2 -24) = -22;
A12 = (-1)3•(1+ 18) = -19;
A13 = (-1)^4•(4 + 6) = 10;
A21 = (-1)3•(9 + 4) = -13;
A22 = (-1)^4•(5 - 3) = 2;
A23 = (-1)^5•(20 + 27);
A31 = (-1)^4•(54 + 2) = 56;
A32 = (-1)^5•(30 + 1) = -31;
A33 = (-1)^6•(10 - 9) = 1.
8
Make final the attached matrix of the resulting algebraic additions.