Using determinants, it is possible to express the solution to a system of equations whose coefficient matrix is invertible:

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1 Cramer s Rule and the Adjugate Using determinants, it is possible to express the solution to a system of equations whose coefficient matrix is invertible: Theorem [Cramer s Rule] If A is an invertible n n matrix, then the solution to the matrix equation Ax = b has entries given by the formulas x i = det A i ( b) det A, i = 1,2,,n, where A i ( b) is the matrix obtained from A by replacing its ith column with the vector b. Proof Since we have A i ( b) = a 1 L b L a n = Ae 1 L Ax L Ae n = A [ e 1 L x L e n ], det A i (b) = det A det[ e 1 L x L e n ] = det A x i and the result follows immediately as det A 0. //

2 While it is possible to use Cramer s Rule to solve a system of equations, it should be noted that the row reduction methods we have been using are far more computationally efficient. If A is invertible, then we can interpret the equation AA 1 = I as saying that the matrix equation Ax = e i has as its solution the ith column vector of A 1. Applying Cramer s Rule, we conclude that (A 1 ) ij = det A i (e j ) det A. But we can compute det A i (e j ) by cofactor expansion along its ith column; this shows that det A i (e j ) = ( 1) i+ j det A ji = C ji is the ( j, i) cofactor of A. So (A 1 ) ij = C ji det A. This means that if we define a matrix whose (i, j) entry is C ji this matrix is denoted adj A and is called the adjugate of A, then we have the Theorem If A is invertible then A 1 = 1 adj A.// det A

3 Area and Volume Theorem If A is a 2 2 matrix, then its two column vectors a 1 and a 2 determine a parallelogram in R 2 whose area is deta. Proof The line through the origin in the direction of a 1 contains one side of this parallelogram as well as all the vectors in Span{ a 1}. The opposite side of the parallelogram is contained in the parallel line determined by all the vectors in a 2 + Span{ a 1}, namely all vectors of the form ca 1 + a 2. (See Figure 2, p. 205). In fact, the parallelogram with sides a 1 and ca 1 + a 2 for any choice of c has the same base and height, so they all have the same area. In particular, the area of the parallelogram formed by any two vectors is unaffected by swapping them or by replacing one of them by its sum with a multiple of the other. This shows that the area of the parallelogram formed by the columns of A is the same as the area of the parallelogram formed by the columns of any matrix obtained from A by performing column operations of the swap or replacement types (no scalings allowed). Therefore, performing column operations of these types to a 2 2 matrix does not change the area of the associated parallelogram.

4 But if the columns of A are scalar multiples of each other, then the associated parallelogram is degenerate and has area 0. Likewise, deta = 0. If the columns of A are linearly independent, the associated parallelogram is not degenerate and we can perform column operations of the two required types that bring A to diagonal form a 0. Since 0 d the columns of this last matrix are sides of a rectangle of dimensions a and d, the area of the parallelogram is ad = det a 0 = det A. // 0 d The corresponding result for a 3 3 matrix A is the Theorem If A is a 3 3 matrix, then its three column vectors a 1,a 2 and a 3 determine a parallelepiped in R 3 whose volume is deta. Proof The volume of a parallelepiped equals the area of its base times its height. The plane determined by Span {a 1,a 2 } contains one face of the solid and the opposite face lies in the parallel plane determined by a 3 + Span {a 1,a 2 }. The four edges lying between these faces are copies of the vector a 3 (as in Figure 4, p. 206). But the volume

5 is unchanged if these edges are replaced by vectors of the form a 3 + ra 1 + sa 2 since this does not change the base or the height of the solid. This shows that the volume of the parallelepiped formed by the columns of A is the same as the volume of the parallelepiped formed by the columns of any matrix obtained from A by performing column operations of the swap or replacement types (no scalings allowed). Therefore, performing column operations of these types to a 3 3 matrix does not change the volume of the associated parallelepiped. If the columns of A are linearly dependent, then the associated parallelepiped is degenerate and has volume 0. Likewise, deta = 0. If the columns of A are linearly independent, then the associated parallelepiped is not degenerate and we can perform appropriate column operations of these two types that bring A to diagonal form. Since the columns of this last matrix are sides of a rectangular parallelepiped with dimensions a 11, a 22, a 33, the volume of the solid is thus a a 11 a 22 a 33 = det 0 a 22 0 = det A. // 0 0 a 33

6 As corollaries to these two geometric theorems, we get Corollary Let T :R 2 R 2 be a linear transformation with standard matrix A. If Π is some parallelogram in R 2 then so is T (Π), and Area of T (Π) = deta (Area of Π). Proof Suppose adjacent sides of Π are determined by the two vectors p 1 and p 2. Then every point in Π corresponds to the vector s 1 p 1 + s 2 p 2 where 0 s 1,s 2 1. It follows that every point in T (Π) has the form T (s 1 p 1 + s 2 p 2 ) = s 1 T (p 1 ) + s 2 T (p 2 ) where 0 s 1,s 2 1; that is, adjacent sides of T (Π) are determined by the two vectors T (p 1 ) and T (p 2 ), the columns of the matrix [ T (p 1 ) T (p 2 )] = [ Ap 1 Ap 2 ] = A [ p 1 p 2 ]. So ( ). // Area of T (Π) = det T (p 1 ) T (p 2 ) = det A det p 1 p 2 = det A Area of Π

7 The corresponding result in R 3 is similarly proved: Corollary Let T :R 3 R 3 be a linear transformation with standard matrix A. If Π is some parallelepiped in R 3 then so is T (Π), and Volume of T (Π) = deta (Volume of Π). //