Page 98 - Elementary_Linear_Algebra_with_Applications_Anton__9_edition

EXAMPLE 2 Solving Two Linear Systems at Once
Solve the systems

(a)

(b)

Solution

The two systems have the same coefficient matrix. If we augment this coefficient matrix with the columns of constants on the right
sides of these systems, we obtain

Reducing this matrix to reduced row-echelon form yields (verify)

It follows from the last two columns that the solution of system (a) is , , and the solution of system (b) is ,
,.

Properties of Invertible Matrices matrix B such that

Up to now, to show that an matrix A is invertible, it has been necessary to find an

The next theorem shows that if we produce an matrix B satisfying either condition, then the other condition holds
automatically.

THEOREM 1.6.3

Let A be a square matrix. , then .
(a) If B is a square matrix satisfying

(b) If B is a square matrix satisfying , then .