806 CHAPTER 11 Systems of Equations and Inequalities Figure 17 Seeing the Concept Compute the determinant of A in Example 14 using a graphing utility. What is the result? Are you surprised? → ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 1 0 0 0 1 0 0 0 1 − − − − ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 7 4 3 4 1 3 4 3 4 1 1 1 1 The matrix [ ] A I3 is now in reduced row echelon form, and the identity matrix I3 is on the left of the vertical bar. The inverse of A is = − − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ −A 7 4 3 4 1 3 4 3 4 1 1 1 1 1 You should verify that this is the correct inverse by showing that = = − − AA A A I . 1 1 3 ↑ = + =− + R r r R r r 1 1 3 1 2 3 2 Now Work PROBLEM 37 If transforming the matrix [ ] A In into reduced row echelon form does not result in the identity matrix In to the left of the vertical bar, A is singular and has no inverse. Showing That a Matrix Has No Inverse Show that the matrix = ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ A 4 6 2 3 has no inverse. EXAMPLE 14 Algebraic Solution Begin by writing the matrix [ ] A I . 2 A I 4 6 2 3 1 0 0 1 2 [ ] = ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ Then use row operations to transform A I2 [ ] into reduced row echelon form. [ ] = ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ → ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ → − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ A I 4 6 2 3 1 0 0 1 1 3 2 2 3 1 4 0 0 1 1 3 2 0 0 1 4 0 1 2 1 2 R r 1 4 1 1 ↑ = R r r 2 2 1 2 ↑ =− + The matrix [ ] A I2 is sufficiently reduced for it to be clear that the identity matrix cannot appear to the left of the vertical bar, so A is singular and has no inverse. Graphing Solution Enter the matrix A. Figure 17 shows the result of trying to find its inverse using a TI-84 Plus CE. The ERROR comes about because A is singular. Now Work PROBLEM 65
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