Inverse Linear Substitutions and Matrices

We generalize Example 3 (on page 5 of Anton–Rorres) by solving the linear system

\{ \begin{matrix}x & +y & +2z & = & u & & & (1) \\ 2x & +4y & −3z & = & & v & & (2) \\ 3x & +6y & −5z & = & & & w & (3) \end{matrix} with matrix A = [ \begin{matrix}1 & 1 & 2 \\ 2 & 4 & −3 \\ 3 & 6 & −5 \end{matrix} ]

for x, y, and z in terms of u, v, and w. In the book, u = 9, v = 1, and w = 0. The point is that we can solve the more general problem with very little extra work, by following exactly the same steps as before (annotated symbolically).

2y −7z = −2u +v  (4) = (2) − 2(1)
3y −11z = −3u  +w (5) = (3) − 3(1)
y −\frac{7}{2}z = −u +\frac{1}{2}v  (6) = \frac{1}{2}(4)
−\frac{1}{2}z =  −\frac{3}{2}v +w (7) = (5) − 3(6)
z =  3v −2w (8) = −2(7)
x  +\frac{11}{2}z = 2u −\frac{1}{2}v  (9) = (1) − (6)
x  = 2u −17v +11w (10) = (9) − \frac{11}{2}(8)
y  = −u +11v −7w (11) = (6) + \frac{7}{2}(8)

Thus the desired solution consists of equations (10), (11), and (8),

\{ \begin{matrix}x & = & 2u & −17v & +11w \\ y & = & −u & +11v & −7w \\ z & = & & 3v & −2w \end{matrix} with matrix B = [ \begin{matrix}2 & −17 & 11 \\ −1 & 11 & −7 \\ 0 & 3 & −2 \end{matrix} ]

This is the inverse linear substitution, whose matrix B is the inverse matrix A^{−1} to A. In matrix notation, we solved the equation Ax = u and obtained the solution x = Bu, where we write

x = [ \begin{matrix}x \\ y \\ z \end{matrix} ] , u = [ \begin{matrix}u \\ v \\ w \end{matrix} ]

Computationally, we in effect applied row reduction to the 3×6 partitioned matrix [A|I] to produce the reduced row echelon matrix [I|B],

[ \begin{matrix}1 & 1 & 2 & 1 & 0 & 0 \\ 2 & 4 & −3 & 0 & 1 & 0 \\ 3 & 6 & −5 & 0 & 0 & 1 \end{matrix} ] ⟶ [ \begin{matrix}1 & 0 & 0 & 2 & −17 & 11 \\ 0 & 1 & 0 & −1 & 11 & −7 \\ 0 & 0 & 1 & 0 & 3 & −2 \end{matrix} ]

(If it had failed to reduce to this form, the message would have been that A was not invertible.)

110.201 Linear Algebra JMB File: lsinv, Revision A; 27 Aug 2001; Page 1