## Linear Algebra and Its Applications, Exercise 1.5.3

Exercise 1.5.3. From equations (6) and (3) respectively we have $L = E^{-1}F^{-1}G^{-1} = \begin{bmatrix} 1&0&0 \\ 2&1&0 \\ -1&-1&1 \end{bmatrix} \quad GFE = \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ -1&1&1 \end{bmatrix}$

Multiply the two matrices, in both orders. Explain the two answers. $\begin{bmatrix} 1&0&0 \\ 2&1&0 \\ -1&-1&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ -2&1&0 \\ -1&1&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$

and $\begin{bmatrix} 1&0&0 \\ -2&1&0 \\ -1&1&1 \end{bmatrix} \begin{bmatrix} 1&0&0 \\ 2&1&0 \\ -1&-1&1 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$

In other words, the product of the matrices is the same in both cases, and is the identity matrix I. This can be explained in two ways:

First, GFE is the matrix that embodies the elimination steps to take the matrix A to the upper triangular matrix U, while L takes U back to A: (GFE)A = U and LU = A. So applying GFE to A followed by L will take us back to A: $(E^{-1}F^{-1}G^{-1})(GFE)A = (E^{-1}F^{-1}G^{-1})U = LU = A$

Thus $(E^{-1}F^{-1}G^{-1})(GFE)$ is the identity matrix I.

Similarly, applying L to U followed by GFE takes us back to U: $(GFE)(E^{-1}F^{-1}G^{-1})U = (GFE)LU = (GFE)A = U$

Thus $(GFE)(E^{-1}F^{-1}G^{-1})$ is also the identity matrix I.

Second, from the definition of the inverse of a matrix and the associative property of matrix multiplication we have $(E^{-1}F^{-1}G^{-1})(GFE) = (E^{-1}F^{-1})(G^{-1}G)(FE) = (E^{-1}F^{-1})I(FE) = (E^{-1}F^{-1})(FE) = E^{-1}(F^{-1}F)E = E^{-1}IE = E^{-1}E = I$

and $(GFE)(E^{-1}F^{-1}G^{-1}) = (GF)(EE^{-1})(F^{-1}G^{-1}) = (GF)I(F^{-1}G^{-1}) = (GF)(F^{-1}G^{-1}) = G(FF^{-1})G^{-1} = GIG^{-1} = GG^{-1} = I$

Thus $(E^{-1}F^{-1}G^{-1})(GFE) = (GFE)(E^{-1}F^{-1}G^{-1}) = I$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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