## 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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