Back to Gaussian elimination revisited

Activity 5.1.3

Avg difficulty 2.5

We will consider the matrix A=[121−2−3−2374]A = \left[\begin{array}{rrr} 1 & 2 & 1 \\ -2 & -3 & -2 \\ 3 & 7 & 4 \\ \end{array}\right] and begin performing Gaussian elimination without using partial pivoting.

  1. Perform two row replacement operations to find the row equivalent matrix A′=[121010011].A' = \left[\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end{array}\right]\text{.} Find elementary matrices E1E_1 and E2E_2 that perform these two operations so that E2E1A=A′E_2E_1 A = A'.

  2. Perform a third row replacement to find the upper triangular matrix U=[121010001].U = \left[\begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} Find the elementary matrix E3E_3 such that E3E2E1A=UE_3E_2E_1A = U.

  3. We can write A=E1−1E2−1E3−1UA=E_1^{-1}E_2^{-1}E_3^{-1} U. Find the inverse matrices E1−1E_1^{-1}, E2−1E_2^{-1}, and E3−1E_3^{-1} and the product L=E1−1E2−1E3−1L = E_1^{-1}E_2^{-1}E_3^{-1}. Then verify that A=LUA=LU.

  4. Suppose that we want to solve the equation Ax=b=[4−712]A\xvec = \bvec = \threevec4{-7}{12}. We will write Ax=LUx=L(Ux)=bA\xvec = LU\xvec = L(U\xvec) = \bvec and introduce an unknown vector c\cvec such that Ux=cU\xvec = \cvec. Find c\cvec by noting that Lc=bL\cvec = \bvec and solving this equation.

  5. Now that we have found c\cvec, find x\xvec by solving Ux=cU\xvec = \cvec.

  6. Using the factorization A=LUA=LU and this two-step process, solve the equation Ax=[2−27]A\xvec = \threevec{2}{-2}{7}.

Austin, Understanding Linear Algebra, §5.1 Gaussian elimination revisited, Activity 5.1.3

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Adapted from Understanding Linear Algebra © 2023 David Austin. Licensed under CC BY 4.0. Source: Understanding Linear Algebra. Imported at revision 3e6dced.