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Chapter 3: Problem 33

Show that every elementary matrix \(E\) is invertible, and its inverse is an elementary matrix.

Short Answer

Expert verified
In summary, every elementary matrix E can be obtained from the identity matrix I by performing a single elementary row operation, such as swapping two rows, multiplying a row by a nonzero scalar, or adding a multiple of one row to another row. To show that E is invertible and its inverse is an elementary matrix, we can reverse the initial operation on E to obtain the identity matrix I. This demonstrates that E is invertible, and the inverse matrix E鈦宦� is also an elementary matrix since it is formed by applying a single elementary row operation on the identity matrix.

Step by step solution

01

Recall the definition of an elementary matrix

An elementary matrix is a matrix that can be obtained from the identity matrix by performing a single elementary row operation. There are three types of elementary row operations: 1. Swap two rows. 2. Multiply a row by a nonzero scalar. 3. Add a multiple of one row to another row.
02

Type 1: Swap two rows

Let E be an elementary matrix formed by swapping the i-th row and j-th row of the identity matrix I (where i 鈮� j). We can obtain the inverse of E by performing the same row operation on E (swapping i-th and j-th rows again). This transformation results in the identity matrix I, meaning that E is indeed invertible and E鈦宦� = E.
03

Type 2: Multiply a row by a nonzero scalar

Now let E be an elementary matrix formed by multiplying the k-th row of the identity matrix I by a nonzero scalar, say c. To find the inverse of E, we can multiply the k-th row of E by the reciprocal of c (1/c). This operation will return the identity matrix I, indicating that E is invertible. The inverse matrix is E鈦宦�, which is also an elementary matrix since this operation is the same type as the one used to create E.
04

Type 3: Add a multiple of one row to another row

Lastly, let E be an elementary matrix created by adding a multiple of the m-th row to the n-th row of the identity matrix I (where m 鈮� n). To find the inverse matrix E鈦宦�, we can perform the reverse operation - subtract the same multiple of the m-th row from the n-th row. This transformation results in the identity matrix I, establishing that E is invertible. The inverse is E鈦宦�, which is also an elementary matrix since it is obtained from the identity matrix through a single elementary row operation. In conclusion, all elementary matrices are invertible, and their inverses are elementary matrices as well.

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Most popular questions from this chapter

For each of the following homogeneous systems of linear equations, find the dimension of and a basis for the solution space. (a) \(x_{1}+3 x_{2}=0\) \(2 x_{1}+6 x_{2}=0\) (b) \(x_{1}+x_{2}-x_{3}=0\) \(4 x_{1}+x_{2}-2 x_{3}=0\) (c) \(x_{1}+2 x_{2}-x_{3}=0\) \(2 x_{1}+x_{2}+x_{3}=0\) (d) \(\quad x_{1}-x_{2}+x_{3}=0\) \(x_{1}+2 x_{2}-2 x_{3}=0\) (e) \(x_{1}+2 x_{2}-3 x_{3}+x_{4}=0\) (f) \(x_{1}+2 x_{2}=0\) \(x_{1}-x_{2}=0\) (g) \(x_{1}+2 x_{2}+x_{3}+x_{4}=0\) \(x_{2}-x_{3}+x_{4}=0\)

Find the \(L U\) factorization of the matrix \(A=\left[\begin{array}{rrr}1 & 2 & 1 \\ 2 & 3 & 3 \\ -3 & -10 & 2\end{array}\right]\).

Find the LU factorization of \((\mathrm{a}) \quad A=\left[\begin{array}{rrr}1 & -3 & 5 \\ 2 & -4 & 7 \\ -1 & -2 & 1\end{array}\right]\) (b) \(B=\left[\begin{array}{rrr}1 & 4 & -3 \\ 2 & 8 & 1 \\ -5 & -9 & 7\end{array}\right]\)

Use the proof of Theorem \(3.2\) to obtain the inverse of each of the following elementary matrices. (a) $\left(\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right)$ (b) $\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\end{array}\right)$ (c) $\left(\begin{array}{rrr}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1\end{array}\right)$

It can be shown that the vectors \(u_{1}=(2,-3,1), u_{2}=(1,4,-2), u_{3}=\) \((-8,12,-4), u_{4}=(1,37,-17)\), and \(u_{5}=(-3,-5,8)\) generate \(\mathrm{R}^{3}\). Find a subset of $\left\\{u_{1}, u_{2}, u_{3}, u_{4}, u_{5}\right\\}\( that is a basis for \)\mathrm{R}^{3}$.
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