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Chapter 3: Problem 5

Prove that \(E\) is an elementary matrix if and only if \(E^{t}\) is.

Short Answer

Expert verified
Proof: We analyzed each of the three types of elementary matrices (Type 1, Type 2, and Type 3) and found that for each type, if E is an elementary matrix, its transpose \(E^{t}\) is also an elementary matrix. Therefore, this holds for all elementary matrices, and we can conclude that E is an elementary matrix if and only if \(E^{t}\) is an elementary matrix.

Step by step solution

01

Define an elementary matrix

An elementary matrix is a matrix that can be obtained from the identity matrix by performing exactly one elementary row operation. There are three types of elementary matrices: 1. Type 1: Obtained by swapping two rows of the identity matrix 2. Type 2: Obtained by multiplying a row of the identity matrix by a non-zero constant 3. Type 3: Obtained by adding/subtracting a multiple of one row of the identity matrix to another row Now, let's check each type of elementary matrix and see if their transposes also form elementary matrices.
02

Analyze Type 1 elementary matrix and its transpose

A Type 1 elementary matrix is created by swapping two rows of the identity matrix. Let E be a Type 1 elementary matrix, and let \(E^{t}\) be its transpose. To obtain \(E^{t}\), we swap the corresponding columns of E. But since E was obtained from the identity matrix by swapping rows, swapping columns will result in another identity matrix with a pair of swapped columns. Thus, \(E^{t}\) is also an elementary matrix of Type 1.
03

Analyze Type 2 elementary matrix and its transpose

A Type 2 elementary matrix is created by multiplying a row of the identity matrix by a non-zero constant. Let E be a Type 2 elementary matrix, and let \(E^{t}\) be its transpose. To obtain \(E^{t}\), we replace the corresponding row of E with the same row multiplied by the non-zero constant. Since E was obtained from the identity matrix by multiplying a row by a non-zero constant, replacing the row with the multiplied row will result in another identity matrix with one row multiplied by the same non-zero constant. Thus, \(E^{t}\) is also an elementary matrix of Type 2.
04

Analyze Type 3 elementary matrix and its transpose

A Type 3 elementary matrix is created by adding/subtracting a multiple of one row of the identity matrix to another row. Let E be a Type 3 elementary matrix, and let \(E^{t}\) be its transpose. To obtain \(E^{t}\), we add/subtract the corresponding columns of E scaled by the same multiple. But since E was obtained from the identity matrix by adding/subtracting a multiple of one row to another row, adding/subtracting the scaled columns will result in another identity matrix with the addition/subtraction of the same multiple of one column to another column. Thus, \(E^{t}\) is an elementary matrix of Type 3.
05

Conclusion

In conclusion, we have proven that if E is an elementary matrix, then its transpose \(E^{t}\) is also an elementary matrix. This holds for all three types of elementary matrices. Therefore, E is an elementary matrix if and only if \(E^{t}\) is an elementary matrix.

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Most popular questions from this chapter

A system \(\mathscr{L}\) of linear equations is said to be consistent if no linear combination of its equations is a degenerate equation \(L\) with a nonzero constant. Show that \(\mathscr{L}\) is consistent if and only if \(\mathscr{L}\) is reducible to echelon form.

Find the rank of the following matrices. (a) $\left(\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0\end{array}\right)$ (b) $\left(\begin{array}{lll}1 & 1 & 0 \\ 2 & 1 & 1 \\ 1 & 1 & 1\end{array}\right)$ (c) \(\left(\begin{array}{lll}1 & 0 & 2 \\ 1 & 1 & 4\end{array}\right)\) (d) \(\left(\begin{array}{lll}1 & 2 & 1 \\ 2 & 4 & 2\end{array}\right)\) (e) $\left(\begin{array}{rrrrr}1 & 2 & 3 & 1 & 1 \\ 1 & 4 & 0 & 1 & 2 \\ 0 & 2 & -3 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0\end{array}\right)$ (f) $\left(\begin{array}{rrrrr}1 & 2 & 0 & 1 & 1 \\ 2 & 4 & 1 & 3 & 0 \\ 3 & 6 & 2 & 5 & 1 \\ -4 & -8 & 1 & -3 & 1\end{array}\right)$ (g) $\left(\begin{array}{llll}1 & 1 & 0 & 1 \\ 2 & 2 & 0 & 2 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1\end{array}\right)$

Let \(A\) be an \(m \times n\) matrix. Prove that if \(B\) can be obtained from \(A\) by an elementary row [column] operation, then \(B^{t}\) can be obtained from \(A^{t}\) by the corresponding elementary column [row] operation.

Let the reduced row echelon form of \(A\) be $$ \left(\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -2 \\ 0 & 1 & -5 & 0 & -3 \\ 0 & 0 & 0 & 1 & 6 \end{array}\right) $$ Determine \(A\) if the first, second, and fourth columns of \(A\) are $$ \left(\begin{array}{r} 1 \\ -1 \\ 3 \end{array}\right), \quad\left(\begin{array}{r} 0 \\ -1 \\ 1 \end{array}\right), \quad \text { and } \quad\left(\begin{array}{r} 1 \\ -2 \\ 0 \end{array}\right) $$ respectively.Let the reduced row echelon form of \(A\) be $$ \left(\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -2 \\ 0 & 1 & -5 & 0 & -3 \\ 0 & 0 & 0 & 1 & 6 \end{array}\right) $$ Determine \(A\) if the first, second, and fourth columns of \(A\) are $$ \left(\begin{array}{r} 1 \\ -1 \\ 3 \end{array}\right), \quad\left(\begin{array}{r} 0 \\ -1 \\ 1 \end{array}\right), \quad \text { and } \quad\left(\begin{array}{r} 1 \\ -2 \\ 0 \end{array}\right) $$ respectively.Let the reduced row echelon form of \(A\) be $$ \left(\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -2 \\ 0 & 1 & -5 & 0 & -3 \\ 0 & 0 & 0 & 1 & 6 \end{array}\right) $$ Determine \(A\) if the first, second, and fourth columns of \(A\) are $$ \left(\begin{array}{r} 1 \\ -1 \\ 3 \end{array}\right), \quad\left(\begin{array}{r} 0 \\ -1 \\ 1 \end{array}\right), \quad \text { and } \quad\left(\begin{array}{r} 1 \\ -2 \\ 0 \end{array}\right) $$ respectively.Let the reduced row echelon form of \(A\) be $$ \left(\begin{array}{rrrrr} 1 & 0 & 2 & 0 & -2 \\ 0 & 1 & -5 & 0 & -3 \\ 0 & 0 & 0 & 1 & 6 \end{array}\right) $$ Determine \(A\) if the first, second, and fourth columns of \(A\) are $$ \left(\begin{array}{r} 1 \\ -1 \\ 3 \end{array}\right), \quad\left(\begin{array}{r} 0 \\ -1 \\ 1 \end{array}\right), \quad \text { and } \quad\left(\begin{array}{r} 1 \\ -2 \\ 0 \end{array}\right) $$ respectively.

Solve (a) \(\quad e x=\pi\) (b) \(3 x-4-x=2 x+3\) (c) \(7+2 x-4=3 x+3-x\)
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