In Exercises 1–2, find the rank and nullity of the matrix \(A\) by reducing it to row echelon form. (a) \(A=\left[\begin{array}{rrrr}1 & 2 & -1 & 1 \\2 & 4 & -2 & 2 \\3 & 6 & -3 & 3 \\4 & 8 & -4 & 4\end{array}\right]\) (b) \(A=\left[\begin{array}{rrrrr}1 & -2 & 2 & 3 & -1 \\-3 & 6 & -1 & 1 & -7 \\2 & -4 & 5 & 8 & -4\end{array}\right]\) Equation Transcription: [] [] Text Transcription: A A=[_4 8 -4 4 ^3 6 -3 3 ^2 4 -2 2 ^1 2 -1 1] A=[_ 2 -4 5 8 -4 ^-3 6 -1 1 -7 ^1 -2 2 3 -1]
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Textbook Solutions for Elementary Linear Algebra
Question
It can be proved that a nonzero matrix ???? has rank k if and only if some \(k \times k\) submatrix has a nonzero determinant and all square submatrices of larger size have determinant zero. Use this fact to find the rank of
\(A=\left|\begin{array}{ccccc} 3 & -1 & 3 & 2 & 5 \\ 5 & -3 & 2 & 3 & 4 \\ 1 & -3 & -5 & 0 & -7 \\ 7 & -5 & 1 & 4 & 1 \end{array}\right| \)
Check your result by computing the rank of ???? in a different way.
Solution
The first step in solving 4.9 problem number trying to solve the problem we have to refer to the textbook question: It can be proved that a nonzero matrix ???? has rank k if and only if some \(k \times k\) submatrix has a nonzero determinant and all square submatrices of larger size have determinant zero. Use this fact to find the rank of \(A=\left|\begin{array}{ccccc} 3 & -1 & 3 & 2 & 5 \\ 5 & -3 & 2 & 3 & 4 \\ 1 & -3 & -5 & 0 & -7 \\ 7 & -5 & 1 & 4 & 1 \end{array}\right| \)Check your result by computing the rank of ???? in a different way.
From the textbook chapter Rank, Nullity, and the Fundamental Matrix Spaces you will find a few key concepts needed to solve this.
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