Show that the rows of the matrix \(\Mtrx{A}\) below are linearly independent.
\[ \Mtrx{A} = \left[ \begin{array}{rrrr} 5 & 2 & 0 & 1 \\ 4 & -1 & 3 & 0 \end{array} \right] \]According to the determinant test for linear independence it is enough if we find two columns in \(\Mtrx{A}\) such that the determinant of the resulting \((2,2)\)-matrix is not 0. – A convenient choice are the 3rd and 4th columns:
\[ \det\, \left[ \begin{array}{rr} 0 & 1 \\ 3 & 0 \end{array} \right] = -3 \neq 0 \]Therefore the rows of \(\Mtrx{A}\) are linearly independent.
Show that the rows of the matrix \(\Mtrx{B}\) below are linearly independent.
\[ \Mtrx{A} = \left[ \begin{array}{rrrrrr} 1 & 4 & -5 & 2 & 9 & 1 \\ 2 & 3 & 2 & 3 & 4 & 2 \\ 21& -2& -9 & -4& 14 & 7 \\ \end{array} \right] \]According to the determinant test for linear independence it is enough if we find three columns in \(\Mtrx{B}\) such that the determinant of the resulting \((3,3)\)-matrix is not 0. – The first column has a large number in it. when computing determinants this may result in a higher computational effort. So let us try the 2nd, 3rd, and 4th columns:
\[ \det\, \left[ \begin{array}{rrr} 4 & -5 & 2 \\ 3 & 2 & 3 \\ -2 & -9 & -4 \end{array} \right] = 4\cdot(-8+27) -3\cdot(20 +18) -2\cdot(-15 -4) = 0 \]Ooops! – This is not what we were hoping to find. Unfortunately, we can not conclude anything from this result: the fact that some columns produce a matrix whose determinant is 0 does not mean that all columns produce a matrix with 0-determinant. Let use try columns 1, 2, 3:
\[ \det\, \left[ \begin{array}{rrr} 1 & 4 & -5 \\ 2 & 3 & 2 \\ 21 & -2 & -9 \end{array} \right] = 1\cdot(-27 +4) -2\cdot(-36-10) + 21\cdot(8+15) = 342 \neq 0 \]