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Question

# Find the rank by theorem "Rank in Terms of Determinants " and check by row reduction. (Show details.)$\left[\begin{array}{rrr}2 & 1 & 0 \\ 13 & -13 & 12 \\ -3 & 5 & -4\end{array}\right]$

Solution

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We have to find the determinant of the given matrix:

\begin{align*} \begin{vmatrix} 2 & 1 & 0 \\ 13 & -13 & 12 \\ -3 & 5 & -4 \end{vmatrix} &= 2(-1)^{1+1} \begin{vmatrix} -13 & 12 \\ 5 & -4 \end{vmatrix} + 1 (-1)^{1+2} \begin{vmatrix} 13 & 12 \\ -3 & -4 \end{vmatrix} \\ &= 2 ( -13 (-4) - 12 \cdot 5) - (13(-4)-12(-3)) \\ &= 2( 52 - 60) - ( -52+36) \\ &=2(-8) - (-16) \\ &=-16+16 =0 \end{align*}

Now, calculate the determinant of one submatrix of the given matrix:

\begin{align*} \begin{vmatrix} 2 & 1 \\ 13 & -13 \end{vmatrix} = 2(-13) - 1 \cdot13 = -26-13 = - 39 \ne 0 \end{align*}

Since there is at least one $2\times2$ submatrix that has the non zero determinant, the rank of the given matrix is 2.

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