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I have a system of equations to solve $Ax = b$, but the determinant of matrix $A$ is zero.

$$A=\begin{bmatrix}1&1&0&0\\0&0&1&1\\1&0&1&0\\0&1&0&1\end{bmatrix}$$

To me, none of the rows/columns of this matrix look dependent. I am wondering why the determinant of matrix $A$ is zero?

Also, how should I solve this system or estimate $x$?

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    $\begingroup$ Note that adding the first two rows and subtracting the third yields the fourth, so the row (and column) vectors are not linearly independent. $\endgroup$ – user455343 Jul 28 '17 at 19:37
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    $\begingroup$ I think I see why you're confused -- there is a nice symmetry in this matrix that would indeed seem to suggest that it is full-rank, since all components of each basis vector contribute equally or "fairly". It definitely defies my intuition as well, so I guess this is yet another lesson that you shouldn't trust your intuition! $\endgroup$ – Mehrdad Jul 28 '17 at 23:43
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    $\begingroup$ The sum of the first two is equal to the sum of the last two. $\endgroup$ – Alfred Yerger Jul 29 '17 at 4:28
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Suppose $b=\begin{bmatrix}b_1&b_2&b_3&b_4\end{bmatrix}^T$ and consider the Gaussian elimination on the complete matrix: \begin{align} \left[\begin{array}{cccc|c} 1&1&0&0&b_1\\ 0&0&1&1&b_2\\ 1&0&1&0&b_3\\ 0&1&0&1&b_4 \end{array}\right] &\to \left[\begin{array}{cccc|c} 1&1&0&0&b_1\\ 0&0&1&1&b_2\\ 0&-1&1&0&b_3-b_1\\ 0&1&0&1&b_4 \end{array}\right] &&R_3\gets R_3-R_1 \\[6px]&\to \left[\begin{array}{cccc|c} 1&1&0&0&b_1\\ 0&1&-1&0&b_1-b_3\\ 0&0&1&1&b_2\\ 0&1&0&1&b_4 \end{array}\right] &&R_3\leftrightarrow R_2,\quad R_2\gets-R_2 \\[6px]&\to \left[\begin{array}{cccc|c} 1&1&0&0&b_1\\ 0&1&-1&0&b_1-b_3\\ 0&0&1&1&b_2\\ 0&0&1&1&b_4-b_1+b_3 \end{array}\right] &&R_4\gets R_4-R_2 \\[6px]&\to \left[\begin{array}{cccc|c} 1&1&0&0&b_1\\ 0&1&-1&0&b_1-b_3\\ 0&0&1&1&b_2\\ 0&0&0&0&b_4-b_1+b_3-b_2 \end{array}\right] &&R_4\gets R_4-R_3 \tag{*} \\[6px]&\to \left[\begin{array}{cccc|c} 1&1&0&0&b_1\\ 0&1&0&1&b_1-b_3+b_2\\ 0&0&1&1&b_2\\ 0&0&0&0&b_4-b_1+b_3-b_2 \end{array}\right] &&R_2\gets R_2+R_3 \\[6px]&\to \left[\begin{array}{cccc|c} 1&0&0&-1&b_3-b_2\\ 0&1&0&1&b_1-b_3+b_2\\ 0&0&1&1&b_2\\ 0&0&0&0&b_4-b_1+b_3-b_2 \end{array}\right] &&R_1\gets R_1-R_2 \end{align} From the step marked as (*) we deduce that

  1. the matrix $A$ has rank $3$ (hence zero determinant);
  2. the system has solution if and only if $b_4-b_1+b_3-b_2=0$.

When $b_4-b_1+b_3-b_2=0$, the system has infinitely many solutions in the form $$ \begin{bmatrix} b_3-b_2+h\\ b_1-b_3+b_2-h\\ b_2-h\\ h \end{bmatrix} = \begin{bmatrix} b_3-b_2\\ b_1-b_3+b_2\\ b_2\\ 0 \end{bmatrix} + h\begin{bmatrix} 1 \\ -1 \\ -1 \\ 1 \end{bmatrix} $$ where $h$ is any scalar.

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$$ (1,1,0,0)+(0,0,1,1)-(1,0,1,0)-(0,1,0,1)=(0,0,0,0) $$ so the rows are linearly dependent.

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Consider the following block matrix: $$ M=\left[\begin{array}[cc]\\A&B\\ C&D\end{array}\right] $$ where $A$,$B$,$C$ and $D$ have the same order. If $D$ be invertible then we have: $$ \det(M)=\det(D)\cdot \det(A-BD^{-1}C) $$ In your question $C=D=I_2$,the identity matrix of order $2$, which results that: $$ \det(M)=\underbrace{\det(D)}_1\cdot \det(A-B)=\det( \left[ \begin{array}[cc] \\1&1\\ -1&-1 \end{array} \right])=0 $$

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