Find $x$ such that the rank of the matrix is $2$

Question

Given the matrix $$\begin{pmatrix} 1 & 3 & -3 & x\\ 2 & 2 & x & -4 \\ 1 & 1-x & 2x+1 & -5-3x\\ \end{pmatrix}$$ Find $x$ such that the rank of the matrix is $2$

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First, I note that I cannot use the determinant method since this is not a square matrix. So I move on to the minor method. Since the given rank is $2$, determinant of any $x \times 3$ minor should be zero. So I take

$$\begin{vmatrix} 1 & 3 & -3\\ 2 & 2 & x \\ 1 & 1-x & 2x+1 \\ \end{vmatrix} =0.$$

My calculation yields $x=\pm 2$. However, if I substitute the values in the matrix, the rank comes out to be $3$. What am I doing wrong?

Answer 1

$\begin{pmatrix} 1 & 3 & -3 & x\\ 2 & 2 & x & -4 \\ 1 & 1-x & 2x+1 & -5-3x\\ \end{pmatrix}$ $\overset{R_2-2R_1\\R_3-R_1}\rightarrow\begin{pmatrix} 1 & 3 & -3 & x\\ 0 & -4 & x+6 & -4-2x \\ 0 & -2-x & 2x+4 & -5-4x\\ \end{pmatrix}$

$\overset{\frac{R_2}{4}}\rightarrow\begin{pmatrix} 1 & 3 & -3 & x\\ 0 & -1 & \frac{1}{4}(x+6)& -1-\frac{x}{2} \\ 0 & -2-x & 2x+4 & -5-4x\\ \end{pmatrix}$

$\overset{R_3-(x+2)R_2}\rightarrow\begin{pmatrix} 1 & 3 & -3 & x\\ 0 & -1 & \frac{1}{4}(x+6)& -1-\frac{x}{2} \\ 0 & 0 & \frac{-x^2}{4}+1 & \frac{x^2}{4}-\frac{5x}{4}-4\\ \end{pmatrix}$

To be a rank $2$ matrix, $3$rd row should be zero.But $\frac{-x^2}{4}+1=0\\ \frac{x^2}{4}-\frac{5x}{4}-4=0\tag{1}$

has no solution.

So the rank of the matrix will be $3$ and independent of the choice of $x$ .

Answer 2

There are several equivalent definitions of the "rank" of a matrix. If you wish to use determinants, use the definition:'The rank of a matrix $A$ is the largest $n$ such that $A$ contains an $n \times n$ sub-matrix whose determinant is non-zero.' Since your matrix contains 2x2 sub-matrices with non-zero determinant, the rank must be 2 or 3. Look at all the 3x3 sub-matrices. If you can find $x$ such that all 4 of these sub-matrices have determinant 0, then for that value of $x$ the rank of your matrix is 2. If there is no such $x$, the rank is 3.

Answer 3

Matlab yields this reduced echelon form $$ \text{rref}(A) = \begin{bmatrix} 1 & 0 & 0 & -\frac{-x^2+8\,x+3}{x-2}\\ 0 & 1 & 0 & \frac{13\,x+14}{x^2-4}\\ 0 & 0 & 1 & \frac{2\,\left(-x^2+4\,x+6\right)}{x^2-4} \end{bmatrix} $$ Rank 2 requires the last row to be zero, which is not possible. Notice the 1 in the third column in the last row. This cannot be eliminated. As a result, $\text{rank}(A)=3$.

syms x real
A=[1 3 -3 x;2 2 x -4;1 1-x 2*x+1 -5-3*x];
rref(A)