Finding x in matrix algebra

Question

The question asked me to calculate $|A|$ for the given matrix below and state whether or not it is invertible.

$$ \left( \begin{matrix} 1 & 0 & 2 & 1\\ 0 & x & x & x\\ x & x & 3x & 1\\ 0 & x & x & 0\\ \end{matrix} \right) $$

I got my answer to be $-2x^3 + 2x^3 = 0$, indicating that the matrix is not invertible because the determinant is a zero, regardless of the value of $x$.

However, I'm not certain whether I'm right after doing all the calculations. Could you help me verify?

Answer 1

Okay, here's every step:

$$\left|\begin{matrix} x & x & x \\ x & 3x & 1 \\ x & x & 0\end{matrix}\right|+2\left|\begin{matrix} 0 & x & x \\ x & x & 1 \\ 0 & x & 0\end{matrix}\right| - \left|\begin{matrix} 0 & x & x \\ x & x & 3x \\ 0 & x & x\end{matrix}\right| \\ = x^2\left|\begin{matrix} 1 & 1 & x \\ 1 & 3 & 1 \\ 1 & 1 & 0\end{matrix}\right|+2x^2\left|\begin{matrix} 0 & 1 & x \\ 1 & 1 & 1 \\ 0 & 1 & 0\end{matrix}\right| - x^3\left|\begin{matrix} 0 & 1 & 1 \\ 1 & 1 & 3 \\ 0 & 1 & 1\end{matrix}\right| \\ = x^2\left(\left|\begin{matrix} 3 & 1 \\ 1 & 0\end{matrix}\right| - \left|\begin{matrix} 1 & 1 \\ 1 & 0\end{matrix}\right| + x\left|\begin{matrix} 1 & 3 \\ 1 & 1\end{matrix}\right|\right) + 2x^2\left(- \left|\begin{matrix} 1 & 1 \\ 0 & 0\end{matrix}\right| + x\left|\begin{matrix} 1 & 1 \\ 0 & 1\end{matrix}\right|\right) -x^3\left(- \left|\begin{matrix} 1 & 3 \\ 0 & 1\end{matrix}\right| + \left|\begin{matrix} 1 & 1 \\ 0 & 1\end{matrix}\right|\right) \\ = -2x^3+2x^3-0 \\ =0$$

So you did it correctly.

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Or, or course, for the easier way, just recognize that twice the first column plus the second column equals the third column of your matrix so the determinant must be $0$.

Answer 2

Recall that performing row replacement operations does not affect the determinant. Hence, if we replace row $3$ with (row $3$ + $(-x) \cdot $ row $1$), then replace row $3$ with (row $3$ - row$2$) and replace row $4$ with (row $4$ - row $2$), we obtain: $$ \begin{vmatrix} 1 & 0 & 2 & 1\\ 0 & x & x & x\\ x & x & 3x & 1\\ 0 & x & x & 0\\ \end{vmatrix} = \begin{vmatrix} 1 & 0 & 2 & 1\\ 0 & x & x & x\\ 0 & x & x & 1-x\\ 0 & x & x & 0\\ \end{vmatrix} = \begin{vmatrix} 1 & 0 & 2 & 1\\ 0 & x & x & x\\ 0 & 0 & 0 & 1-2x\\ 0 & 0 & 0 & -x\\ \end{vmatrix} $$ but since this last matrix is upper triangular, its determinant is the product of the diagonal entries. But one of the diagonal entries is a zero, so $\det A = 0$, as you computed.

Answer 3

$A$ has zero determinant.

If you row reduce based on the first row first column, then the 2nd and 3rd columns of the new matrix become $$ \begin{array}{cc} x&x \\ x&x \\x&x \end{array} $$ Then when you divide the second row by $x$ and row reduce based on the 2nd row 2nd column you get the matrix $$\left( \begin{array}{cccc} 1&0&2&1 \\ 0&1&1&1 \\ 0&0&0&1-2x \\ 0&0&0&-x \end{array} \right) $$ The last two rows of this matrix are clearly multiples of one another, thus linearly depended, so the determinant will be zero.