Matrix Algebra Determinant: What's Wrong Here?

Question

Trying to find det(ATA), and I get 313110. But, the online calculator says the answer is 4900. I can't see where I've gone wrong in my working, can anyone help?

A:

Answer 1

Expanding on my answer in the comments:

An easier way of doing this might be to observe $$\det(A^T A) = \det(A^T) \det(A)$$ since determinant is multiplicative, and then that $$\det(A^T) = \det(A)$$ since the determinant of a transpose is exactly the determinant. So, $\det(A^T A) = \det(A)^2$, which could well be easier to calculate.

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The issue with your attempt is that your "row operations" weren't valid row operations. In particular, $R_1 \to 3R_1 + 22R_3$ isn't allowed, but $R_1 \to R_1 + \frac{22}{3} R_3$ is. You can add a scalar multiple of one row to another, but you can't just scale up a row. In fact, the determinant on your second line is $27$ times the determinant on the first line.

The other issue was mentioned already by grand_chat, that you wrote a $2074$ where you meant $2072$. Other than these two mistakes, everything was correct.

However, the best method for solving this is undoubtedly to simplify the expression first, as I outlined above. This avoids arithmetic with large numbers where copying errors are more likely.

Answer 2

$A^TA$ appears correct. One error: Changing $2072$ to $2074$.

The other errors occur when you do row operations. For example, in your answer you replace $R1$ by $3R1 + 22R3$. If you replace $R1$ by $R1 + \frac{22}3R3$ (i.e., if you add a multiple of another row to $R1$), then the determinant is unchanged; but the move you've made requires that you carry a factor of $\frac13$ in front. Similarly if you replace $R2$ as you've done, you need to carry another factor of $\frac13$. In the end you'll need to adjust your answer by $\frac1{27}$ to obtain $4900$ as the determinant.

Answer 3

Let use two fact

• $\det(A^T A) = \det(A)^2$
• det(A) doesn't change adding a scalar multiple of one column/row to another column/row, thus the expression can be simplified

Notably

$$det\begin{bmatrix}1&3&2&4\\1&-1&1&-3\\-2&3&7&-1\\4&6&2&8 \end{bmatrix}=det\begin{bmatrix}1&3&2&4\\0&-4&-1&-7\\-2&3&7&-1\\0&12&16&6 \end{bmatrix}=det\begin{bmatrix}1&3&2&4\\0&-4&-1&-7\\0&9&11&7\\0&12&16&6 \end{bmatrix}$$

then use Laplace on first column and

$$detA=det\begin{bmatrix}-4&-1&-7\\9&11&7\\12&16&6 \end{bmatrix}=70$$

Answer 4

The determinant is not preserved when you do those transformation. Precisely, you're multiplying the determinant by $3^3$.

Indeed, if I ask Pari-GP to compute the determinant of $$ \begin{bmatrix} 896&1267&413\\ 925&1274&460\\ 1510&2072&760 \end{bmatrix} $$ the answer is $-44100$. And $$ \frac{(-3)(-44100)}{3^3}=4900 $$

As noted in comments you have mistakenly written $2074$ instead of $2072$.