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If

$$\det \underbrace{\begin{bmatrix} a&b\\ c&d\end{bmatrix}}_{=: A} = -3$$

calculate the determinant

$$\det \underbrace{\begin{bmatrix} 2&-2&0\\ c+1&-1&2a\\ d-2&2&2b\end{bmatrix}}_{=: B}$$

I have no idea how to approach this problem. I can see that the matrix $A$ is contained in $B$ and transposed with a row switch.

$$\begin{bmatrix}2&-2&0\\c+1&-1&2a\\d-2&2&2b\end{bmatrix}^T = \begin{bmatrix}2&c+1&d-2\\-2&-1&2\\0&2a&2b\end{bmatrix}$$

Then after some row switches: $R_2 \leftrightarrow R_3$, then $R_1 \leftrightarrow R_3$

$$\begin{bmatrix}-2&-1&2\\0&2a&2b\\2&c+1&d-2\end{bmatrix}$$

And with row operation $R_3+R_1$..

$$\begin{bmatrix}-2&-1&2\\0&2a&2b\\0&c&d\end{bmatrix}$$

I can calculate the determinant as $(-2) \cdot (-3) \cdot (2) \cdot (-1)^2$ using cofactor expansion and the properties of determinants. The determinant of the transposed matrix is equal to the determinant of the matrix, the row switches are accounted for by the $(-1)^2$, the row multiplied by 2 is accounted for by $(2)$.

So my answer is 12. Please help me if there's a mistake. Thank you!

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    $\begingroup$ I think it's funny that you say you have "no idea" how to approach the problem, then proceed to give a solution. $\endgroup$ – Cheerful Parsnip May 24 '18 at 0:11
  • $\begingroup$ As I was typing the question, I came up with a "solution" but I'm not confident that it's right.. $\endgroup$ – Shawn S May 24 '18 at 0:13
  • $\begingroup$ @V K Your answer looks right to me, and seems to agree with @Bernard's answer below, since $bc-ad=-\det(A)$. $\endgroup$ – Cheerful Parsnip May 24 '18 at 0:19
  • $\begingroup$ I see that now. Thanks $\endgroup$ – Shawn S May 24 '18 at 0:22
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A determinant is an alternate multilinear function of its columns (and of its rows as well). So \begin{alignat}{2} \begin{vmatrix}2&-2&0\\c+1&-1&2a\\d-2&2&2b\end{vmatrix}&=\begin{vmatrix}0&-2&0\\c &-1&2a\\d &2&2b\end{vmatrix} +\underbrace{\begin{vmatrix}2&-2&0\\1&-1&2a\\-2&2&2b\end{vmatrix}}_{=0 \text{ since columns 1}\\\text{and 2 are collinear}}\\ &=2\begin{vmatrix}0&-2&0\\c &-1&a\\d &2&b\end{vmatrix}=2\cdot 2\begin{vmatrix}c &a\\d&b\end{vmatrix}&\quad&\text{(expanding by the 1st row)}\\&=4(bc-ad). \end{alignat}

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  • $\begingroup$ Since the rows in the minor are switched, I should negate the determinant of matrix A? $\endgroup$ – Shawn S May 24 '18 at 0:21
  • $\begingroup$ Yes, taking into account the determinant of a matrix and the determinant of the transpose are equal. $\endgroup$ – Bernard May 24 '18 at 0:24
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Add the second column to the first column, the determinant remains the same. Then in the first line there is only one entry $\ne 0$. One can also "factorize" a $2$ in the third column. This leads to a quick answer...

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