# Given a determinant of a 2x2 matrix, calculate the determinant of 3x3 matrix?

If the determinant of a matrix A, $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ = -3, calculate the determinant of matrix B, $\begin{bmatrix}2&-2&0\\c+1&-1&2a\\d-2&2&2b\end{bmatrix}$.

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 $(2)$.

• I think it's funny that you say you have "no idea" how to approach the problem, then proceed to give a solution. – Cheerful Parsnip May 24 '18 at 0:11
• As I was typing the question, I came up with a "solution" but I'm not confident that it's right.. – Shawn S May 24 '18 at 0:13
• @V K Your answer looks right to me, and seems to agree with @Bernard's answer below, since $bc-ad=-\det(A)$. – Cheerful Parsnip May 24 '18 at 0:19
• I see that now. Thanks – Shawn S May 24 '18 at 0:22

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}
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...