How to solve this? Advice?

Question

Edit: My original problem includes a matrix, I'm not sure how to format it here so it'll be a bit messy. I was asked to find det(I+M) and someone told me to add all the columns to the last row? Didn't really get what they meant. R1 -> | a+1 b c d | R2 -> | a b+1 c d | R3 -> | a b c+1 d | R4 -> | a b c d+1|

Answer 1

Alternatively, given: $$M=\begin{pmatrix}a&b&c&d\\a&b&c&d\\a&b&c&d\\a&b&c&d\end{pmatrix},\\ $$ the required determinant is: $$\det(I+M)=\begin{vmatrix}a+1&b&c&d\\a&b+1&c&d\\a&b&c+1&d\\a&b&c&d+1\end{vmatrix}\stackrel{C_1+C_2+C_3+C_4\to C_1}{=}\\ \begin{vmatrix}a+b+c+d+1&b&c&d\\a+b+c+d+1&b+1&c&d\\a+b+c+d+1&b&c+1&d\\a+b+c+d+1&b&c&d+1\end{vmatrix}=\\ (a+b+c+d+1)\begin{vmatrix}1&b&c&d\\1&b+1&c&d\\1&b&c+1&d\\1&b&c&d+1\end{vmatrix}\stackrel{-R_1+R_2\to R_2;\\-R_1+R_3\to R_3;\\-R_1+R_4\to R_4}{=}\\ (a+b+c+d+1)\begin{vmatrix}1&b&c&d\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{vmatrix}=a+b+c+d+1.$$ Note: This is what "someone" suggested to you.

Answer 2

Let $M= ev^T$ where $v=\begin{bmatrix}a \\ b\\ c \\ d \end{bmatrix}$ and $e$ is the all-one vector.

By matrix determinant lemma, we have $$\det(I+M)=\det(I+ev^T)=(1+v^TI^{-1}e)\det(I)=1+a+b+c+d$$

As a practice, perhaps try to perform elementary row/ column operations and see if you can get the same answer. (most likely this is the expected approach and you should try it.)

Remark:

• Adding all the columns to the last row doesn't make much sense to me.
• Perhaps they mean adding all the columns to the last column.