Determinant of 4x4 w/ all entries unknowns

Question

Any help with this problem would be greatly appreciated.

If $A$ is the matrix $$ A = \begin{bmatrix} a & b & c & d+1\\ a & b & c+1 & d \\ a & b+1 & c & d \\ a+1 & b & c & d \end{bmatrix} $$ find the determinant of $B = 3A^5 A^t A^{-1}.$

I see that $\det\left( A^t A^{-1} \right) = 1$ but I'm having difficulty computing $\det\left(3A^5\right)$. Laplace expansion seems like it will be too time consuming as this is an exam review question(I don't think we would be expected to do Laplace expansion with a 4x4 matrix on an exam). And my apologies for the link. The site won't let me post images as I am new here. Thanks!

Answer 1

HINT

• The rule is $\det(AB) = \det(A) \cdot \det(B)$.Hence, $\det(A^2) = \det(A)^2$, for example.
• Multiplying every entry in a matrix by a constant is equivalent to multiplying the matrix by a diagonal matrix, with all entries equal to that constant, i.e. $5A = (5I)A$.

Can you finish now?

Answer 2

Essentially, we only need to calculate $\det{A}$. For the first 3 rows, substract the next one. The determinant is unchanged and we get $$\begin{pmatrix} 0 & 0 & -1 & 1 \\ 0 & -1 & 1 & 0 \\ -1 & 1 & 0 & 0 \\ a+1 & b & c & d \end{pmatrix}.$$

Then expand the last column. And we get \begin{align*} \det{A} &= d\cdot(1) +(-1)\cdot \begin{vmatrix} 0 & -1 & 1 \\ -1 & 1 & 0 \\ a+1 & b & c \end{vmatrix}\\ &=d-\left( c\cdot (-1) + \begin{vmatrix} -1 & 1 \\ a+1& b \end{vmatrix} \right)\\ &= d +c -(-b - (a+1))\\ &=a+b+c+d+1. \end{align*} For the determinant of $3A^5 A^\top A^{-1}$ just use that $\det$ is multiplicative. That is, $\det (3A^5 A^\top A^{-1})=3^4 \det (A^5)\det (A^\top)\det (A^{-1})=3^4 (a+b+c+d+1)^5$.

Answer 3

Flipping the matrix left and right, the determinant remains unchanged (because the matrix has an even number of columns) but the matrix becomes $I+ev^T$, where $e=(1,1,1,1)^T$ and $v=(d,c,b,a)^T$. Now you may use Sylvester's determinant theorem to obtain $\det(A)=\det(I+ev^T)=1+v^Te=1+a+b+c+d$.