Question

If $k \in R$, then det$\{adj(kI_n)\}$ is equal to

Answer 1

k^{n(n-1)}

Answer 2

To find $\det(\text{adj}(kI_n))$, we use the properties of determinants and adjoint matrices.

Let $A = kI_n$. Here, $I_n$ is the $n \times n$ identity matrix, and $k \in R$ is a scalar.

Step 1: Calculate the determinant of $A$. The matrix $A = kI_n$ is a diagonal matrix with $k$ along the main diagonal: $A = \begin{pmatrix} k & 0 & \dots & 0 \\ 0 & k & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & k \end{pmatrix}_{n \times n}$ The determinant of $A$ is given by: $\det(A) = \det(kI_n)$ Using the property that for a scalar $c$ and an $n \times n$ matrix $M$, $\det(cM) = c^n \det(M)$: $\det(kI_n) = k^n \det(I_n)$ Since the determinant of an identity matrix is 1 ($\det(I_n) = 1$): $\det(A) = k^n \cdot 1 = k^n$

Step 2: Use the property of the determinant of an adjoint matrix. For any $n \times n$ matrix $M$, the determinant of its adjoint is given by the formula: $\det(\text{adj}(M)) = (\det(M))^{n-1}$ This formula is generally valid for invertible matrices (i.e., $\det(M) \neq 0$).

Step 3: Substitute $\det(A)$ into the formula. Substitute $\det(A) = k^n$ into the formula for $\det(\text{adj}(A))$: $\det(\text{adj}(kI_n)) = (\det(kI_n))^{n-1} = (k^n)^{n-1}$ Using the exponent rule $(a^b)^c = a^{bc}$: $\det(\text{adj}(kI_n)) = k^{n(n-1)}$

Step 4: Verify for the special case when $k=0$. The formula $\det(\text{adj}(M)) = (\det(M))^{n-1}$ is typically derived assuming $\det(M) \neq 0$. Let's check if our result holds when $k=0$.

If $k=0$, then $A = 0 \cdot I_n = O_n$ (the $n \times n$ zero matrix). We need to find $\det(\text{adj}(O_n))$.

• Case 1: $n=1$ $A = [0]$. The adjoint of a $1 \times 1$ matrix $[a]$ is defined as $[1]$. So, $\text{adj}([0]) = [1]$. $\det(\text{adj}([0])) = \det([1]) = 1$. Our formula $k^{n(n-1)}$ gives $0^{1(1-1)} = 0^0$. If we adopt the convention $0^0=1$ (which is common in such contexts), the formula holds.

• Case 2: $n > 1$ $A = O_n$. The adjoint matrix $\text{adj}(O_n)$ is the transpose of the cofactor matrix. For $n > 1$, any $(n-1) \times (n-1)$ submatrix of $O_n$ will be a zero matrix of size $(n-1) \times (n-1)$. The determinant of such a submatrix is 0. Therefore, all minors $M_{ij}$ are 0, which implies all cofactors $C_{ij} = (-1)^{i+j} M_{ij}$ are also 0. Thus, $\text{adj}(O_n) = O_n$. $\det(\text{adj}(O_n)) = \det(O_n) = 0$. Our formula $k^{n(n-1)}$ gives $0^{n(n-1)}$. Since $n > 1$, $n(n-1)$ is a positive integer (e.g., for $n=2$, $n(n-1)=2$; for $n=3$, $n(n-1)=6$). Any positive integer power of 0 is 0. So, $0^{n(n-1)} = 0$. The formula holds for $n>1$ and $k=0$.

Since the formula $k^{n(n-1)}$ holds for all $k \in R$ and $n \in N$ (with the standard convention $0^0=1$), it is the general solution.