Question

Let 𝑀 = \(\begin{bmatrix} K & 1 & 1 \\ 1 & K & 1 \\ 1 & 1 & K \end{bmatrix}\)and I₃ be the identity matrix of order 3. If the rank of the matrix 10 𝐼₃ − 𝑀 is 2 then 𝑘 is equal to _____ (in integer).

Answer 1

Correct Answer: 8

Given:
\[ M=\begin{bmatrix}K&1&1\\[4pt]1&K&1\\[4pt]1&1&K\end{bmatrix},\qquad A:=10I_3-M. \]

 Step 1 — rewrite \(M\) using the all-ones matrix \(J\) 
Let \(J\) be the \(3\times3\) matrix of all ones. Then \[ M=(K-1)I_3+J, \] so \[ A=10I_3-M=10I_3-(K-1)I_3-J=(11-K)I_3-J. \] Put \(a:=11-K\). Then \(A=aI_3-J\). 

Step 2 — eigenvalues of \(J\)
For \(J\) (size \(3\)), the eigenvalues are: \[ \lambda_J=3\ (\text{once, eigenvector }(1,1,1)),\quad 0\ (\text{multiplicity }2). \] 

Step 3 — eigenvalues of \(A=aI_3-J\)
Eigenvalues of \(A\) are \(a-3\) (once) and \(a\) (twice). The rank of \(A\) equals the number of nonzero eigenvalues. We want \(\operatorname{rank}(A)=2\). That happens exactly when one eigenvalue is zero and the other two are nonzero. This requires \(a-3=0\) while \(a\neq0\). So \(a=3\). 

Step 4 — find \(K\)
\[ a=11-K=3 \quad\Rightarrow\quad K=11-3=8. \] 

Answer: \(\boxed{\,K=8\,}\)