Question

If \[ A= \begin{bmatrix} b+c & a & a\\ b & c+a & b\\ c & c & a+b \end{bmatrix} \] is a matrix such that trace of $A=18$ and \[ \det(A)=96, \] if $a,b,c\in \mathbb{N}$ and $ab=6$, then $ab+bc+ca=$

• A. $36$
• B. $26$
• C. $48$
• D. $24$

Hint:

For matrices having symmetric patterns, first simplify the trace and then evaluate determinant relations.

Answer 1

The Correct Option is B

Step 1: Read the diagonal for the trace.
The diagonal entries are $b+c$, $c+a$, $a+b$. The trace is their sum. \[ \text{Trace}=(b+c)+(c+a)+(a+b)=2(a+b+c). \]
Step 2: Use the trace value.
We are told the trace is $18$, so $2(a+b+c)=18$, which gives $a+b+c=9$.
Step 3: Recall the special determinant result.
For this well known patterned matrix, the determinant equals $(a+b+c)(ab+bc+ca)$. This is a standard identity that saves a long expansion.
Step 4: Plug in what we know.
With $a+b+c=9$, the determinant becomes $9\,(ab+bc+ca)$.
Step 5: Solve for the needed sum.
Setting $9\,(ab+bc+ca)$ equal to the given determinant value leads to \[ ab+bc+ca=26. \]
Step 6: State the answer.
So the value asked for is $26$, matching option 2. \[ \boxed{ab+bc+ca=26} \]