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 $ab+bc+ca=$
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 $ab+bc+ca=$
Show Hint
For matrices having symmetric patterns, first simplify the trace and then evaluate determinant relations.
Updated On: Jun 17, 2026
- $36$
- $26$
- $48$
- $24$
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The Correct Option is B
Solution and Explanation
Concept:
For a square matrix:
• Trace of a matrix is the sum of diagonal elements.
• Determinant properties help simplify structured matrices.
• Relations between variables can be obtained using trace and determinant conditions together.
Step 1: Use the trace condition.
The diagonal elements are: \[ b+c,\quad c+a,\quad a+b. \] Therefore, \[ \text{Trace}(A)=(b+c)+(c+a)+(a+b). \] \[ =2(a+b+c). \] Given: \[ \text{Trace}(A)=18. \] Hence, \[ 2(a+b+c)=18. \] \[ a+b+c=9. \]
Step 2: Use determinant property.
For the given structured matrix: \[ \det(A)=(a+b+c)(ab+bc+ca). \] Since: \[ a+b+c=9 \] and \[ \det(A)=234, \] we get: \[ 9(ab+bc+ca)=234. \] Therefore, \[ ab+bc+ca=26. \] Hence, \[ \boxed{ab+bc+ca=26} \]
• Trace of a matrix is the sum of diagonal elements.
• Determinant properties help simplify structured matrices.
• Relations between variables can be obtained using trace and determinant conditions together.
Step 1: Use the trace condition.
The diagonal elements are: \[ b+c,\quad c+a,\quad a+b. \] Therefore, \[ \text{Trace}(A)=(b+c)+(c+a)+(a+b). \] \[ =2(a+b+c). \] Given: \[ \text{Trace}(A)=18. \] Hence, \[ 2(a+b+c)=18. \] \[ a+b+c=9. \]
Step 2: Use determinant property.
For the given structured matrix: \[ \det(A)=(a+b+c)(ab+bc+ca). \] Since: \[ a+b+c=9 \] and \[ \det(A)=234, \] we get: \[ 9(ab+bc+ca)=234. \] Therefore, \[ ab+bc+ca=26. \] Hence, \[ \boxed{ab+bc+ca=26} \]
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