Question

Let $A=\begin{bmatrix} x^{2} & 6 & 9 \\ 3 & y^{2} & 9 \\ 4 & 5 & z^{2} \end{bmatrix},B=\begin{bmatrix} 2x & 3 & 5 \\ 2 & 2y & 6 \\ 1 & 4 & 2z-3 \end{bmatrix}$ if $traceA=traceB,$ then $x+y+z$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. None of those

Answer 1

Answer: (C)

Trace $A=x^{2}+y^{2}+z^{2}$
Trace $B=2x+2y+2z-3$
$x^{2}+y^{2}+z^{2}=2x+2y+2z-3$
$x^{2}+y^{2}+z^{2}-2x-2y-2z+3=0$
$\left(x - 1\right)^{2}+\left(y - 1\right)^{2}+\left(z - 1\right)^{2}=0$
It is true if $x=1,y=1,z=1$
So, $x+y+z=3$