Let a -- 2b + c = 1.

Question

If $$f(x)=\left| {\begin{pmatrix} {x + a} & {x + 2} & {x + 1}  \
   {x + b} & {x + 3} & {x + 2}  \
   {x + c} & {x + 4} & {x + 3}  \ \end{pmatrix} } \right|$$, then:

Accepted Answer

R1 $$ \to $$ R1 + R3 -- 2R2

f(x) = $$\left| {\begin{pmatrix} {a + c - 2b} & 0 & 0  \
   {x + b} & {x + 3} & {x + 2}  \
   {x + c} & {x + 4} & {x + 3}  \

\end{pmatrix} } \right|$$

= (a + c -- 2b) ((x + 3)2 -- (x + 2)(x + 4))

= x2 + 6x + 9 -- x2 -- 6x -- 8 = 1

$$ \therefore $$ f(x) = 1

$$ \Rightarrow $$  f(50) = 1

Solutions · 1