If A=[1 -2 4 5] and f(t)=t2-3 t+7, then f(A)+[3 6 -12 -9] is equal to

Question

If A=[1 -2 4 5] and f(t)=t2-3 t+7, then f(A)+[3 6 -12 -9] is equal to (A) [1 0 0 1] (B) [0 0 0 0] (C) [0 1 1 0] (D) [1 1 0 0]

Tardigrade Admin · Accepted Answer

Given that A=[1 -2 4 5] and f(t)=t2-3 t+7, then Now ,A2=[1 -2 4 5]A=[1 -2 4 5] =[-7 -12 24 17] Now, f(A)=A2-3 A+7 =[-7 -12 24 17]-3[1 -2 4 5]+7[1 0 0 1] =[-3 -6 12 9] ∴ f(A)+[3 6 -12 -9]=[-3 -6 12 9]+[3 6 -12 -9] =[0 0 0 0]

Q. If $A = [14 - 2 5]$ and $f (t) = t^{2} - 3 t + 7$ , then $f (A) + [3 - 12 6 - 9]$ is equal to

$[1001]$

$[0000]$

$[0110]$

$[1010]$

Given that $A = [14 - 2 5]$ and $f (t) = t^{2} - 3 t + 7$ , then
Now , $A^{2} = [14 - 2 5] A = [14 - 2 5]$
$= [- 7 24 - 12 17]$
Now, $f (A) = A^{2} - 3 A + 7$
$= [- 7 24 - 12 17] - 3 [14 - 2 5] + 7 [1001]$
$= [- 3 12 - 6 9]$
$∴ f (A) + [3 - 12 6 - 9] = [- 3 12 - 6 9] + [3 - 12 6 - 9]$
$= [0000]$

Q. If A=[14​−25​] and f(t)=t2−3t+7, then f(A)+[3−12​6−9​] is equal to

[10​01​]

[00​00​]

[01​10​]

[10​10​]