Let A+2B=[ 2 4 0 6 -3 3 -5 3 5 ] and 2A-B=[ 6 -2 4 6 1 5 6 3 4 ] , then tr(A)-tr(B) is equal to (where, tr(A)= trace of matrix A i.e. sum of the principal diagonal elements of matrix A )

Question

Let A+2B=[ 2 4 0 6 -3 3 -5 3 5 ] and 2A-B=[ 6 -2 4 6 1 5 6 3 4 ] , then tr(A)-tr(B) is equal to (where, tr(A)= trace of matrix A i.e. sum of the principal diagonal elements of matrix A ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let, x=[ 2 4 0 6 -3 3 -5 3 5 ] y=[ 6 -2 4 6 1 5 2 3 4 ] A+2B=x 2A-B=y A=(1/5)(x + 2 y),B=(1/5)(2 x - y) A-B=(1/5)(- x + 3 y)=(1/5)[ 16 -10 12 12 6 12 1 6 7 ] tr(A)-tr(B)=tr(A - B) =(29/5)

Q. Let $A + 2 B = ⎣ ⎡ 26 - 5 4 - 3 3 035 ⎦ ⎤$ and $2 A - B = ⎣ ⎡ 666 - 2 13 454 ⎦ ⎤$ , then $t r (A) - t r (B)$ is equal to

(where, $t r (A) =$ trace of matrix $A$ i.e. sum of the principal diagonal elements of matrix $A$ )

Let, $x = ⎣ ⎡ 26 - 5 4 - 3 3 035 ⎦ ⎤$
$y = ⎣ ⎡ 662 - 2 13 454 ⎦ ⎤$
$A + 2 B = x$
$2 A - B = y$
$A = \frac{1}{5} (x + 2 y), B = \frac{1}{5} (2 x - y)$
$A - B = \frac{1}{5} (- x + 3 y) = \frac{1}{5} ⎣ ⎡ 16121 - 10 66 12127 ⎦ ⎤$
$t r (A) - t r (B) = t r (A - B)$
$= \frac{29}{5}$

Q. Let A+2B=⎣⎡​26−5​4−33​035​⎦⎤​ and 2A−B=⎣⎡​666​−213​454​⎦⎤​ , then tr(A)−tr(B) is equal to (where, tr(A)= trace of matrix A i.e. sum of the principal diagonal elements of matrix A )

Let, x=⎣⎡​26−5​4−33​035​⎦⎤​ y=⎣⎡​662​−213​454​⎦⎤​ A+2B=x 2A−B=y A=51​(x+2y),B=51​(2x−y) A−B=51​(−x+3y)=51​⎣⎡​16121​−1066​12127​⎦⎤​ tr(A)−tr(B)=tr(A−B) =529​

Q. Let $A + 2 B = ⎣ ⎡ 26 - 5 4 - 3 3 035 ⎦ ⎤$ and $2 A - B = ⎣ ⎡ 666 - 2 13 454 ⎦ ⎤$ , then $t r (A) - t r (B)$ is equal to

(where, $t r (A) =$ trace of matrix $A$ i.e. sum of the principal diagonal elements of matrix $A$ )