If A is a 2× 2 matrix such that A[ 1 -1 ]=[ -1 2 ] and A2[ 1 -1 ]=[ 1 0 ] , then trace of A is (where, the trace of the matrix is the sum of all principal diagonal elements of the matrix)

Question

If A is a 2× 2 matrix such that A[ 1 -1 ]=[ -1 2 ] and A2[ 1 -1 ]=[ 1 0 ] , then trace of A is (where, the trace of the matrix is the sum of all principal diagonal elements of the matrix) (A) 1 (B) 0 (C) 2 (D) 5

Tardigrade Admin · Accepted Answer

Let, A=[ a b c d ] A[ 1 -1 ]=[ -1 2 ]⇒ [ a b c d ][ 1 -1 ]=[ -1 2 ] On solving, we get, a-b=-1, c-d=2 Also, A2[ 1 -1 ]=A[ A [ 1 -1 ] ]=A[ -1 2 ]=[ 1 0 ] This gives, -a+2b=1, -c+2d=0 ⇒ b=0, a=-1, d=2, c=4

Q. If $A$ is a $2 \times 2$ matrix such that $A [1 - 1] = [- 1 2]$ and $A^{2} [1 - 1] = [10]$ , then trace of $A$ is (where, the trace of the matrix is the sum of all principal diagonal elements of the matrix)

$1$

$0$

$2$

$5$

Let, $A = [a c b d]$
$A [1 - 1] = [- 1 2] \Rightarrow [a c b d] [1 - 1] = [- 1 2]$
On solving, we get,
$a - b = - 1, c - d = 2$
Also,
$A^{2} [1 - 1] = A [A [1 - 1]] = A [- 1 2] = [10]$
This gives, $- a + 2 b = 1, - c + 2 d = 0$
$\Rightarrow b = 0, a = - 1, d = 2, c = 4$

Q. If A is a 2×2 matrix such that A[1−1​]=[−12​] and A2[1−1​]=[10​] , then trace of A is (where, the trace of the matrix is the sum of all principal diagonal elements of the matrix)

1

0

2

5

Let, A=[ac​bd​] A[1−1​]=[−12​]⇒[ac​bd​][1−1​]=[−12​] On solving, we get, a−b=−1,c−d=2 Also, A2[1−1​]=A[A[1−1​]​]=A[−12​]=[10​] This gives, −a+2b=1,−c+2d=0 ⇒b=0,a=−1,d=2,c=4

Q. If $A$ is a $2 \times 2$ matrix such that $A [1 - 1] = [- 1 2]$ and $A^{2} [1 - 1] = [10]$ , then trace of $A$ is (where, the trace of the matrix is the sum of all principal diagonal elements of the matrix)

$1$

$0$

$2$

$5$

Let, $A = [a c b d]$
$A [1 - 1] = [- 1 2] \Rightarrow [a c b d] [1 - 1] = [- 1 2]$
On solving, we get,
$a - b = - 1, c - d = 2$
Also,
$A^{2} [1 - 1] = A [A [1 - 1]] = A [- 1 2] = [10]$
This gives, $- a + 2 b = 1, - c + 2 d = 0$
$\Rightarrow b = 0, a = - 1, d = 2, c = 4$