Let A= 1 -1 2 -1 and B= a 1 b -1 are two matrices. If A+B^2=A^2+B^2, then the value of 3a+4b is equal to

Correct answer: (C) 19. Given, A+B^2=A^2+B^2 A^2+B^2+AB+BA=A^2+B^2⇒AB+BA=O AB= 1 -1 2 -1 a 1 b -1 = a-b 2 2a-b 3 BA= a 1 b -1 1 -1 2 -1 = a+2 -a-1 b-2 -b+1 AB+BA= 2a-b+2 -a+1 2a-2 -b+4 =O 2a-b+2=0 -a+1=0⇒a=1 2a-2=0⇒a=1 -b+4=0⇒b=4

Q30 / 75 · NTA Abhyas (JEE Main) 2020 · Mathematics · Matrices

Let $A = [\begin{array}{l} 1 & - 1 \\ 2 & - 1 \end{array}]$ and $B = [\begin{array}{l} a & 1 \\ b & - 1 \end{array}]$ are two matrices. If ${(A + B)}^{2} = A^{2} + B^{2}$ , then the value of $3 a + 4 b$ is equal to

A
$15$
B
$17$
Correct answer
C
$19$
D
$21$

Solution

Given,

{(A + B)}^{2} = A^{2} + B^{2}

A^{2} + B^{2} + A B + B A = A^{2} + B^{2} \Rightarrow A B + B A = O

A B = [\begin{array}{l} 1 & - 1 \\ 2 & - 1 \end{array}] [\begin{array}{l} a & 1 \\ b & - 1 \end{array}] = [\begin{array}{l} a - b & 2 \\ 2 a - b & 3 \end{array}]

B A = [\begin{array}{l} a & 1 \\ b & - 1 \end{array}] [\begin{array}{l} 1 & - 1 \\ 2 & - 1 \end{array}] = [\begin{array}{l} a + 2 & - a - 1 \\ b - 2 & - b + 1 \end{array}]

A B + B A = [\begin{array}{l} 2 a - b + 2 & - a + 1 \\ 2 a - 2 & - b + 4 \end{array}] = O

2 a - b + 2 = 0

- a + 1 = 0 \Rightarrow a = 1

2 a - 2 = 0 \Rightarrow a = 1

- b + 4 = 0 \Rightarrow b = 4

Hence, the correct answer is option: 'C'.

Let A=1-12-1 and B=a 1b-1 are two matrices. If A+B2=A2+B2, then the value of 3a+4b is equal to

Solution

More Matrices Questions

Let $A = [\begin{array}{l} 1 & - 1 \\ 2 & - 1 \end{array}]$ and $B = [\begin{array}{l} a & 1 \\ b & - 1 \end{array}]$ are two matrices. If ${(A + B)}^{2} = A^{2} + B^{2}$ , then the value of $3 a + 4 b$ is equal to