Let A= x 1 1 0 , x ε R and A^4=a_ij. If a_11=109, then a_22 is equal to_____________.

Correct answer: 10. A^2= x 1 1 0 × x 1 1 0 = x^2+1 x x 1 A^4= x^2+1 x x 1 × x^2+1 x x 1 = x^4+3x^2+1 x^3+2x x^3+2x x^2+1 Given a_11=109⇒x^4+3x^2+1=109 x^2+12x^2-9=0⇒x^2=9 ∴a_22=x^2+1=10

Q46 / 75 · JEE Main 2020 (03 Sep Shift 1) · Mathematics

Let $A = [\begin{matrix} x & 1 \\ 1 & 0 \end{matrix}], x ε R$ and $A^{4} = [a_{i j}] .$ If $a_{11} = 109,$ then $a_{22}$ is equal to_____________.

Correct answer: 10

Solution

$A^{2} = [\begin{matrix} x & 1 \\ 1 & 0 \end{matrix}] \times [\begin{matrix} x & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} x^{2} + 1 & x \\ x & 1 \end{matrix}]$

$A^{4} = [\begin{matrix} x^{2} + 1 & x \\ x & 1 \end{matrix}] \times [\begin{matrix} x^{2} + 1 & x \\ x & 1 \end{matrix}] = [\begin{matrix} x^{4} + 3 x^{2} + 1 & x^{3} + 2 x \\ x^{3} + 2 x & x^{2} + 1 \end{matrix}]$

Given $a_{11} = 109 \Rightarrow x^{4} + 3 x^{2} + 1 = 109$

$(x^{2} + 12) (x^{2} - 9) = 0 \Rightarrow x^{2} = 9$

$∴ a_{22} = x^{2} + 1 = 10$

Hence, the correct answer is 10.

Let A=x110, x ε R and A4=aij. If a11=109, then a22 is equal to_____________.

Solution

Relevant Questions

Let $A = [\begin{matrix} x & 1 \\ 1 & 0 \end{matrix}], x ε R$ and $A^{4} = [a_{i j}] .$ If $a_{11} = 109,$ then $a_{22}$ is equal to_____________.