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Matrices and Determinants question

2020 · 3 Sep · Shift 1 · Q41
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Matrices and Determinants question

2020 · 3 Sep · Shift 1 · Q41

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
Let A = [x110]\left[ {\begin{matrix} x & 1 \\ 1 & 0 \\ \end{matrix} } \right][x1​10​], x ∈\in∈ R and A4 = [aij]. If a11 = 109, then a22 is equal to ‾\underline{\hspace{2cm}}​ .
Numerical answer
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Correct answer: 10

  1. We are given A=[x110]A=\begin{bmatrix}x&1\\1&0\end{bmatrix}A=[x1​10​] and A4=[aij]A^4=[a_{ij}]A4=[aij​] with a11=109.a_{11}=109.a11​=109. We need to find a22a_{22}a22​.

  2. First compute powers of AAA.

\begin{bmatrix}x^2+1&x\\x&1\end{bmatrix}$$ 3. Now compute $A^4=(A^2)^2$. Let $$A^2=\begin{bmatrix}x^2+1&x\\x&1\end{bmatrix}.$$ Then $$A^4=\begin{bmatrix}x^2+1&x\\x&1\end{bmatrix} \begin{bmatrix}x^2+1&x\\x&1\end{bmatrix}.$$ Now multiply: - Entry $(1,1)$: $$a_{11}=(x^2+1)^2+x^2=x^4+3x^2+1$$ - Entry $(2,2)$: $$a_{22}=x^2+1$$ 4. Given $a_{11}=109$, so $$x^4+3x^2+1=109$$ $$x^4+3x^2-108=0$$ Let $y=x^2$. Then $$y^2+3y-108=0$$ $$ (y+12)(y-9)=0 $$ So $$y=9\quad\text{or}\quad y=-12.$$ But $y=x^2\ge 0$, hence $$x^2=9.$$ 5. Therefore, $$a_{22}=x^2+1=9+1=10.$$ So the required integer is $$\boxed{10}$$
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