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Matrices and Determinants question

2021 · 25 Feb · Shift 1 · Q40
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Matrices and Determinants question

2021 · 25 Feb · Shift 1 · Q40

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
If A=[0−tan⁡(θ2)tan⁡(θ2)0]A = \left[ {\begin{matrix} 0 & { - \tan \left( {{\theta \over 2}} \right)} \\ {\tan \left( {{\theta \over 2}} \right)} & 0 \\ \end{matrix} } \right]A=[0tan(2θ​)​−tan(2θ​)0​] and (I2+A)(I2−A)−1=[a−bba]({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\begin{matrix} a & { - b} \\ b & a \\ \end{matrix} } \right](I2​+A)(I2​−A)−1=[ab​−ba​], then 13(a2+b2)13({a^2} + {b^2})13(a2+b2) is equal to
Numerical answer
View written solutionFree

Correct answer: 13

  1. Let t=tan⁡(θ2).t=\tan\left(\frac{\theta}{2}\right).t=tan(2θ​). Then A=[0−tt0].A=\begin{bmatrix}0&-t\\ t&0\end{bmatrix}.A=[0t​−t0​].

  2. Compute I2+AI_2+AI2​+A and I2−AI_2-AI2​−A: I2+A=[1−tt1],I2−A=[1t−t1].I_2+A=\begin{bmatrix}1&-t\\ t&1\end{bmatrix},\qquad I_2-A=\begin{bmatrix}1&t\\ -t&1\end{bmatrix}.I2​+A=[1t​−t1​],I2​−A=[1−t​t1​].

  3. Find (I2−A)−1(I_2-A)^{-1}(I2​−A)−1. For a matrix [1t−t1],\begin{bmatrix}1&t\\ -t&1\end{bmatrix},[1−t​t1​], its determinant is 1⋅1−t(−t)=1+t2.1\cdot 1-t(-t)=1+t^2.1⋅1−t(−t)=1+t2. So, (I2−A)−1=11+t2[1−tt1].(I_2-A)^{-1}=\frac{1}{1+t^2}\begin{bmatrix}1&-t\\ t&1\end{bmatrix}.(I2​−A)−1=1+t21​[1t​−t1​].

  4. Now compute (I2+A)(I2−A)−1.(I_2+A)(I_2-A)^{-1}.(I2​+A)(I2​−A)−1. Thus,

=\begin{bmatrix}1&-t\\ t&1\end{bmatrix}\cdot \frac{1}{1+t^2}\begin{bmatrix}1&-t\\ t&1\end{bmatrix}. $$ Multiply the matrices: $$\begin{bmatrix}1&-t\\ t&1\end{bmatrix} \begin{bmatrix}1&-t\\ t&1\end{bmatrix} = \begin{bmatrix}1-t^2&-2t\\ 2t&1-t^2\end{bmatrix}.$$ Hence, $$(I_2+A)(I_2-A)^{-1}=\frac{1}{1+t^2}\begin{bmatrix}1-t^2&-2t\\ 2t&1-t^2\end{bmatrix}.

  1. Compare with the given form [a−bba].\begin{bmatrix}a&-b\\ b&a\end{bmatrix}.[ab​−ba​]. Therefore, a=1−t21+t2,b=2t1+t2.a=\frac{1-t^2}{1+t^2},\qquad b=\frac{2t}{1+t^2}.a=1+t21−t2​,b=1+t22t​.

  2. Compute a2+b2a^2+b^2a2+b2: a2+b2=(1−t21+t2)2+(2t1+t2)2.a^2+b^2=\left(\frac{1-t^2}{1+t^2}\right)^2+\left(\frac{2t}{1+t^2}\right)^2.a2+b2=(1+t21−t2​)2+(1+t22t​)2. Taking common denominator, a2+b2=(1−t2)2+4t2(1+t2)2.a^2+b^2=\frac{(1-t^2)^2+4t^2}{(1+t^2)^2}.a2+b2=(1+t2)2(1−t2)2+4t2​. Now, (1−t2)2+4t2=1−2t2+t4+4t2=1+2t2+t4=(1+t2)2.(1-t^2)^2+4t^2=1-2t^2+t^4+4t^2=1+2t^2+t^4=(1+t^2)^2.(1−t2)2+4t2=1−2t2+t4+4t2=1+2t2+t4=(1+t2)2. So, a2+b2=(1+t2)2(1+t2)2=1.a^2+b^2=\frac{(1+t^2)^2}{(1+t^2)^2}=1.a2+b2=(1+t2)2(1+t2)2​=1.

  3. Therefore, 13(a2+b2)=13⋅1=13.13(a^2+b^2)=13\cdot 1=13.13(a2+b2)=13⋅1=13.

So the required integer is 13.\boxed{13}.13​.

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