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  • Let <img alt="A=\left[\begin{array}{ll} 2 & 3 \\ a & 0 \end{array}\right], a \in \mathbf{R}" src="/latex-image/?A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bll%7D%202%20%26%203%20%5C%5C%20a%20%26%200%20%5Cend%7Barray%7D%5Cright%5D%2C%20a%20%5Cin%20%5Cmathb

Let A=\left[\begin{array}{ll} 2 & 3 \\ a & 0 \end{array}\right], a \in \mathbf{R} be written as P+Q where P is a symmetric matrix and Q is skew symmetric matrix. If det(Q) = 9, then the modulus of the sum of all possible values of determinant of P is equal to :
     
Option: 1 36
Option: 2 24
Option: 3 45  
Option: 4 18

Answers (1)

best_answer

Using property of matrices,

A= P+Q

A can be written sum of a symmetric and a skew symmetric matrix.

Where P=\frac{1}{2}\left(A+A^{\prime}\right) \; and \; Q=\frac{1}{2}\left(A-A^{\prime}\right)

\begin{aligned} \therefore \quad Q &=\frac{1}{2}\left(A-A^{\prime}\right) \\ &=\frac{1}{2}\left(\left[\begin{array}{cc} 2 & 3 \\ a & 0 \end{array}\right]-\left[\begin{array}{cc} 2 & a \\ 3 & 0 \end{array}\right]\right) \\ &=\frac{1}{2}\left[\begin{array}{cc} 0 & 3-a \\ a-3 & 0 \end{array}\right] \\ &=\left[\begin{array}{cc} 0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0 \end{array}\right] \end{aligned}

Given  \left | Q \right |= 9

\begin{aligned} &\Rightarrow 0-\left(\frac{a-3}{2}\right)\left(\frac{3-a}{2}\right)=9 \\ &\Rightarrow(a-3)^{2}=36 \\ &\Rightarrow a-3=6 \text { or } a-3=-6 \\ &\Rightarrow a=9 \text { or } a=-3 . \end{aligned}

So,

\begin{array}{ll} A=\left[\begin{array}{ll} 2 & 3 \\ 9 & 0 \end{array}\right] \text { or } A=\left[\begin{array}{cc} 2 & 3 \\ -3 & 0 \end{array}\right] \\ P=\frac{1}{2}\left(\left[\begin{array}{ll} 2 & 3 \\ 9 & 0 \end{array}\right]+\left[\begin{array}{cc} 2 & 9 \\ 3 & 0 \end{array}\right]\right) \text { or } P=\frac{1}{2}\left(\left[\begin{array}{cc} 2 & 3 \\ -3 & 0 \end{array}\right]+\left[\begin{array}{cc} 2 & -3 \\ 3 & 0 \end{array}\right]\right) \end{array}

\begin{aligned} P=\left[\begin{array}{ll} 2 & 6 \\ 6 & 0 \end{array}\right] \quad \text { or } P &=\left[\begin{array}{ll} 2 & 0 \\ 0 & 0 \end{array}\right] \\\Rightarrow|P| &=-36 \quad \text { or } \quad|P|=0 \end{aligned}

Sum of possible |P|=-36+0=-36 .

Mod of this value = 36

Hence, the correct answer is option (1)

Posted by

Kuldeep Maurya

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