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Matrices and Determinants question

2021 · 20 Jul · Shift 1 · Q26
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  5. /2021 · 20 Jul · Shift 1 · Q26

Matrices and Determinants question

2021 · 20 Jul · Shift 1 · Q26

JEE MainMathematicsMatrices and DeterminantsMCQ+4 / −1
Let A=[23a0]A = \left[ {\begin{matrix} 2 & 3 \\ a & 0 \\ \end{matrix} } \right]A=[2a​30​], a ∈\in∈ 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 :
  1. A
    36
  2. B
    24
  3. C
    45
  4. D
    18
View written solutionFree

Correct answer: A

  1. Write the decomposition

Any square matrix can be uniquely written as A=P+QA = P + QA=P+Q where P=A+AT2(symmetric),Q=A−AT2(skew-symmetric).P = \frac{A + A^T}{2} \quad \text{(symmetric)}, \qquad Q = \frac{A - A^T}{2} \quad \text{(skew-symmetric)}.P=2A+AT​(symmetric),Q=2A−AT​(skew-symmetric).

Given A=[23a0],AT=[2a30].A=\begin{bmatrix}2&3\\ a&0\end{bmatrix}, \qquad A^T=\begin{bmatrix}2&a\\ 3&0\end{bmatrix}.A=[2a​30​],AT=[23​a0​].

So,

=\frac{1}{2}\begin{bmatrix}4&3+a\\ a+3&0\end{bmatrix} =\begin{bmatrix}2&\frac{a+3}{2}\\ \frac{a+3}{2}&0\end{bmatrix}.$$ And $$Q=\frac{1}{2}\left(\begin{bmatrix}2&3\\ a&0\end{bmatrix}-\begin{bmatrix}2&a\\ 3&0\end{bmatrix}\right) =\frac{1}{2}\begin{bmatrix}0&3-a\\ a-3&0\end{bmatrix} =\begin{bmatrix}0&\frac{3-a}{2}\\ \frac{a-3}{2}&0\end{bmatrix}.$$ 2. **Use the condition $\det(Q)=9$** For a matrix of the form $$\begin{bmatrix}0&x\\ -x&0\end{bmatrix},$$ its determinant is $x^2$. Here, $$Q=\begin{bmatrix}0&\frac{3-a}{2}\\ -\frac{3-a}{2}&0\end{bmatrix},$$ so $$\det(Q)=\left(\frac{3-a}{2}\right)^2=9.$$ Thus, $$\frac{(3-a)^2}{4}=9 \implies (3-a)^2=36 \implies 3-a=\pm 6.$$ Hence, $$a=-3 \quad \text{or} \quad a=9.$$ 3. **Find $\det(P)$ for both values** We have $$P=\begin{bmatrix}2&\frac{a+3}{2}\\ \frac{a+3}{2}&0\end{bmatrix}.$$ Therefore, $$\det(P)=2\cdot 0-\left(\frac{a+3}{2}\right)^2=-\frac{(a+3)^2}{4}.$$ - For $a=-3$: $$\det(P)=-\frac{0^2}{4}=0.$$ - For $a=9$: $$\det(P)=-\frac{12^2}{4}=-36.$$ 4. **Sum all possible values of $\det(P)$** The possible values are $0$ and $-36$. Their sum is $$0+(-36)=-36.$$ Its modulus is $$|-36|=36.$$ 5. **Final answer** Thus the required value is $$\boxed{36}.$$
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