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If $\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}}$ and $\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{~A}$, then $\operatorname{det} \mathrm{X}$ is equal to :

JEE · Math · previous-year question

  1. A.243
  2. B.729correct
  3. C.27
  4. D.891

Answer

B. 729

Explanation

The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant. Given $\mathrm{A}$, $\mathrm{B}$, and that $\mathrm{C} = \mathrm{ABA}^{\mathrm{T}}$, and $\mathrm{X} = \mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}$, we find $\operatorname{det} \mathrm{X}$ as follows: First, we express $|\mathrm{C}|$, the determinant of $\mathrm{C}$, in terms of the determinants of $\mathrm{A}$ and $\mathrm{B}$, using the property that $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$: $\begin{aligned} |\mathrm{C}| &= |\mathrm{ABA}^{\mathrm{T}}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{A}^{\mathrm{T}}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{B}| \quad \text{since } |\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|. \end{aligned}$ Next, we find $|\mathrm{X}|$, using the property $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ and substituting $|\mathrm{C}|$: $\begin{aligned} |\mathrm{X}| &= |\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}| = |\mathrm{A}^{\mathrm{T}}| \cdot |\mathrm{C}|^2 \cdot |\mathrm{A}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{C}|^2. \end{aligned}$ Substituting the expression for $|\mathrm{C}|$ obtained earlier: $\begin{aligned} |\mathrm{X}| &= |\mathrm{A}|^2 \cdot (|\mathrm{A}|^2 \cdot |\mathrm{B}|)^2 \\\\ &= (|\mathrm{A}|^3 \cdot |\mathrm{B}|)^2. \end{aligned}$ The determinants of $\mathrm{A}$ and $\mathrm{B}$ are calculated as: $$ |A|=\left|\begin{array}{cc} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{array}\right|=2+1=3 $$ $$ |B|=\left|\begin{array}{ll} 1 & 0 \\ 1 & 1 \end{array}\right|=1 $$ Finally, substituting these values into our expression for $|\mathrm{X}|$: $\begin{aligned} |\mathrm{X}| &= (3^3 \cdot 1)^2 \\\\ &= 729. \end{aligned}$ Therefore, $\operatorname{det} \mathrm{X} = 729$.

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