If A=[array|c|c| √2 & 1 -1 & √2 array],B=[array|c|c| 1 & 0 1 & 1 array],C=ABA^T and X=A^TC…

Question

If $A=[\begin{array}{|c|c|} \sqrt{2} & 1 \ -1 & \sqrt{2} \end{array}],B=[\begin{array}{|c|c|} 1 & 0 \ 1 & 1 \end{array}],C={ABA}^{T}$ and $X={A}^{T}{C}^{2}A$, then $\det X$ is equal to :

Accepted Answer

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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