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  • Find x, y, z if A = [0 2y z] satisfies A’ = A^-1

Find x, y, z if A=\begin{bmatrix} 0 &2y &z \\x &y &-z \\x &-y &z \end{bmatrix} satisfies A'= A^{-1}

Answers (1)

We are given the following matrix A such that,

A=\begin{bmatrix} 0 &2y &z \\x &y &-z \\x &-y &z \end{bmatrix}

We need to find the values of x, y and z such that A'= A\textsuperscript{-1}

If A' = A\textsuperscript{-1}

Pre-multiplying A on both sides, we get

AA' = AA\textsuperscript{-1}

$ \Rightarrow $ AA'= Iwhere I is the identity matrix.

\begin{aligned} &\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]^{T}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right]\left[\begin{array}{ccc} 0 & x & x \\ 2 y & y & -y \\ z & -z & z \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &\text { By the rule of matrix multiplication we have: }\\ \end{aligned}

\Rightarrow\left[\begin{array}{ccc} 4 y^{2}+z^{2} & 2 y^{2}-z^{2} & -2 y^{2}+z^{2} \\ 2 y^{2}-z^{2} & x^{2}+y^{2}+z^{2} & x^{2}-y^{2}-z^{2} \\ -2 y^{2}+z^{2} & x^{2}-y^{2}+z^{2} & x^{2}+y^{2}+z^{2} \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]

On equating the corresponding elements of matrix as the matrix is equal to each other.

We need basically 3 equations as we have 3 variables to solve for. You can pick any three elements and equate them.

We have the following equations,

\\4y\textsuperscript{2} + z\textsuperscript{2} = 1 $ \ldots $ (1) \\x\textsuperscript{2} + y\textsuperscript{2} + z\textsuperscript{2} = 1 $ \ldots $ (2) \\2y\textsuperscript{2} - z\textsuperscript{2} = 0 $ \ldots $ (3)

By Adding equation 2 and 3, we get,

\\6y\textsuperscript{2} = 1 \\$ \Rightarrow $ y\textsuperscript{2} = 1/6

\\ y=\pm \frac{1}{\sqrt{6}}\\$ From equation $3,$ we get, $z^{2}=2 y^{2}$\\ $\Rightarrow z^{2}=2(1 / 6)$ \\$\therefore z^{2}=1 / 3$ \\$z=\pm \frac{1}{\sqrt{3}}$ \\From equation $2,$ we get, \\$x^{2}=1-y^{2}-z^2$ \\$\Rightarrow x^{2}=1-(1 / 6)-(1 / 3)$ \\$\Rightarrow x^{2}=1-1 / 2=1 / 2$ \\$x=\pm \frac{1}{\sqrt{2}}$ \\Thus, we get that, \\$\mathrm{x}=\pm \frac{1}{\sqrt{2} ;} \mathrm{y}=\pm \frac{1}{\sqrt{6} }\text { and } \mathrm{z}=\pm \frac{1}{\sqrt{3}}$

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