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# Find the value of x, y, and z if $\left( {\begin{array}{*{20}{c}} 0&{2y}&z \\ x&y&{ - z} \\ x&{ - y}&z \end{array}} \right)$ satisfies $A' = {A^{ - 1}}$

Last updated date: 17th Sep 2024
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Hint: According to given in the question we have to find the values of x, y, and z if $\left( {\begin{array}{*{20}{c}} 0&{2y}&z \\ x&y&{ - z} \\ x&{ - y}&z \end{array}} \right)$that satisfies $A' = {A^{ - 1}}$. So, first of all we have to solve the $A' = {A^{ - 1}}$ to determine the value of x, y, and z but before that we have to rearrange the terms of $A' = {A^{ - 1}}$ by taking the ${A^{ - 1}}$ to the left hand side of the expression to solve it.
Now in the right hand side we will obtain the identity matrix which is as given below:
Identity matrix $I = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$………………….(A)
Now we have to solve that$A'A$, but first of all we have to obtain the matrix $A'$and after it we have to multiply the obtained matrix $A'$with$A$.
Now, we have to compare the multiplication of $A'$and $A$which is $A'A$with the right hand of which is an identity matrix.

Complete step-by-step solution:
Step 1: First of all we have to rearrange the terms of the matrix as mentioned in the solution hint. Hence,
$\Rightarrow \dfrac{{A'}}{{{A^{ - 1}}}} = I \\ \Rightarrow A'A = I................(1) \\$
Step 2: Now, we have to obtain the value of $A'$which can be obtained as,
$A' = \left( {\begin{array}{*{20}{c}} 0&x&x \\ {2y}&y&{ - y} \\ z&{ - z}&z \end{array}} \right)$
Step 3: Now we have to substitute the value of A’ as obtained in the step 2 in the expression (1). Hence,
$\Rightarrow \left( {\begin{array}{*{20}{c}} 0&x&x \\ {2y}&y&{ - y} \\ z&{ - z}&z \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 0&{2y}&z \\ x&y&{ - z} \\ x&{ - y}&z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right)$
Step 4: Now, we have to multiply all the terms of the obtained matrix,
$\Rightarrow \left( {\begin{array}{*{20}{c}} {0 \times 0 + x \times x + x \times x}&{0 \times 2y + x \times y - x \times y}&{0 \times z - x \times z + x \times z} \\ {2y \times 0 + y \times x - x \times y}&{2y \times 2y + y \times y + y \times y}&{2y \times z - y \times z - y \times z} \\ {z \times 0 - z \times x + z \times x}&{z \times 2y - z \times y - z \times y}&{z \times z + ( - z) \times ( - z) + z \times z} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right) \\ \Rightarrow \left( {\begin{array}{*{20}{c}} {2{x^2}}&0&0 \\ 0&{4{y^2} + 2{y^2}}&0 \\ 0&0&{3{z^2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right) \\$
Step 5: Now, we have to compare the obtained matrix as in solution step 4 to obtain the value of x, y, and z. Hence,
$\Rightarrow 2{x^2} = 1 \\ \Rightarrow {x^2} = \dfrac{1}{2} \\ \Rightarrow x = \sqrt {\dfrac{1}{2}} \\$
And,
$\Rightarrow 6{y^2} = 1 \\ \Rightarrow {y^2} = \dfrac{1}{6} \\ \Rightarrow y = \sqrt {\dfrac{1}{6}} \\$
And same as the value of z can be obtained as below:
$\Rightarrow 3{z^2} = 1 \\ \Rightarrow {z^2} = \dfrac{1}{3} \\ \Rightarrow z = \sqrt {\dfrac{1}{3}} \\$

Hence, we have obtained the value of x, y, and z for the given matrix $\left( {\begin{array}{*{20}{c}} 0&{2y}&z \\ x&y&{ - z} \\ x&{ - y}&z \end{array}} \right)$ which are $x = \sqrt {\dfrac{1}{2}} ,y = \sqrt {\dfrac{1}{6}} ,$ and $z = \sqrt {\dfrac{1}{3}}$

Note: For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix and the result matrix has the number of rows of the first and the number of columns of the second matrix.
Two matrices can be added and subtracted only if they have the same dimensions, that is they must have the same number of rows and columns.