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Sir, We have given $A= \begin{bmatrix}q_1 & q_2&q_3 \\ q_4 & q_5&q_6\\ q_7 & q_8&q_9 \end{bmatrix} \tag 1$.

A is a matrix with determinant 1,orthogonal , invertible and not necessary that all $q_i$s are equal. All entries of the A are constants cant alter

$C= \begin{bmatrix}p_1 & p_2&p_3 \\ p_4 & p_5&p_6\\ p_7 & p_8&p_9 \end{bmatrix} \tag 2$

C is a matrix which is non-invertible and not necessary that all $p_i$s are equal. All entries of the C are constants cant alter

$B= \left( \begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \\ \end{array} \right)\tag 3$

B is a matrix which is non-invertible. It is the varible matrix,and the variables are x,y,z

Question

If we have the equation $A=BC\tag 4$

Can we find the solutions of x,y,z interms of constants $p_i$s and $q_i$s ?

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In fact the system does not have any solution. Note that from $A$ being invertible, $C$ being not invertible and $A=BC$, one have $$0 \ne \det A = \det B \cdot \det C = \det B\cdot 0 = 0.$$ So there there is no solution.

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The equation has no solution because $1 \stackrel!= \det(A) = \det(BC) = \det(B) \det(C) \stackrel!= 0$

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