Question

Find the value of given matrix $\left( \begin{matrix} ax & by & cz \\\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\\ 1 & 1 & 1 \\\ \end{matrix} \right)$
(a) $\left( \begin{matrix} a & b & c \\\ x & y & z \\\ yz & xz & xy \\\ \end{matrix} \right)$
(b) $\left( \begin{matrix} x & y & z \\\ a & b & c \\\ yz & xz & xy \\\ \end{matrix} \right)$
(c) $\left( \begin{matrix} {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\\ 1 & 1 & 1 \\\ \end{matrix} \right)$
(d) $~\left( \begin{matrix} {{x}^{{}}} & {{y}^{{}}} & {{z}^{{}}} \\\ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \\\ 1 & 1 & 1 \\\ \end{matrix} \right)$

Answer 1

Hint: We will solve this question which is given n matrix form by using the properties of determinant and matrices. We will make necessary arrangements in between rows and columns of the given matrix and apply properties of the matrices to obtain the solution of the given question.

Complete step-by-step answer:
We have to find the value of $\left( \begin{matrix} ax & by & cz \\\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\\ 1 & 1 & 1 \\\ \end{matrix} \right)$.
Taking x common from Column 1- C1, y common from Column2-C2 and z common from Column 3-C3 from the above given matrix we have,

ax & by & cz \\\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\\ 1 & 1 & 1 \\\ \end{matrix} \right)=xyz\left( \begin{matrix} a & b & c \\\ x & y & z \\\ \dfrac{1}{x} & \dfrac{1}{y} & \dfrac{1}{z} \\\ \end{matrix} \right)$$ We need to make necessary arrangements to eliminate the xyz part which is common on the right hand side matrix. To do it we multiply xyz inside only with R3, that is the third row, we get $$\left( \begin{matrix} ax & by & cz \\\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\\ 1 & 1 & 1 \\\ \end{matrix} \right)=\left( \begin{matrix} a & b & c \\\ x & y & z \\\ \dfrac{xyz}{x} & \dfrac{xyz}{y} & \dfrac{xyz}{z} \\\ \end{matrix} \right)$$ Now we solve the right-hand side of the obtained equation by cancelling x obtained at the position (3,1), i.e. third row and first column , by cancelling y at the position (3,2) i.e. third row second column and by cancelling z at the position (3,3) i.e. third row and third column we get, $$\left( \begin{matrix} ax & by & cz \\\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\\ 1 & 1 & 1 \\\ \end{matrix} \right)=\left( \begin{matrix} a & b & c \\\ x & y & z \\\ yz & xz & xy \\\ \end{matrix} \right)$$ Thus, we got $$\left( \begin{matrix} ax & by & cz \\\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\\ 1 & 1 & 1 \\\ \end{matrix} \right)=\left( \begin{matrix} a & b & c \\\ x & y & z \\\ yz & xz & xy \\\ \end{matrix} \right)$$ This is the required solution of the question. Thus, we have the correct answer as (a) Note: Always remember that doing calculations where we are multiplying xyz or any other terms to the matrix or to one column or one row of the matrix does not change the value of the matrix because these all come under the properties of matrices and determinants. The point of error would be multiplying any term to all rows and columns simultaneously at one time which would be wrong, arriving at the incorrect solution.