Question

Using properties of determinant then show that

1&{{{\log }_x}y}&{{{\log }_x}z} \\\ {{{\log }_y}x}&1&{{{\log }_y}z} \\\ {{{\log }_z}x}&{{{\log }_z}y}&1 \end{array}} \right| = 0$$

Answer 1

Here we will use the matrix determinant formula which states that for a matrix with dimensions $2 \times 2$ , the determinant will be defined as:

a&b; \\\ c&d; \end{array}} \right| = ad - bc$$ Similarly, for a matrix with dimensions $$3 \times 3$$ , the determinant will be defined as: $$\left| {\text{A}} \right| = \left| {\begin{array}{*{20}{c}} a&d;&g; \\\ b&e;&h; \\\ c&f;&i; \end{array}} \right| = a\left| {\begin{array}{*{20}{c}} e&h; \\\ f&i; \end{array}} \right| - b\left| {\begin{array}{*{20}{c}} d&g; \\\ f&i; \end{array}} \right| + c\left| {\begin{array}{*{20}{c}} d&g; \\\ e&h; \end{array}} \right|$$ $$ \Rightarrow \left| {\text{A}} \right| = a\left( {ei - hf} \right) - b\left( {di - gf} \right) + c\left( {dh - ge} \right)$$ $$ \Rightarrow \left| {\text{A}} \right| = aei - ahf - bdi + bgf + cdh - cge$$ The symbol for determinant is shown as $$\left| {\text{A}} \right|$$ or $$\Delta {\text{A}}$$. **Complete step-by-step answer:** Step 1: We need to prove that the determinant $$\left| {\begin{array}{*{20}{c}} 1&{{{\log }_x}y}&{{{\log }_x}z} \\\ {{{\log }_y}x}&1&{{{\log }_y}z} \\\ {{{\log }_z}x}&{{{\log }_z}y}&1 \end{array}} \right|$$ is equal to zero. Let, $$\Delta = \left| {\begin{array}{*{20}{c}} 1&{{{\log }_x}y}&{{{\log }_x}z} \\\ {{{\log }_y}x}&1&{{{\log }_y}z} \\\ {{{\log }_z}x}&{{{\log }_z}y}&1 \end{array}} \right|$$ We can write the above matrix as below: $$ \Rightarrow \Delta = \left| {\begin{array}{*{20}{c}} 1&{\dfrac{{\log y}}{{\log x}}}&{\dfrac{{\log z}}{{\log x}}} \\\ {\dfrac{{\log x}}{{\log y}}}&1&{\dfrac{{\log z}}{{\log y}}} \\\ {\dfrac{{\log x}}{{\log z}}}&{\dfrac{{\log y}}{{\log z}}}&1 \end{array}} \right|$$ …………………….. (1) $$\left( {\because {{\log }_n}m = \dfrac{{\log m}}{{\log n}}} \right)$$ Step 2: Writing $$1 = \dfrac{{\log x}}{{\log x}}$$ in the first row, $$1 = \dfrac{{\log y}}{{\log y}}$$ in the second row, and $$1 = \dfrac{{\log z}}{{\log z}}$$ the last row in the expression (1). Step 3: By substituting the values mentioned in step number 2, we get: $$ \Rightarrow \Delta = \left| {\begin{array}{*{20}{c}} {\dfrac{{\log x}}{{\log x}}}&{\dfrac{{\log y}}{{\log x}}}&{\dfrac{{\log z}}{{\log x}}} \\\ {\dfrac{{\log x}}{{\log y}}}&{\dfrac{{\log y}}{{\log y}}}&{\dfrac{{\log z}}{{\log y}}} \\\ {\dfrac{{\log x}}{{\log z}}}&{\dfrac{{\log y}}{{\log z}}}&{\dfrac{{\log z}}{{\log z}}} \end{array}} \right|$$ ……………. (2) In the above expression (2), by taking $$\log x$$ common from the first row, $$\log y$$ common from the second row, and $$\log z$$ common from the last row, we get: $$ \Rightarrow \Delta = \dfrac{1}{{\log x\log y\log z}}\left| {\begin{array}{*{20}{c}} {\log x}&{\log y}&{\log z} \\\ {\log x}&{\log y}&{\log z} \\\ {\log x}&{\log y}&{\log z} \end{array}} \right|$$ Step 4: By using the property of determinant of matrices that if any two or more rows of the matrix are equal then the determinant of that matrix will be equals to zero. Now in the expression $$\dfrac{1}{{\log x\log y\log z}}\left| {\begin{array}{*{20}{c}} {\log x}&{\log y}&{\log z} \\\ {\log x}&{\log y}&{\log z} \\\ {\log x}&{\log y}&{\log z} \end{array}} \right|$$ , the determinant will be equals to zero, so the final answer will be equals to as below: $$ \Rightarrow \Delta = 0$$ **Hence proved that $$\left| {\begin{array}{*{20}{c}} 1&{{{\log }_x}y}&{{{\log }_x}z} \\\ {{{\log }_y}x}&1&{{{\log }_y}z} \\\ {{{\log }_z}x}&{{{\log }_z}y}&1 \end{array}} \right| = 0$$.** **Note:** There are many properties of the determinant of the matrix but students’ needs to remember the important ones as shown below: If all the elements of a row or column of any matrix are zero, then the determinant will always be equal to zero. If all the elements of a row or column of a determinant are multiplied by any non-zero constant value, then the determinant will also be multiplied by the same constant value. If all the elements of a row or column are proportional or we can say identical to the elements of another row or column, then the determinant will be equal to zero.