Question

Using the properties of determinant, prove the following
$\left| \begin{matrix} {{x}^{2}}+1 & xy & xz \\\ xy & {{y}^{2}}+1 & yz \\\ xz & yz & {{z}^{2}}+1 \\\ \end{matrix} \right|=1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}}$

Answer 1

To solve this we will first multiply ${{R}_{1}},{{R}_{2}},{{R}_{3}}$ with x, y, z respectively. And then take x, y, z common from ${{C}_{1}},{{C}_{2}},{{C}_{3}}$ respectively. Then we will use elementary row and column transformation to simplify the determinant to a lower triangular determinant.

Complete step by step answer:
Now to prove the given equation we will first simplify left hand side
Now consider left hand side of the given equation $\left| \begin{matrix} {{x}^{2}}+1 & xy & xz \\\ xy & {{y}^{2}}+1 & yz \\\ xz & yz & {{z}^{2}}+1 \\\ \end{matrix} \right|$
Let us multiply ${{R}_{1}}$ by x, ${{R}_{2}}$ by y and ${{R}_{3}}$ by z. Since we are multiplying xyz we will have to divide xyz outside determinant too, hence we get

x\left( {{x}^{2}}+1 \right) & {{x}^{2}}y & {{x}^{2}}z \\\ x{{y}^{2}} & y\left( {{y}^{2}}+1 \right) & {{y}^{2}}z \\\ x{{z}^{2}} & y{{z}^{2}} & z\left( {{z}^{2}}+1 \right) \\\ \end{matrix} \right|$$ Now let us take x common from ${{C}_{1}}$, y common from ${{C}_{2}}$ and z common from ${{C}_{3}}$ Hence we get. $$\begin{aligned} & =\dfrac{xyz}{xyz}\left| \begin{matrix} \left( {{x}^{2}}+1 \right) & {{x}^{2}} & {{x}^{2}} \\\ {{y}^{2}} & \left( {{y}^{2}}+1 \right) & {{y}^{2}} \\\ {{z}^{2}} & {{z}^{2}} & \left( {{z}^{2}}+1 \right) \\\ \end{matrix} \right| \\\ & =\left| \begin{matrix} \left( {{x}^{2}}+1 \right) & {{x}^{2}} & {{x}^{2}} \\\ {{y}^{2}} & \left( {{y}^{2}}+1 \right) & {{y}^{2}} \\\ {{z}^{2}} & {{z}^{2}} & \left( {{z}^{2}}+1 \right) \\\ \end{matrix} \right| \\\ \end{aligned}$$ Now let we will use Row transformation ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$ $=\left| \begin{matrix} 1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}} & 1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}} & 1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}} \\\ {{y}^{2}} & \left( {{y}^{2}}+1 \right) & {{y}^{2}} \\\ {{z}^{2}} & {{z}^{2}} & \left( {{z}^{2}}+1 \right) \\\ \end{matrix} \right|$ Now let us take the $1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}}$ common from1st row we get $$=(1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}})\left| \begin{matrix} 1 & 1 & 1 \\\ {{y}^{2}} & {{y}^{2}}+1 & {{y}^{2}} \\\ {{z}^{2}} & {{z}^{2}} & {{z}^{2}}+1 \\\ \end{matrix} \right|$$ Now let us use column transformation ${{C}_{2}}={{C}_{2}}-{{C}_{1}}$ Using this we get. $=(1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}})\left| \begin{matrix} 1 & 0 & 1 \\\ {{y}^{2}} & 1 & {{y}^{2}} \\\ {{z}^{2}} & 0 & {{z}^{2}}+1 \\\ \end{matrix} \right|$ Now let us use column transformation ${{C}_{3}}={{C}_{3}}-{{C}_{1}}$ $=(1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}})\left| \begin{matrix} 1 & 0 & 0 \\\ {{y}^{2}} & 1 & 0 \\\ {{z}^{2}} & 0 & 1 \\\ \end{matrix} \right|$ Now we know that the determinant of a triangular matrix is the product of its diagonal. Here the product of all diagonal entries is 1 × 1 × 1 = 1 Hence the value of determinant = 1 Hence we get $\begin{aligned} & =(1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}})(1) \\\ & =1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}} \\\ \end{aligned}$ Now $1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}}$ is right hand side of the given equation Hence, Given equation is proved. **Note:** Now while multiplying a scalar to a determinant note that it can be multiplied only to one row or one column and not all the entries of the determinant. Similarly if we are taking something common it can be taken from only one row or column at a time and not from all the entries. Also row transformation and column transformation does not change the value of determinant.