Question

Consider three matrices given as $A=\left[ \begin{matrix} x & y & z \\\ \end{matrix} \right],B=\left[ \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right],C=\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right].$ Then ABC = 0, if
$\left( a \right)\left[ a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}+2gxy+2fyz+2czx \right]=0$
$\left( b \right)\left[ a{{x}^{2}}+c{{y}^{2}}+b{{z}^{2}}+xy+yz+zx \right]=0$
$\left( c \right)\left[ a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}+2hxy+2by+2cx \right]=0$
$\left( d \right)\left[ a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}+2gzx+2fyz+2hxy \right]=0$

Answer 1

To solve this question, just multiply the matrix AB first and then multiply it by C to get ABC. The multiplication of two matrices M and N of the order $M=m\times {{m}^{'}}$ and $N=n\times {{n}^{'}}$ is valid if the number of columns of $M={{m}^{'}}$ is equal to the number of rows of N = n.

Complete step-by-step solution
We are given that $A=\left[ \begin{matrix} x & y & z \\\ \end{matrix} \right],B=\left[ \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right],C=\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right].$ So, basically, we have to calculate ABC now. Let us first compute AB. This is possible as the order of $A=1\times 3$ and order of $B=3\times 3.$ So, as the row of B = column of A.
$\Rightarrow 3=3$
So, the matric multiplication AB product is possible.
We must know how to multiply two matrices. So, let us consider two matrices as given below.

a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right]\left[ \begin{matrix} {{a}^{'}} & {{b}^{'}} & {{c}^{'}} \\\ {{d}^{'}} & {{e}^{'}} & {{f}^{'}} \\\ {{g}^{'}} & {{h}^{'}} & {{i}^{'}} \\\ \end{matrix} \right]$$ We will multiply the elements of the first row of the first matrix with the elements of the first column of the second matrix and then add them to get the elements in the resultant matrix. So, we will have first element as $$a{{a}^{'}}+b{{d}^{'}}+c{{g}^{'}}.$$ Similarly, we can compute the other elements too. Now, coming back to AB and performing the matrix multiplication, we get, $$AB=\left[ \begin{matrix} x & y & z \\\ \end{matrix} \right]\left[ \begin{matrix} a & h & g \\\ h & b & f \\\ g & f & c \\\ \end{matrix} \right]$$ $$\Rightarrow AB=\left[ \begin{matrix} ax+hy+gz & hx+by+fz & gx+fy+cz \\\ \end{matrix} \right]$$ Now, let us compute ABC. Now order of $$AB=1\times 3$$ and of $$C=3\times 1.$$ The number of columns of A = 3 = number of rows of C. So, multiplication ABC is valid. $$ABC=\left( AB \right)C=AB\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]$$ $$\Rightarrow ABC=\left[ \begin{matrix} ax+hy+gz & hx+by+fz & gx+fy+cz \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]$$ $$\Rightarrow ABC=\left[ a{{x}^{2}}+hyx+gzx+hxy+b{{y}^{2}}+fzy+gxz+fyz+c{{z}^{2}} \right]$$ Now, given that ABC = 0. $$\Rightarrow a{{x}^{2}}+b{{y}^{2}}+c{{z}^{2}}+2hxy+2gzx+2fzy=0$$ **Hence, option (d) is the right answer.** **Note:** Another way to solve this can be, first multiply BC, which is valid as the order of $$B=3\times 3$$ and order of $$C=3\times 1=n$$ and the number of rows of c = number of columns of A = 3. Therefore, the number of columns of A = 3. Therefore, BC is valid. Now, we have ordered BC as $$3\times 1.$$ Again, the order of $$A=1\times 3.$$ Similarly, the product of A(BC) is also valid.