Question

If $\det A=0$, then the matrix equation $AX=B$, has

& \text{(A) infinity solutions} \\\ & \text{(B) unique solution} \\\ & \text{(C) no solution} \\\ & \text{(D) infinity solutions or no solutions} \\\ \end{aligned}$$

Answer 1

Hint : Let us assume $A=\left( \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right)$, $X=\left( \begin{aligned} & x \\\ & y \\\ & z \\\ \end{aligned} \right)$ and $B=\left( \begin{aligned} & {{d}_{1}} \\\ & {{d}_{2}} \\\ & {{d}_{3}} \\\ \end{aligned} \right)$. Now we should find the value of AX. We know that according to multiplication rule of matrices “ Two matrices ${{A}_{m\times n}}$ and ${{B}_{p\times q}}$ can be multiplied if $n=p$ and the obtained matrix on multiplication is equal to ${{C}_{m\times q}}$”. By this rule, we can find the order of matrix AX.

Complete step-by-step answer :
From the question, a matrix equation $AX=B$ is given.
Let us assume $A=\left( \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right)$, $X=\left( \begin{aligned} & x \\\ & y \\\ & z \\\ \end{aligned} \right)$ and $B=\left( \begin{aligned} & {{d}_{1}} \\\ & {{d}_{2}} \\\ & {{d}_{3}} \\\ \end{aligned} \right)$.
Now we have to find the value of AX.

{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right)\left( \begin{aligned} & x \\\ & y \\\ & z \\\ \end{aligned} \right)$$ We know that according to multiplication rule of matrices “ Two matrices $${{A}_{m\times n}}$$ and $${{B}_{p\times q}}$$ can be multiplied if $$n=p$$ and the obtained matrix on multiplication is equal to $${{C}_{m\times q}}$$”. We know that the order of matrix A is $$3\times 3$$ and the order of matrix X is $$3\times 1$$. Hence, the order of matrix AX is $$3\times 1$$. Now we should compare each and every element of AX and D. Now we can obtain three equations of x, y and z. In this way, we can find the nature of solutions of linear equations. $$\Rightarrow AX=\left( \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right)\left( \begin{aligned} & x \\\ & y \\\ & z \\\ \end{aligned} \right)$$ $$\Rightarrow AX=\left( \begin{aligned} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z \\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z \\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z \\\ \end{aligned} \right)$$ We know that AX is equal to D. So, we get $$\left( \begin{matrix} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z \\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z \\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}z \\\ \end{matrix} \right)=\left( \begin{matrix} & {{d}_{1}} \\\ & {{d}_{2}} \\\ & {{d}_{3}} \\\ \end{matrix} \right)$$ From this, we get $$\begin{aligned} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}={{d}_{1}} \\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}={{d}_{2}} \\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}={{d}_{3}} \\\ \end{aligned}$$ Let us assume $$\begin{aligned} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}={{d}_{1}}.....(1) \\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}={{d}_{2}}....(2) \\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}={{d}_{3}}.....(3) \\\ \end{aligned}$$ It is clear that equation (1), equation (2) and equation (3) are set of linear equations. In the question, it was given that the value of det A is equal to zero. $$\det A=\det \left( \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right)=0.....(4)$$ So, from equation (4) it is clear that the determinant of coefficients of given linear equations is equal to zero. We know that if the determinant of the coefficient matrix is equal to 0, then the given set of linear equations will have infinite solutions (or) no solutions. Hence, option D is correct. **Note** : If the below set of linear equations $$\begin{aligned} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}={{d}_{1}} \\\ & {{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}={{d}_{2}} \\\ & {{a}_{3}}x+{{b}_{3}}y+{{c}_{3}}={{d}_{3}} \\\ \end{aligned}$$has solutions. Let us assume $${{\Delta }_{1}}=\left( \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{d}_{1}} \\\ {{a}_{2}} & {{b}_{2}} & {{d}_{2}} \\\ {{a}_{3}} & {{b}_{3}} & {{d}_{3}} \\\ \end{matrix} \right),{{\Delta }_{2}}=\left( \begin{matrix} {{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\\ {{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\\ {{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\\ \end{matrix} \right),{{\Delta }_{3}}=\left( \begin{matrix} {{a}_{1}} & {{d}_{1}} & {{c}_{1}} \\\ {{a}_{2}} & {{d}_{2}} & {{c}_{2}} \\\ {{a}_{3}} & {{d}_{3}} & {{c}_{3}} \\\ \end{matrix} \right)$$. The solutions are said to be unique, if the determinant of coefficient matrix, $${{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}$$ should not be equal to zero. The solutions are infinitely many if the determinant of coefficient matrix is equal to zero and $${{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}$$ should not be equal to zero. If there is no solution for the given set of linear equations, then the determinant of coefficient matrix and $${{\Delta }_{1}},{{\Delta }_{2}},{{\Delta }_{3}}$$ is equal to zero.