Question

If the system of equations $x+y+z=5,x+2y+3z=9,x+3y+\alpha z=\beta$ has infinitely many solutions, then $\beta -\alpha$ equals?
(a) 5
(b) 18
(c) 21
(d) 8

Answer 1

We are asked to find the value of $\beta -\alpha .$ We will start by converting the linear equations $x+y+z=5,x+2y+3z=9,x+3y+\alpha z=\beta$ into the matrix form given as AX = B. We are given that the system of equations has infinitely many solutions then we know that the determinant of the coefficient matrix A and the matrix obtained by changing column of A by B is always 0, i.e. $D\left( A \right)={{D}_{X}}={{D}_{Y}}={{D}_{Z}}=0.$ Then using D (A) = 0 we get the value of $\alpha .$ Then once we have $\alpha ,$ we will use ${{D}_{X}}\text{ or }{{D}_{Y}}\text{ or }{{D}_{Z}}=0$ to get the value of $\beta .$ At last, we will subtract $\alpha$ from $\beta .$

Complete step-by-step solution:
We are given three linear equations in three variables as

& x+y+z=5 \\\ & x+2y+3z=9 \\\ & x+3y+\alpha z=\beta \\\ \end{aligned}$$ Firstly, we will change the given linear equation in the matrix form AX = B where A is called the coefficient matrix. So, we get, $$\left[ \begin{matrix} 1 & 1 & 1 \\\ 1 & 2 & 3 \\\ 1 & 3 & \alpha \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 5 \\\ 9 \\\ \beta \\\ \end{matrix} \right]$$ Now, as our system has infinite solutions, so we know that the determinant of the coefficient matrix is zero when the system has infinite solutions along with the matrix to obtain by changing column A with B. Now as $$A=\left[ \begin{matrix} 1 & 1 & 1 \\\ 1 & 2 & 3 \\\ 1 & 3 & \alpha \\\ \end{matrix} \right]$$ We will find the determinant of A by expanding row 1. $$\left| \begin{matrix} 1 & 1 & 1 \\\ 1 & 2 & 3 \\\ 1 & 3 & \alpha \\\ \end{matrix} \right|=1\left( 2\alpha -3\times 3 \right)-1\left( \alpha -3 \right)+1\left( 3-2 \right)$$ Simplifying, we get, $$\left| \begin{matrix} 1 & 1 & 1 \\\ 1 & 2 & 3 \\\ 1 & 3 & \alpha \\\ \end{matrix} \right|=2\alpha -9-\alpha +3+1$$ So, we get, $$\det \left( A \right)=\alpha -5$$ As, det (A) = 0, we get, $$\Rightarrow \alpha -5=0$$ So, we get, $$\Rightarrow \alpha =5$$ We also have that the determinant of the matrix obtained by changing one column of A with B is also 0. Now, we will form that matrix and solve its determinant. Now, replacing the first column of $$A=\left[ \begin{matrix} 1 & 1 & 1 \\\ 1 & 2 & 3 \\\ 1 & 3 & \alpha \\\ \end{matrix} \right]$$ with $$B=\left[ \begin{matrix} 5 \\\ 9 \\\ \beta \\\ \end{matrix} \right].$$ We get the new matrix as, $$\left[ \begin{matrix} 5 & 1 & 1 \\\ 9 & 2 & 3 \\\ \beta & 3 & \alpha \\\ \end{matrix} \right].$$ Now, we will find the determinant of this matrix to find $$\beta .$$ $${{D}_{X}}=\left| \begin{matrix} 5 & 1 & 1 \\\ 9 & 2 & 3 \\\ \beta & 3 & 5 \\\ \end{matrix} \right|\left[ As\text{ }\alpha =5 \right]$$ Expanding along row 1, we get, $${{D}_{X}}=5\left( 2\times 5-3\times 3 \right)-1\left( 9\times 5-3\times \beta \right)+1\left( 9\times 3-2\beta \right)$$ Simplifying further, we get, $$\Rightarrow {{D}_{X}}=5-45+3\beta +27-2\beta $$ $$\Rightarrow {{D}_{X}}=\beta -13$$ As, $${{D}_{X}}=0,$$ as the system has infinite solutions, we get, $$\Rightarrow \beta -13=0$$ $$\Rightarrow \beta =13$$ Now, we got $$\beta $$ as 13 and $$\alpha $$ as 5. So, $$\beta -\alpha =13-5=8$$ **Hence, option (d) is the right answer.** **Note:** Remember we do not have to replace B with each column of A one by one and find the determinant, i.e we do not have to evaluate $${{D}_{X}},{{D}_{Y}},{{D}_{Z}}.$$ Just solving either one of them and equating as zero will give us our solution for finding $$\beta .$$ Also, we can directly expand the matrix to find the determinant or we apply certain row operations to simplify our matrix. In both the case, the determinant will be the same.