The system of equations $3 x+2 y+z=6$, $3 x+4 y+3 z=14$ and $6 x+10 y+8 z=a$, has infinite number of…
The system of equations $3 x+2 y+z=6$, $3 x+4 y+3 z=14$ and $6 x+10 y+8 z=a$, has infinite number of solutions, if $a$ is equal to
- $8$
- $12$
- $24$
- $36$
Solution
Given system of equation is
$
\begin{gathered}
3 x+2 y+z=6 \\
3 x+4 y+3 z=14 \\
6 x+10 y+8 z=a \\
A=\left[\begin{array}{rrr}
3 & 2 & 1 \\
3 & 4 & 3 \\
6 & 10 & 8
\end{array}\right], B=\left[\begin{array}{r}
6 \\
14 \\
a
\end{array}\right]
\end{gathered}
$
Here, $\quad A=\left[\begin{array}{rrr}3 & 2 & 1 \\ 3 & 4 & 3 \\ 6 & 10 & 8\end{array}\right], B=\left[\begin{array}{r}6 \\ 14 \\ a\end{array}\right]$
$
\begin{aligned}
& C_{11}=(32-30)=2, C_{12}=-(24-18)=-6, \\
& C_{13}=(30-24)=6 \\
& C_{21}=-(16-10)=-6, C_{22}=(24-6)=18, \\
& C_{23}=-(30-12)=-18 \\
& C_{31}=(6-4)=2, C_{32}=-(9-3)=-6, \\
& C_{33}=(12-6)=6
\end{aligned}
$
$
\begin{aligned}
\operatorname{adj} A & =\left[\begin{array}{lll}
C_{11} & C_{12} & C_{13} \\
C_{21} & C_{22} & C_{23} \\
C_{31} & C_{32} & C_{33}
\end{array}\right]^{\prime}=\left[\begin{array}{rrr}
2 & -6 & 6 \\
-6 & 18 & -18 \\
2 & -6 & 6
\end{array}\right]^{\prime} \\
& =\left[\begin{array}{rrr}
2 & -6 & 2 \\
-6 & 18 & -6 \\
6 & -18 & 6
\end{array}\right]
\end{aligned}
$
So,
$
\begin{aligned}
& \text { A) } B=\left[\begin{array}{rrr}
2 & -6 & 2 \\
-6 & 18 & -6 \\
6 & -18 & 6
\end{array}\right]\left[\begin{array}{r}
6 \\
14 \\
a
\end{array}\right] \\
& =\left[\begin{array}{c}
12-84+2 a \\
-36+252-6 a \\
36-252+6 a
\end{array}\right]=\left[\begin{array}{c}
-72+2 a \\
216-6 a \\
-216+6 a
\end{array}\right] \\
& |A|=\left|\begin{array}{rrr}
3 & 2 & 1 \\
3 & 4 & 3 \\
6 & 10 & 8
\end{array}\right| \\
& =3(32-30)-2(24-18)+1(30-24) \\
& =3(2)-2(6)+6=6-12+6=0 \\
&
\end{aligned}
$
and
We know that, if $|A|=0$ and $(\operatorname{adj} A) \cdot B=0$, then the system of equations is consistent and has an infinite number of solutions.
$
\begin{gathered}
(\operatorname{adj} A) \cdot B=0 \\
\Rightarrow \quad\left[\begin{array}{c}
-72+2 a \\
216+6 a \\
-216+6 a
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
\end{gathered}
$
On comparing, we get
$
\begin{aligned}
2 a-72 & =0 \\
a & =36
\end{aligned}
$
$
\therefore \quad a=36
$
Asked in: AP EAMCET 2013
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