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3 and 4 .Determinants and Matrices
medium
${x_1} + 2{x_2} + 3{x_3} = a\ \ ;\ \ 2{x_1} + 3{x_2} + {x_3} =b\ \ ;\ \ $ $3{x_1} + {x_2} + 2{x_3} = c$ this system of equations has
AInfinite solution
BNo solution
CUnique solution
DNone of these
Solution
(c) We have, ${x_1} + 2{x_2} + 3{x_3} = c$
$2a{x_1} + 3{x_2} + {x_3} = c$
$3b{x_1} + {x_2} + 2{x_3} = c$
Let $a = b = c = 1$.
Then $D = \left| {\,\begin{array}{*{20}{c}}1&2&3\\2&3&1\\3&1&2\end{array}\,} \right|$ = $1\,(5) – 2\,(1) + 3\,( – 7) = – 18 \ne 0$
${D_x} = \left| {\,\begin{array}{*{20}{c}}1&2&3\\1&3&1\\1&1&2\end{array}\,} \right| = – 3$
Similarly ${D_y} = {D_z} = – 3$. Now, $x = \frac{{{D_z}}}{D}$ = $\frac{1}{6}$
Hence $D \ne 0$, $x = y = z$, i.e., unique solution.
$2a{x_1} + 3{x_2} + {x_3} = c$
$3b{x_1} + {x_2} + 2{x_3} = c$
Let $a = b = c = 1$.
Then $D = \left| {\,\begin{array}{*{20}{c}}1&2&3\\2&3&1\\3&1&2\end{array}\,} \right|$ = $1\,(5) – 2\,(1) + 3\,( – 7) = – 18 \ne 0$
${D_x} = \left| {\,\begin{array}{*{20}{c}}1&2&3\\1&3&1\\1&1&2\end{array}\,} \right| = – 3$
Similarly ${D_y} = {D_z} = – 3$. Now, $x = \frac{{{D_z}}}{D}$ = $\frac{1}{6}$
Hence $D \ne 0$, $x = y = z$, i.e., unique solution.
Standard 12
Mathematics
Similar Questions
Let $\alpha, \beta$ and $\gamma$ be real numbers. consider the following system of linear equations
$x+2 y+z=7$
$x+\alpha z=11$
$2 x-3 y+\beta z=\gamma$
Match each entry in List – $I$ to the correct entries in List-$II$
List – $I$ | List – $II$ |
($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has | ($1$) a unique solution |
($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has | ($2$) no solution |
($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$, then the system has |
($3$) infinitely many solutions |
($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has | ($4$) $x=11, y=-2$ and $z=0$ as a solution |
($5$) $x=-15, y=4$ and $z=0$ as a solution |
Then the system has