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Solve system of linear equations, using matrix method. $x-y+z=4$ ; $2 x+y-3 z=0$ ; $x+y+z=2$
$x=-2, y=-1,z=1$
$x=-2, y=-1,z=-1$
$x=2, y=-1,z=1$
$x=-2, y=1,z=1$
Solution
The given system of equation can be written in the form of $A X=B$, where
$A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
Now,
$|A|=1(1+3)+1(2+3)+1(2-1)=4+5+1=10 \neq 0$
Thus $A$ is non-singular. Therefore, its inverse exists.
Now,
$A_{11}=4, A_{12}=-5, A_{13}=1$
$A_{21}=2, A_{22}=0, A_{33}=-2$
$A_{31}=2, A_{32}=5, A_{33}=3$
$\therefore A^{-1}=\frac{1}{|A|}(a d j A)=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{10}\left[\begin{array}{c}16+0+4 \\ -20+0+10 \\ 4+0+6\end{array}\right]$
$=\frac{1}{10}\left[\begin{array}{c}20 \\ -10 \\ 10\end{array}\right]$
$=\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]$
Hence, $x=2, y=-1$ and $z=1$
Similar Questions
Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1, \alpha, \beta\}$. For a $3 \times 3$ matrix $M=\left(a_{\ell}\right) 3 \times 3_3$, define $R_l=a_{l 1}+a_{l 2}+a_\beta$ and $C_j=a_{1 j}+a_{2 l}+a_{3 j}$ for $i=1,2,3$ and $j=1,2,3$
Match each entry in $List-I$ to the correct entry in $List-II$.
| $List-I$ | $List-II$ |
| ($P$) The number of matrices $M=\left(a_{i j}\right)_3 \times 3$ with all entries in $T$ such that $R_i=C_j=0$ for all $i, j$ is | ($1$) ($1$) |
| ($Q$) The number of symmetric matrices $M=\left(a_{i j}\right) 3 \times 3$ with all entries in $T$ such that $C_j=0$ for all $j$ is | ($2$) ($2$) |
| ($R$) Let $M=\left(a_{i j}\right) 3 \times 3$ be a skew symmetric matrix such that $a_{i j} \in T$ for $i>j$. Then the number of elements in the set $\left\{\left(\begin{array}{l}x \\ y \\ z\end{array}\right): x, y \cdot z \in R, M\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{c}a_{12} \\ 0 \\ -a_{23}\end{array}\right)\right\}$ is is | ($3$) Infinite |
| ($S$) Let $M=\left(a_{i j}\right)_3 \times 3$ be a matrix with all entries in $T$ such that $R_i=0$ for all $i$. Then the absolute value of the determinant of $M$ is | ($4$) ($6$) |
| ($5$) ($0$) |
The correct option is
