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Solve system of linear equations, using matrix method. $x-y+2 z=7$ ; $3 x+4 y-5 z=-5$ ; $2 x-y+3 z=12$
$x=2, y=1,z=3$
$x=-2, y=-1,z=3$
$x=-2, y=-1,z=3$
$x=2, y=1,z=-3$
Solution
The given system of equation can be written in the form of $A X=B$, where
$A=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and
$B=\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$
Now,
$|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 \neq 0$
Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$A_{11}=7, A_{12}=-19, A_{13}=11$
$A_{11}=1, A_{22}=-1, A_{23}=-1$
$A_{31}=-3, A_{32}=11, A_{33}=7$
$\therefore A^{-1}=|A|^{(a d j A)}=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]$
$\therefore X=A^{-1} B=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{4}\left[\begin{array}{c}49-5-36 \\ -133+5+132 \\ -77+5+84\end{array}\right]$
$=\frac{1}{4}\left[\begin{array}{l}8 \\ 4 \\ 12\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]$
Hence, $x=2, y=1$ and $z=3$
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
