Find the value of $x, y$, and $z$ from the following equation:
(i) $\left[\begin{array}{ll}4 & 3 \\ x & 5\end{array}\right]=\left[\begin{array}{ll}y & z \\ 1 & 5\end{array}\right]$
(ii) $\left[\begin{array}{ll}x+y & 2 \\ 5+z & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$
(iii) $\left[\begin{array}{c}x+y+z \\ x+z \\ y+z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]$
(i) $\left[\begin{array}{ll}4 & 3 \\ x & 5\end{array}\right]=\left[\begin{array}{ll}y & z \\ 1 & 5\end{array}\right]$
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x = 1, y = 4, and z = 3
(ii) $\left[\begin{array}{ll}x+y & 2 \\ 5+z & x y\end{array}\right]=\left[\begin{array}{ll}6 & 2 \\ 5 & 8\end{array}\right]$
As the given matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
x + y = 6, xy = 8, 5 + z = 5
Now, 5 + z = 5 ⇒ z = 0
We know that:
$(x-y)^{2}=(x+y)^{2}-4 x y$
$\Rightarrow(x-y)^{2}=36-32=4$
$\Rightarrow x-y=\pm 2$
Now, when x − y = 2 and x + y = 6, we get x = 4 and y = 2
When x − y = − 2 and x + y = 6, we get x = 2 and y = 4
∴x = 4, y = 2, and z = 0 or x = 2, y = 4, and z = 0
(iii) $\left[\begin{array}{c}x+y+z \\ x+z \\ y+z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]$
As the two matrices are equal, their corresponding elements are also equal.
Comparing the corresponding elements, we get:
$x+y+z=9 \ldots(1)$
$x+z=5 \ldots(2)$
$y+z=7 \ldots$ (3)
From (1) and (2), we have:
$y+5=9$
$\Rightarrow y=4$
Then, from (3), we have:
$4+z=7$
$\Rightarrow z=3$
$\therefore x+z=5$
$\Rightarrow x=2$
$\therefore x=2, y=4$, and $z=3$