# Find the value of $x, y$, and $z$ from the following equation:

Question:

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) $\left[\begin{array}{c}x+y+z \\ x+z \\ y+z\end{array}\right]=\left[\begin{array}{l}9 \\ 5 \\ 7\end{array}\right]$

Solution:

(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 + = 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 − = − 2 and x + y = 6, we get x = 2 and = 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 $z=3$