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# Find the value of $x$, $y$ and $z$ from the following equations.(i) $\left[ \begin{matrix} 4 & 3 \\ x & 5 \\\end{matrix} \right]=\left[ \begin{matrix} y & z \\ 1 & 5 \\\end{matrix} \right]$(ii) $\left[ \begin{matrix} x+y & 2 \\ 5+z & xy \\\end{matrix} \right]=\left[ \begin{matrix} 6 & 2 \\ 5 & 8 \\\end{matrix} \right]$(iii) $\left[ \begin{matrix} x+y+z \\ x+z \\ y+z \\\end{matrix} \right]=\left[ \begin{matrix} 9 \\ 5 \\ 7 \\\end{matrix} \right]$

Last updated date: 20th Jun 2024
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Hint: We will use the law of equality and equate each element which is at the same position corresponding to its matrix. After equating we will get equations and by solving those equations, we will get the required values.

(i)
Given $\left[ \begin{matrix} 4 & 3 \\ x & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} y & z \\ 1 & 5 \\ \end{matrix} \right]$
Equating elements in first row first column of both the matrix, then we will get
$y=4$
Equating elements in first row second column of both the matrix, then we will get
$z=3$
Equating elements in second row and first column of both the matrix, then we will get
$x=1$

(ii)
Given $\left[ \begin{matrix} x+y & 2 \\ 5+z & xy \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & 2 \\ 5 & 8 \\ \end{matrix} \right]$
Equating elements in first row first column of both the matrix, then we will get
$x+y=6...\left( \text{i} \right)$
Equating elements in second row first column of both the matrix, then we will get
\begin{align} & z+5=5 \\ & \Rightarrow z=0 \\ \end{align}
Equating elements in second row and second column of both the matrix, then we will get
$xy=8...\left( \text{ii} \right)$
Solving equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$ by substituting the value $y=6-x$ from equation $\left( \text{i} \right)$, then we will get
\begin{align} & x\left( 6-x \right)=8 \\ & \Rightarrow 6x-{{x}^{2}}=8 \\ & \Rightarrow {{x}^{2}}-6x+8=0 \\ & \Rightarrow {{x}^{2}}-4x-2x+8=0 \\ & \Rightarrow x\left( x-4 \right)-2\left( x-4 \right)=0 \\ & \Rightarrow \left( x-4 \right)\left( x-2 \right)=0 \\ & \Rightarrow x=4\text{ or }x=2 \\ \end{align}
If $x=4$, then the value of $y$ is $6-x=6-4=2$.
If $x=2$, then the value of $y$ is $6-x=6-2=4$.

(iii)
Given $\left[ \begin{matrix} x+y+z \\ x+z \\ y+z \\ \end{matrix} \right]=\left[ \begin{matrix} 9 \\ 5 \\ 7 \\ \end{matrix} \right]$
Equating the terms in first row first column then we will have
$x+y+z=9....\left( \text{a} \right)$
Equating the terms in second row first column then we will have
$x+z=5....\left( \text{b} \right)$
Equating the terms in third row first column then we will have
$y+z=7....\left( \text{c} \right)$
Reducing the equation $\left( a \right)$ by substituting $x=5-z$ from equation $\left( \text{b} \right)$, then we will get
\begin{align} & x+y+z=9 \\ & \Rightarrow 5-z+y+z=9 \\ & \Rightarrow y=9-5 \\ & \Rightarrow y=4 \\ \end{align}
Now the value of $z$ from equation $\left( \text{c} \right)$ is given by
\begin{align} & y+z=7 \\ & \Rightarrow 4+z=7 \\ & \Rightarrow z=7-4 \\ & \Rightarrow z=3 \\ \end{align}
$\therefore$ $x=5-z=5-3=2$.

Note: Law of equality for matrices only applies when the both the matrices have the same dimensions. So, we need to check the dimensions of the given matrices before going to solve some other problems.