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# Find the value of $x,y,z$ if $\left\{ 3\left[ \begin{matrix} 2 & 0 \\ 0 & 2 \\ 2 & 2 \\\end{matrix} \right]-4\left[ \begin{matrix} 1 & 1 \\ -1 & 2 \\ 3 & 1 \\\end{matrix} \right] \right\}\left[ \begin{matrix} 1 \\ 2 \\\end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\\end{matrix} \right]$

Last updated date: 13th Jul 2024
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Hint: Type of question is based on the matrix. For solving the type of questions we should know the properties based on how to solve these matrix problems. so by using the matrix identity, we will simplify the LHS and RHS side and find out the $x,y,z$.

Complete step-by-step solution:
So moving ahead with question which we have i.e. $\left\{ 3\left[ \begin{matrix} 2 & 0 \\ 0 & 2 \\ 2 & 2 \\ \end{matrix} \right]-4\left[ \begin{matrix} 1 & 1 \\ -1 & 2 \\ 3 & 1 \\ \end{matrix} \right] \right\}\left[ \begin{matrix} 1 \\ 2 \\ \end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\ \end{matrix} \right]$
On solving the LHS side matrix, by using the identity$x\left[ \begin{matrix} a & b \\ c & d \\ e & f \\ \end{matrix} \right]=\left[ \begin{matrix} xa & bx \\ cx & dx \\ ex & fx \\ \end{matrix} \right]$.
we will have;
$\left\{ \left[ \begin{matrix} 6 & 0 \\ 0 & 6 \\ 6 & 6 \\ \end{matrix} \right]-\left[ \begin{matrix} 4 & 4 \\ -4 & 8 \\ 12 & 4 \\ \end{matrix} \right] \right\}\left[ \begin{matrix} 1 \\ 2 \\ \end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\ \end{matrix} \right]$
Now again using the identity of subtraction and addition of matrices in the LHS side i.e. $\left\{ \left[ \begin{matrix} a & b \\ c & d \\ e & f \\ \end{matrix} \right]-\left[ \begin{matrix} g & h \\ i & j \\ k & l \\ \end{matrix} \right] \right\}=\left[ \begin{matrix} a-g & b-h \\ c-i & d-j \\ e-k & f-l \\ \end{matrix} \right]$ , we will have;
\begin{align} & \left\{ \left[ \begin{matrix} 6-4 & 0-4 \\ 0-(-4) & 6-8 \\ 6-12 & 6-4 \\ \end{matrix} \right] \right\}\left[ \begin{matrix} 1 \\ 2 \\ \end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\ \end{matrix} \right] \\ & \left\{ \left[ \begin{matrix} 2 & -4 \\ 4 & -2 \\ -6 & 2 \\ \end{matrix} \right] \right\}\left[ \begin{matrix} 1 \\ 2 \\ \end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\ \end{matrix} \right] \\ \end{align}
As still there are two matrices in the LHS side, so reduce them to a single matrix as we have in the RHS side. So by using the property of multiplication of two matrices i.e. $\left\{ \left[ \begin{matrix} a & b \\ c & d \\ e & f \\ \end{matrix} \right] \right\}\left[ \begin{matrix} g \\ h \\ \end{matrix} \right]=\left[ \begin{matrix} ag+bh \\ cg+dh \\ eg+fh \\ \end{matrix} \right]$, we will have;
\begin{align} & \left[ \begin{matrix} 2 & -4 \\ 4 & -2 \\ -6 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 \\ 2 \\ \end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\ \end{matrix} \right] \\ & \left[ \begin{matrix} 2\times 2+(-4)\times 2 \\ 4\times 1+(-2)\times 2 \\ (-6)\times 1+2\times 2 \\ \end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\ \end{matrix} \right] \\ & \left[ \begin{matrix} 4-8 \\ 4-4 \\ -6+4 \\ \end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\ \end{matrix} \right] \\ & \left[ \begin{matrix} -4 \\ 0 \\ -2 \\ \end{matrix} \right]=\left[ \begin{matrix} x-3 \\ y-1 \\ 2z \\ \end{matrix} \right] \\ \end{align}
On comparing the above result we have $-4=x-3$, $0=y-1v$and $-2=2z$ from where we can calculate the value of $x,y,z$. So
\begin{align} & -4=x-3 \\ &\Rightarrow x= -4+3 \\ &\Rightarrow x= -1 \\ \end{align}
So, we have $x=-1$
Now to find ‘y’ we have $0=y-1$, on simplifying it we have
\begin{align} &\Rightarrow 0=y-1 \\ &\Rightarrow y=1 \\ \end{align}
Hence we got $y=1$
Now to find ‘z’ we have $-2=2z$, on simplifying it we have;
\begin{align} & -2=2z \\ &\Rightarrow z=\dfrac{-2}{2} \\ &\Rightarrow z=-1 \\ \end{align}
Hence, we got $z=-1$
Hence we got the answer i.e. $x= -1$, $y=1$ and $z=-1$

Note: Go through the identity of matrices very well, as they are required to solve the type of question. Avoid making mistakes while multiplying matrices as rows of matrix get multiplied by the column of the 2nd matrix.