
If $I$ is the unit matrix of order $2\times 2$ and $M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ , then find the matrix $M$ .
Answer
529.5k+ views
Hint: Here we have been asked to find the matrix $M$ when $I=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ and $M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ . For that sake we will perform simple arithmetic operations between matrices.
Complete step-by-step solution:
Now considering from the question we have been given that $I$ is an unit matrix of order $2\times 2$ that is $I=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ .
We need to find the matrix $M$ when $M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ .
For doing that, we will perform simple arithmetic operations like addition, subtraction and multiplication between matrices.
By simplifying the matrix we will have $\Rightarrow M-2\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ .
Now we will transfer all other terms except $M$ to the right hand side of the expression.
After further simplifying we will have $\Rightarrow M=2\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ .
Now we will perform further simplifications after that we will have $\begin{align}
& \Rightarrow M=2\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
2 & 0 \\
0 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
-3 & 0 \\
12 & 3 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
2-3 & 0 \\
12 & 2+3 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
-1 & 0 \\
12 & 5 \\
\end{matrix} \right] \\
\end{align}$
Therefore we can conclude that if $I$ is the unit matrix of order $2\times 2$ that is the value of the unit matrix $I$ is mathematically given as $I=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ and $M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ , then the value of the matrix $M$ will be $M=\left[ \begin{matrix}
-1 & 0 \\
12 & 5 \\
\end{matrix} \right]$ .
Note: During the process of answering questions of this type we should read the question carefully. This is a very simple question and can be solved accurately in less span of time. Very few mistakes are possible in questions of this type. The calculations of this type of questions should be done carefully. This type of question does not involve many concepts.
Here it is given that $I$ is a unit matrix if we had not read the question correctly and assumed $I$ as $\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right]$ then we will have
$\begin{align}
& \Rightarrow M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right] \\
& \Rightarrow M=2I+3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right] \\
& \Rightarrow M=2\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
2 & 2 \\
2 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
-3 & 0 \\
12 & 3 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
2-3 & 2 \\
2+12 & 2+3 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
-1 & 2 \\
14 & 5 \\
\end{matrix} \right] \\
\end{align}$ .
This is a wrong answer.
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ and $M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ . For that sake we will perform simple arithmetic operations between matrices.
Complete step-by-step solution:
Now considering from the question we have been given that $I$ is an unit matrix of order $2\times 2$ that is $I=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ .
We need to find the matrix $M$ when $M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ .
For doing that, we will perform simple arithmetic operations like addition, subtraction and multiplication between matrices.
By simplifying the matrix we will have $\Rightarrow M-2\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ .
Now we will transfer all other terms except $M$ to the right hand side of the expression.
After further simplifying we will have $\Rightarrow M=2\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ .
Now we will perform further simplifications after that we will have $\begin{align}
& \Rightarrow M=2\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
2 & 0 \\
0 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
-3 & 0 \\
12 & 3 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
2-3 & 0 \\
12 & 2+3 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
-1 & 0 \\
12 & 5 \\
\end{matrix} \right] \\
\end{align}$
Therefore we can conclude that if $I$ is the unit matrix of order $2\times 2$ that is the value of the unit matrix $I$ is mathematically given as $I=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ and $M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right]$ , then the value of the matrix $M$ will be $M=\left[ \begin{matrix}
-1 & 0 \\
12 & 5 \\
\end{matrix} \right]$ .
Note: During the process of answering questions of this type we should read the question carefully. This is a very simple question and can be solved accurately in less span of time. Very few mistakes are possible in questions of this type. The calculations of this type of questions should be done carefully. This type of question does not involve many concepts.
Here it is given that $I$ is a unit matrix if we had not read the question correctly and assumed $I$ as $\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right]$ then we will have
$\begin{align}
& \Rightarrow M-2I=3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right] \\
& \Rightarrow M=2I+3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right] \\
& \Rightarrow M=2\left[ \begin{matrix}
1 & 1 \\
1 & 1 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-1 & 0 \\
4 & 1 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
2 & 2 \\
2 & 2 \\
\end{matrix} \right]+\left[ \begin{matrix}
-3 & 0 \\
12 & 3 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
2-3 & 2 \\
2+12 & 2+3 \\
\end{matrix} \right] \\
& \Rightarrow M=\left[ \begin{matrix}
-1 & 2 \\
14 & 5 \\
\end{matrix} \right] \\
\end{align}$ .
This is a wrong answer.
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