
If $A = \begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix}$ and $I$ is the unit matrix of order $2$ then find out ${A^2}$
A. $4A - 3I$
B. $3A - 4I$
C. $A - I$
D. $A + I$
Answer
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Hint: Firstly , we will be finding out ${A^2}$ matrix using simple multiplication of same matrix the proceeding with ${A^2}$ we will be finding out relation with the unit matrix in order to get the right answer.
Formula Used:
Cross Product of Two Matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix} \times \begin{bmatrix}e&f\\g&g\end{bmatrix} = \begin{bmatrix}{ae}&{cf}\\{bg}&{dg}\end{bmatrix}$
Unit Matrix $ = \begin{bmatrix}1&0\\0&1\end{bmatrix}$
Complete step by step solution:
For Finding ${A^2}$,
${A^2} = A \times A$
Where matrix $A$ is equal to $\begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix}$.
${A^2} = \begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix} \times \begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix}$
${A^2} = \begin{bmatrix}{4 + 1}&{ - 2 - 2}\\{ - 2 - 2}&{1 + 4}\end{bmatrix}$
${A^2} = \begin{bmatrix}5&{ - 4}\\{ - 4}&5\end{bmatrix}$
After calculating ${A^2}$
We will be checking each option to find out which option is correct.
Lets try with option (A).
A. $4A - 3I$
For calculating $4A - 3I$
$4\begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix} - 3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow \begin{bmatrix}8&{ - 1}\\{ - 4}&2\end{bmatrix} - \begin{bmatrix}3&0\\0&3\end{bmatrix}$
$ \Rightarrow \begin{bmatrix}5&{ - 4}\\{ - 4}&5\end{bmatrix}$
From this we can conclude that the matrix of $4A - 3I$ is equal to ${A^2}$.
Option ‘A’ is correct
Note: 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 while solving the Cross Multiplication of the matrix and while multiplying the unit matrix with the constant.
Formula Used:
Cross Product of Two Matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix} \times \begin{bmatrix}e&f\\g&g\end{bmatrix} = \begin{bmatrix}{ae}&{cf}\\{bg}&{dg}\end{bmatrix}$
Unit Matrix $ = \begin{bmatrix}1&0\\0&1\end{bmatrix}$
Complete step by step solution:
For Finding ${A^2}$,
${A^2} = A \times A$
Where matrix $A$ is equal to $\begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix}$.
${A^2} = \begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix} \times \begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix}$
${A^2} = \begin{bmatrix}{4 + 1}&{ - 2 - 2}\\{ - 2 - 2}&{1 + 4}\end{bmatrix}$
${A^2} = \begin{bmatrix}5&{ - 4}\\{ - 4}&5\end{bmatrix}$
After calculating ${A^2}$
We will be checking each option to find out which option is correct.
Lets try with option (A).
A. $4A - 3I$
For calculating $4A - 3I$
$4\begin{bmatrix}2&{ - 1}\\{ - 1}&2\end{bmatrix} - 3\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$ \Rightarrow \begin{bmatrix}8&{ - 1}\\{ - 4}&2\end{bmatrix} - \begin{bmatrix}3&0\\0&3\end{bmatrix}$
$ \Rightarrow \begin{bmatrix}5&{ - 4}\\{ - 4}&5\end{bmatrix}$
From this we can conclude that the matrix of $4A - 3I$ is equal to ${A^2}$.
Option ‘A’ is correct
Note: 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 while solving the Cross Multiplication of the matrix and while multiplying the unit matrix with the constant.
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