
If $A=\left( \begin{matrix}
0 & 1 \\
0 & 0 \\
\end{matrix} \right)$ , I is the unit matrix of order 2 and a,b are arbitrary constants, then ${{(aI+bA)}^{2}}$ is equal to ?
A . ${{a}^{2}}I+abA$
B. ${{a}^{2}}I+2abA$
C. ${{a}^{2}}I+{{b}^{2}}A$
D. None of these
Answer
163.5k+ views
Hint: In this question, we have given the matrix A which is of $2\times 2$ order and we have to find the value of ${{(aI+bA)}^{2}}$. To solve this question, first, we find the square of matrix A. We find the square of matrix A by multiplying matrix A with matrix A. then by expanding the ${{(aI+bA)}^{2}}$ and solving it, we the desired value.
Complete step by step Solution:
Given $A=\left( \begin{matrix}
0 & 1 \\
0 & 0 \\
\end{matrix} \right)$
And A is of order $2\times 2$ matrix.
We have to find the value of ${{(aI+bA)}^{2}}$
First, we find out ${{A}^{2}}$
It is the multiplication of matrix A with matrix A.
${{A}^{2}}=\left( \begin{matrix}
0 & 1 \\
0 & 0 \\
\end{matrix} \right)$$\left( \begin{matrix}
0 & 1 \\
0 & 0 \\
\end{matrix} \right)$
Now we open the brackets of R.H.S and multiply the terms, we get
${{A}^{2}}=\left( \begin{matrix}
0\times 0+1\times 0 & 0\times 1+1\times 0 \\
0\times 0+0\times 0 & 0\times 1+0\times 0 \\
\end{matrix} \right)$
Adding the terms and solving them, we get
${{A}^{2}}=\left( \begin{matrix}
0+0 & 0+0 \\
0+0 & 0+0 \\
\end{matrix} \right)$
Simplifying further, we get
${{A}^{2}}=\left( \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right)$
Hence the value of ${{A}^{2}}=0$
Now we consider ${{(aI+bA)}^{2}}=(aI+bA)(aI+bA)$
By solving it, we get ${{(aI+bA)}^{2}}={{a}^{2}}I+abIA+baAI+{{b}^{2}}{{A}^{2}}$
Hence ${{(aI+bA)}^{2}}={{a}^{2}}I+2abA$ ( ${{A}^{2}}=0,ab=ba$ for real numbers )
Thus the value of ${{(aI+bA)}^{2}}={{a}^{2}}I+2abA$
Therefore, the correct option is (B).
Note: We know that the given question is in matrix form. A matrix is a set of numbers that are arranged in rows and columns to make a rectangular array. In a matrix, the numbers are called the entries or entities of the matrix.
In Multiplication matrices, the number of column of the first matrix match the number of rows of the second matrix. When we want to multiply the matrices, then the parts of the rows in the first matrix are multiplied by the columns in the second matrix.
Complete step by step Solution:
Given $A=\left( \begin{matrix}
0 & 1 \\
0 & 0 \\
\end{matrix} \right)$
And A is of order $2\times 2$ matrix.
We have to find the value of ${{(aI+bA)}^{2}}$
First, we find out ${{A}^{2}}$
It is the multiplication of matrix A with matrix A.
${{A}^{2}}=\left( \begin{matrix}
0 & 1 \\
0 & 0 \\
\end{matrix} \right)$$\left( \begin{matrix}
0 & 1 \\
0 & 0 \\
\end{matrix} \right)$
Now we open the brackets of R.H.S and multiply the terms, we get
${{A}^{2}}=\left( \begin{matrix}
0\times 0+1\times 0 & 0\times 1+1\times 0 \\
0\times 0+0\times 0 & 0\times 1+0\times 0 \\
\end{matrix} \right)$
Adding the terms and solving them, we get
${{A}^{2}}=\left( \begin{matrix}
0+0 & 0+0 \\
0+0 & 0+0 \\
\end{matrix} \right)$
Simplifying further, we get
${{A}^{2}}=\left( \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right)$
Hence the value of ${{A}^{2}}=0$
Now we consider ${{(aI+bA)}^{2}}=(aI+bA)(aI+bA)$
By solving it, we get ${{(aI+bA)}^{2}}={{a}^{2}}I+abIA+baAI+{{b}^{2}}{{A}^{2}}$
Hence ${{(aI+bA)}^{2}}={{a}^{2}}I+2abA$ ( ${{A}^{2}}=0,ab=ba$ for real numbers )
Thus the value of ${{(aI+bA)}^{2}}={{a}^{2}}I+2abA$
Therefore, the correct option is (B).
Note: We know that the given question is in matrix form. A matrix is a set of numbers that are arranged in rows and columns to make a rectangular array. In a matrix, the numbers are called the entries or entities of the matrix.
In Multiplication matrices, the number of column of the first matrix match the number of rows of the second matrix. When we want to multiply the matrices, then the parts of the rows in the first matrix are multiplied by the columns in the second matrix.
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