
If A = $\left[ \begin{matrix}
2 & 2 \\
a & b \\
\end{matrix} \right]$ and ${{A}^{2}}=0$ then $[a,b]$ =
A . $[-2,-2]$
B . $[2,-2]$
C . $[-2,2]$
D . $[2,2]$
Answer
164.1k+ views
Hint: In this question, we have given the matrix A and ${{A}^{2}}=0$.. A square matrix is a matrix which has the same number of rows and columns. First we find ${{A}^{2}}$by multiply A with A. After finding the ${{A}^{2}}$, we put the matrix equal to zero and by equating the equations, we get the value of a and b and hence our correct option.
Complete Step- by- step Solution:
Given A = $\left[ \begin{matrix}
2 & 2 \\
a & b \\
\end{matrix} \right]$ and ${{A}^{2}}=0$
Given matrix is of $2\times 2$ order.
Top solve this question, first we find ${{A}^{2}}$
${{A}^{2}}$= $\left[ \begin{matrix}
2 & 2 \\
a & b \\
\end{matrix} \right]$ $\left[ \begin{matrix}
2 & 2 \\
a & b \\
\end{matrix} \right]$
Multiplying the above matrices, we get
${{A}^{2}}$= $\left[ \begin{matrix}
2\times 2+2\times a & 2\times 2+2\times b \\
a\times 2+b\times a & a\times 2+b\times b \\
\end{matrix} \right]$
Solving further the above equation, we get
${{A}^{2}}$= $\left[ \begin{matrix}
4+2a & 4+2b \\
2a+ba & 2a+{{b}^{2}} \\
\end{matrix} \right]$
And given that ${{A}^{2}}=0$
Hence $\left[ \begin{matrix}
4+2a & 4+2b \\
2a+ba & 2a+{{b}^{2}} \\
\end{matrix} \right]$ = $\left[ \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right]$
Equating the both sides, we get
$4+2a=0$
And $4+2b=0$
Solving the above two equations, we get
$a=-2,b=-2$
Hence the value of $a=-2$ and $b=-2$
Thus, the Option (A) is the correct answer.
Note: In these types of questions, students make mistakes while multiplying the matrices. In the matrix, the numbers are called the entries or the entities of the matrix. Whenever we solve these types of questions, we must know the multiplication of the matrix. In Multiplication matrices, the number of columns 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 with the columns in the second matrix.
Complete Step- by- step Solution:
Given A = $\left[ \begin{matrix}
2 & 2 \\
a & b \\
\end{matrix} \right]$ and ${{A}^{2}}=0$
Given matrix is of $2\times 2$ order.
Top solve this question, first we find ${{A}^{2}}$
${{A}^{2}}$= $\left[ \begin{matrix}
2 & 2 \\
a & b \\
\end{matrix} \right]$ $\left[ \begin{matrix}
2 & 2 \\
a & b \\
\end{matrix} \right]$
Multiplying the above matrices, we get
${{A}^{2}}$= $\left[ \begin{matrix}
2\times 2+2\times a & 2\times 2+2\times b \\
a\times 2+b\times a & a\times 2+b\times b \\
\end{matrix} \right]$
Solving further the above equation, we get
${{A}^{2}}$= $\left[ \begin{matrix}
4+2a & 4+2b \\
2a+ba & 2a+{{b}^{2}} \\
\end{matrix} \right]$
And given that ${{A}^{2}}=0$
Hence $\left[ \begin{matrix}
4+2a & 4+2b \\
2a+ba & 2a+{{b}^{2}} \\
\end{matrix} \right]$ = $\left[ \begin{matrix}
0 & 0 \\
0 & 0 \\
\end{matrix} \right]$
Equating the both sides, we get
$4+2a=0$
And $4+2b=0$
Solving the above two equations, we get
$a=-2,b=-2$
Hence the value of $a=-2$ and $b=-2$
Thus, the Option (A) is the correct answer.
Note: In these types of questions, students make mistakes while multiplying the matrices. In the matrix, the numbers are called the entries or the entities of the matrix. Whenever we solve these types of questions, we must know the multiplication of the matrix. In Multiplication matrices, the number of columns 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 with the columns in the second matrix.
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