
If $A + B = \left[ {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right]$ and $A - 2B = \left[ {\begin{array}{*{20}{c}}
{ - 1}&1 \\
0&{ - 1}
\end{array}} \right]$ , then $A = $
A. $\left[ {\begin{array}{*{20}{c}}
1&1 \\
2&1
\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}
{\dfrac{2}{3}}&{\dfrac{1}{3}} \\
{\dfrac{1}{3}}&{\dfrac{2}{3}}
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{3}}&{\dfrac{1}{3}} \\
{\dfrac{2}{3}}&{\dfrac{1}{3}}
\end{array}} \right]$
D. None of these
Answer
161.4k+ views
Hint: A matrix is an organised rectangular configuration of real, complex, or function-based values. We will be using the concepts or properties of matrices i.e., addition and scalar multiplication. Then, we will solve the given two equations simultaneously to get the desired answer.
Formula Used: The addition of two matrices, $A$ and $B$ or $A + B$ is defined as: if there are two matrices, $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ of the same order $m \times n$ then:
$C = \left[ {{c_{ij}}} \right]$ and $A + B = C$
$\because $ $\left[ {{c_{ij}}} \right] = \left[ {{a_{ij}}} + {{b_{ij}}} \right]$ .
Complete step-by-step solution:
We have two equations
$A + B = \left[ {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right]$-----(1.1)
And
$A - 2B = \left[ {\begin{array}{*{20}{c}}
{ - 1}&1 \\
0&{ - 1}
\end{array}} \right]$-----(1.2)
Multiplying equation (1.1) by scalar $2$ , we get
$2A + 2B = \left[ {\begin{array}{*{20}{c}}
2&0 \\
2&2
\end{array}} \right]$-----(1.3)
Adding the two equations (1.2) and (1.3), we get
$(A - 2B) + (2A + 2B) = \left[ {\begin{array}{*{20}{c}}
{ - 1}&1 \\
0&{ - 1}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
2&0 \\
2&2
\end{array}} \right]$
$3A = \left[ {\begin{array}{*{20}{c}}
1&1 \\
2&1
\end{array}} \right]$
On dividing both sides by scalar $3$ , we get
$A = \left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{3}}&{\dfrac{1}{3}} \\
{\dfrac{2}{3}}&{\dfrac{1}{3}}
\end{array}} \right]$.
Hence, the correct option will be C.
Note: Only matrices with the same number of dimensions can be added to or removed from one another. Simply add the corresponding entries from the two matrices and insert the result in the appropriate location of the resulting matrix to combine them. Finally, it should be noted that adding matrices in any particular order is irrelevant; therefore, $A + B = B + A$ .
Formula Used: The addition of two matrices, $A$ and $B$ or $A + B$ is defined as: if there are two matrices, $A = \left[ {{a_{ij}}} \right]$ and $B = \left[ {{b_{ij}}} \right]$ of the same order $m \times n$ then:
$C = \left[ {{c_{ij}}} \right]$ and $A + B = C$
$\because $ $\left[ {{c_{ij}}} \right] = \left[ {{a_{ij}}} + {{b_{ij}}} \right]$ .
Complete step-by-step solution:
We have two equations
$A + B = \left[ {\begin{array}{*{20}{c}}
1&0 \\
1&1
\end{array}} \right]$-----(1.1)
And
$A - 2B = \left[ {\begin{array}{*{20}{c}}
{ - 1}&1 \\
0&{ - 1}
\end{array}} \right]$-----(1.2)
Multiplying equation (1.1) by scalar $2$ , we get
$2A + 2B = \left[ {\begin{array}{*{20}{c}}
2&0 \\
2&2
\end{array}} \right]$-----(1.3)
Adding the two equations (1.2) and (1.3), we get
$(A - 2B) + (2A + 2B) = \left[ {\begin{array}{*{20}{c}}
{ - 1}&1 \\
0&{ - 1}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
2&0 \\
2&2
\end{array}} \right]$
$3A = \left[ {\begin{array}{*{20}{c}}
1&1 \\
2&1
\end{array}} \right]$
On dividing both sides by scalar $3$ , we get
$A = \left[ {\begin{array}{*{20}{c}}
{\dfrac{1}{3}}&{\dfrac{1}{3}} \\
{\dfrac{2}{3}}&{\dfrac{1}{3}}
\end{array}} \right]$.
Hence, the correct option will be C.
Note: Only matrices with the same number of dimensions can be added to or removed from one another. Simply add the corresponding entries from the two matrices and insert the result in the appropriate location of the resulting matrix to combine them. Finally, it should be noted that adding matrices in any particular order is irrelevant; therefore, $A + B = B + A$ .
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