Question

Find a $2 \times 2$matrix $A$, such that $A\left( {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  1&4
\end{array}} \right) = 6{I_2}$.

Answer 1

Hint: As we know that a matrix is a set of numbers that are arranged in rows and columns so to form a rectangular array and the numbers inside the matrix are called the elements or entries. We should know that $2 \times 2$ matrix is a square matrix of four numbers in it i.e. two by two matrices. Here in this question we will first assume the matrix $A$ and then we will use this value by applying with the equation and then simplify. Then we will use the corresponding elements of two equal matrices to find the required value.

Complete step-by-step answer:
As per the given question we have $A\left( {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  1&4
\end{array}} \right) = 6{I_2}$ and we have to find the value of the matrix $A$.
Let us assume the matrix $A$ is $\left( {\begin{array}{*{20}{c}}
  w&x \\
  y&z
\end{array}} \right)$,
now by using this value of $A$ we put it in the equation and we get: $\left( {\begin{array}{*{20}{c}}
  w&x \\
  y&z
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  1&4
\end{array}} \right) = 6{I_2}$.
 Also we know that ${I_2}$ is an identity matrix whose value is equal to $\left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)$.
We can further now solve it by applying the addition operation of matrices and putting the value we get:
$\left( {\begin{array}{*{20}{c}}
  {w + x}&{ - 2w + 4x} \\
  {y + z}&{ - 2y + 4z}
\end{array}} \right) = 6\left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)$
We can now put $6$inside the identity matrix by the operation of multiplication :
$\left( {\begin{array}{*{20}{c}}
  {w + x}&{ - 2w + 4x} \\
  {y + z}&{ - 2y + 4z}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  6&0 \\
  0&6
\end{array}} \right)$.
Now we know that the corresponding elements of two equal matrices are equal, we have: $w + x = 6$, we get $w = 6 - x$.
We have another equation i.e. $ - 2w + 4x = 0$, by substituting the value of $w$ in this equation we get;
$ - 2(6 - x) + 4x = 0\\
\Rightarrow - 12 + 2x + 4x = 0$.
It gives us $6x = 12 \Rightarrow x = 2$.
Since we got the value $x$, we will put its value in $w + x = 6$ and then $w = 6 - 2 = 4$.
Similarly we have $y + z = 0$, so we can write $y = - z$.
Another equation is $ - 2y + 4z = 6$, by substituting the value of $y$ we have:
$ - 2( - z) + 4z = 6\\
 \Rightarrow 2z + 4z = 6$.
 Solving this we get $z = 1$.
Putting the value of $z$ in $y = - z$ we get $y = - 1$.
Therefore the values are $w = 4,x = 2,y = - 1$ and $z = 1$.
Hence the matrix $A$ is $\left( {\begin{array}{*{20}{c}}
  4&2 \\
  { - 1}&1
\end{array}} \right)$.

Note: Before solving this kind of questions we should be familiar with matrices, their addition and multiplication properties. One of the important parts in solving this kind of question is by knowing the corresponding elements of two equal matrices are equal and we should be careful while calculating.