Question

Find the values of a and b such that, A=B where \[A = \left[ {\begin{array}{*{20}{c}}
  {a + 4}&{3b} \\
  8&{ - 6}
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
  {2a + 2}&{{b^2} + 2} \\
  8&{{b^2} - 5b}
\end{array}} \right]\]

Answer 1

Hint: This problem is related to matrix. We are given two matrices and a particular condition such that both the matrices are the same. So to find the value of and b we will equate the elements of matrices. And then we will solve the respective equations.

Complete step-by-step answer:
Given are the two matrices A and B. Both are 2 by 2 matrices.
So we will equate the \[{a_{11}}\] elements of both the matrices to get the value of a and \[{a_{12}}\] elements to find the value of b.
So first we will go for the value of a.
We will equate the \[{a_{11}}\] of both the matrices.
\[A = \left[ {\begin{array}{*{20}{c}}
  {a + 4}&{3b} \\
  8&{ - 6}
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
  {2a + 2}&{{b^2} + 2} \\
  8&{{b^2} - 5b}
\end{array}} \right]\]
\[a + 4 = 2a + 2\]
Now taking the a terms on one side,
\[a - 2a = 2 - 4\]
\[ - a = - 2\]
Cancelling minus sign from both the sides,
\[a = 2\]
 this is the value of a.
Now we will go for the value of b.
equating \[{a_{12}}\] element,
\[3b = {b^2} + 2\]
On rearranging the terms we get,
\[{b^2} - 3b + 2 = 0\]
This is a quadratic equation of the form \[a{x^2} + bx + c = 0\]
We will solve this equation by using the factorization method. We will factor the middle term such that the product of the factors will be the third term.
\[{b^2} - 2b - b + 2 = 0\]
Taking b common from first two terms and -1 common from last two terms,
\[b\left( {b - 2} \right) - 1\left( {b - 2} \right) = 0\]
Separating the brackets,
\[\left( {b - 2} \right)\left( {b - 1} \right) = 0\]
Now there are two values of b we have found. But we need only one value and we will get that value by using the other element of the matrix that is in the b form.
So we will equate \[{a_{22}}\] elements \[ - 6 = {b^2} - 5b\] and we will put the values of bone by one.
First we will put b=2:
\[ - 6 = {b^2} - 5b\]
\[ - 6 = 4 - 10\]
\[ - 6 = - 6\]
Thus we came to conclude that b=2 is the correct value.
Now no need to check for other values.
Thus a=2 and b=2 are the correct values.
So, the correct answer is “ a=2 and b=2”.

Note: Note that, we can solve this problem just because the condition A=B is given. Also note that we have not checked for other values of b. but if checked we will get,
b=1
\[ - 6 = {b^2} - 5b\]
\[ - 6 = 1 - 5\]
\[ - 6 \ne - 4\]
Thus b=1 is not the correct value.
The elements of a matrix are arranged in a row and column form.
\[\left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}.....}&{{a_{1n}}} \\
  {{a_{21}}}&{{a_{22}}.....}&{{a_{2n}}} \\
  {{a_{m1}}}&{{a_{m2}}.....}&{{a_{mn}}}
\end{array}} \right]\]