Question

How will you solve \[9a - 4b = - 5\] and \[6a - 2b = - 3\] using matrices?
\[
  A)a = - \dfrac{1}{3},b = \dfrac{1}{2} \\
  B)a = 2,b = 3 \\
  C)a = \dfrac{1}{2},b = \dfrac{1}{2} \\
  D)a = \dfrac{1}{2},b = \dfrac{3}{2} \\
 \]

Answer 1

Hint: Simultaneous equation or system of equation can be solved by basic algebra as well as by matrices. At first the method might be daunting but later on it becomes quick and easy once the preparation portion is mastered and only a few simple calculations are involved.

Complete step-by-step answer:
Since the given equation is
\[9a - 4b = - 5\] and \[6a - 2b = - 3\]
Firstly it will be written as matrices
\[\left( {\begin{array}{*{20}{c}}
  9&{ - 4} \\
  6&{ - 2}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  a \\
  b
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 5} \\
  { - 3}
\end{array}} \right)\]
No we will find the inverse matrix of \[A = \left( {\begin{array}{*{20}{c}}
  9&{ - 4} \\
  6&{ - 2}
\end{array}} \right)\]
\[
   \Rightarrow \left| A \right| = (9( - 2) - (6(4)) = - 18 + 24 = 6 \\
   \Rightarrow {A^{ - 1}} = \dfrac{1}{6}\left( {\begin{array}{*{20}{c}}
  { - 2}&4 \\
  6&9
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - \dfrac{1}{3}}&{\dfrac{2}{3}} \\
  { - 1}&{\dfrac{3}{2}}
\end{array}} \right) \\
 \]


Later on we will multiply each side of the matrix equation by the inverse matrix
\[
   \Rightarrow \left( {\begin{array}{*{20}{c}}
  { - \dfrac{1}{3}}&{\dfrac{2}{3}} \\
  { - 1}&{\dfrac{3}{2}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  9&{ - 4} \\
  6&{ - 2}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  a \\
  b
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - \dfrac{1}{3}}&{\dfrac{2}{3}} \\
  { - 1}&{\dfrac{3}{2}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  { - 5} \\
  { - 3}
\end{array}} \right) \\
   \Rightarrow \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  a \\
  b
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - \dfrac{1}{3}} \\
  {\dfrac{1}{2}}
\end{array}} \right) \\
   \Rightarrow \left( {\begin{array}{*{20}{c}}
  a \\
  b
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - \dfrac{1}{3}} \\
  {\dfrac{1}{2}}
\end{array}} \right) \\
 \]
Therefore the value of \[a = - \dfrac{1}{3}\] and value of \[b = \dfrac{1}{2}\] which means option \[A\]is correct
So, the correct answer is “Option A”.

Note: Remember to always multiply both sides of the matrix by the inverse matrix. In the matrix the numbers, symbols, or expressions are called its entries or its elements. Rows are the horizontal lines and vertical lines of entries in a matrix are called columns respectively.
 If \[A2 \times 2\] matrix is multiplied by unit matrix it remains unchanged
\[\left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)\]
A unit matrix also called an Identity Matrix is taken out if a matrix is multiplied by its inverse.
\[A \times {A^{ - 1}} = I = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\]
In order for finding inverse matrix \[({M^{ - 1}})\]of matrix \[M\]
\[M = \left( {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right)\]
Then find the determinant
\[
   \Rightarrow (|M|) = ad - bc \\
   \Rightarrow {M^{ - 1}} = \dfrac{1}{{(|M|)}}\left( {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right) \\
 \]
Later on \[a\] and \[d\]can be swooped and signs of \[b\] and \[c\]then divide by determinant