Question

If\[A = \left[ {\begin{array}{*{20}{c}}{ - 1}&2\\2&{ - 1}\end{array}} \right]\], \[B = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\], and \[AX = B\]. Then find the value of \[X\].
A. \[\left[ {\begin{array}{*{20}{c}}5&7\end{array}} \right]\]
B. \[\dfrac{1}{3}\left[ {\begin{array}{*{20}{c}}5\\7\end{array}} \right]\]
C. \[\dfrac{1}{3}\left[ {\begin{array}{*{20}{c}}5&7\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}5\\7\end{array}} \right]\]

Answer 1

Hint: First, solve the given equation \[AX = B\] for the value of \[X\] by pre-multiplying both sides by \[{A^{ - 1}}\]. Then calculate the determinant of the matrix \[A\]. If the value of the determinant is non-zero. Then calculate the adjoint matrix of the given matrix and substitute the values in the formula for the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]. Substitute the values of the determinant and adjoint matrix in the equation for \[X\]. Solve the equation by using the scalar and matrix multiplication methods and get the required answer.

Formula used:
The adjoint matrix of a \[2 \times 2\] matrix \[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] is: \[adj A = \left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]
The determinant of a \[2 \times 2\] matrix \[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] is: \[\left| A \right| = ad - bc\]
The inverse matrix: \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]

Complete step by step solution:
The given matrices are \[A = \left[ {\begin{array}{*{20}{c}}{ - 1}&2\\2&{ - 1}\end{array}} \right]\], \[B = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\], and \[AX = B\].

Let’s simplify the given equation \[AX = B\].
Pre-multiply both sides by \[{A^{ - 1}}\].
\[{A^{ - 1}}AX = {A^{ - 1}}B\]
\[ \Rightarrow IX = {A^{ - 1}}B\] , where \[I\] is an identity matrix.
\[ \Rightarrow X = {A^{ - 1}}B\]
\[ \Rightarrow X = \dfrac{1}{{\left| A \right|}}\left( {adj A} \right)B\] \[.....\left( 1 \right)\]
Since the product of a matrix and an identity matrix is an original matrix.

Now calculate the value of \[{A^{ - 1}}\].
To find the inverse matrix, first, calculate the determinant of the matrix \[A\].
Apply the formula for the determinant of a \[2 \times 2\] matrix.
We get,
\[\left| A \right| = \left( { - 1} \right) \times \left( { - 1} \right) - 2 \times 2\]
\[ \Rightarrow \left| A \right| = 1 - 4\]
\[ \Rightarrow \left| A \right| = - 3\] \[.....\left( 2 \right)\]
Since the value of the determinant is non-zero. So, the inverse matrix for the given matrix \[A\] exists.

Now find out the adjoint matrix of the given matrix \[A\].
Apply the rule for the adjoint matrix of a \[2 \times 2\] matrix.
We get,
\[adj A = \left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 2}&{ - 1}\end{array}} \right]\] \[.....\left( 3 \right)\]

Substitute the values of the equation \[\left( 2 \right)\] and \[\left( 3 \right)\] in the equation \[\left( 1 \right)\].
We get,
\[X = \dfrac{1}{{ - 3}}\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\{ - 2}&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\]
Simplify the right-hand side of the above equation by using the scalar multiplication and matrix multiplication methods.
\[X = \left[ {\begin{array}{*{20}{c}}{\dfrac{1}{3}}&{\dfrac{2}{3}}\\{\dfrac{2}{3}}&{\dfrac{1}{3}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\]
\[ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}{\dfrac{1}{3} \times 3 + \dfrac{2}{3} \times 1}\\{\dfrac{2}{3} \times 3 + \dfrac{1}{3} \times 1}\end{array}} \right]\]
\[ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}{1 + \dfrac{2}{3}}\\{2 + \dfrac{1}{3}}\end{array}} \right]\]
 \[ \Rightarrow X = \left[ {\begin{array}{*{20}{c}}{\dfrac{5}{3}}\\{\dfrac{7}{3}}\end{array}} \right]\]
\[ \Rightarrow X = \dfrac{1}{3}\left[ {\begin{array}{*{20}{c}}5\\7\end{array}} \right]\]
Hence the correct option is B.

Note: Students should keep in mind that the product of two matrices is defined if the number of columns of the first matrix is equal to the number of rows of the second matrix. And the adjoint matrix of any matrix is the transpose of its cofactor matrix. But for a \[2 \times 2\] matrix, we don’t need to calculate the cofactor matrix.