Question

The multiplicative inverse of matrix \left[ {\begin{array}{*{20}{l}} 2&1 \\\ 7&4 \end{array}} \right] .
A) \left[ {\begin{array}{*{20}{l}} 4&{ - 1} \\\ 7&2 \end{array}} \right]
B) \left[ {\begin{array}{*{20}{l}} 4&{ - 1} \\\ { - 7}&2 \end{array}} \right]
C) \left[ {\begin{array}{*{20}{l}} 4&{ - 1} \\\ { - 7}&{ - 2} \end{array}} \right]
D) \left[ {\begin{array}{*{20}{l}} { - 4}&{ - 1} \\\ 7&{ - 2} \end{array}} \right]

Answer 1

We can multiply the given matrix with the identity matrix and obtain the same matrix. Then we can apply elementary row operations matrices on both sides of the equation to make it in the form of $A \times {A^{ - 1}} = I$. On comparing the equation we obtained with the equation that gives identity matrix on multiplying a matrix with its inverse, we can obtain the inverse of the given matrix.

Complete step by step solution:
Let A = \left[ {\begin{array}{*{20}{l}} 2&1 \\\ 7&4 \end{array}} \right] and I = \left[ {\begin{array}{*{20}{l}} 1&0 \\\ 0&1 \end{array}} \right]
We know that a matrix multiplied with the identity matrix will give the same matrix.
$\Rightarrow A \times I = A$
\Rightarrow A \times \left[ {\begin{array}{*{20}{l}} 1&0 \\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 2&1 \\\ 7&4 \end{array}} \right]
Now we can do the same elementary operations on both matrices to make the matrix on the RHS the identity matrix.
We can divide the 1st row with 2.
${R_1} \to \dfrac{1}{2}{R_1}$
\Rightarrow A \times \left[ {\begin{array}{*{20}{l}} {\dfrac{1}{2}}&0 \\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1&{\dfrac{1}{2}} \\\ 7&4 \end{array}} \right]
Now we have to make the 7 to zero. For that we can subtract 7 times the 1st row from the 2nd row.
${R_2} \to {R_2} - 7{R_1}$

{\dfrac{1}{2}}&0 \\\ {\dfrac{{ - 7}}{2}}&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1&{\dfrac{1}{2}} \\\ 0&{4 - \dfrac{7}{2}} \end{array}} \right]$$ On simplification, we get $$ \Rightarrow A \times \left[ {\begin{array}{*{20}{l}} {\dfrac{1}{2}}&0 \\\ {\dfrac{{ - 7}}{2}}&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1&{\dfrac{1}{2}} \\\ 0&{\dfrac{1}{2}} \end{array}} \right]$$ Now we can multiply the 2nd row with 2. ${R_2} \to 2{R_2}$ $$ \Rightarrow A \times \left[ {\begin{array}{*{20}{l}} {\dfrac{1}{2}}&0 \\\ { - 7}&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1&{\dfrac{1}{2}} \\\ 0&1 \end{array}} \right]$$ Now we can subtract $\dfrac{1}{2}$ times 2nd row from the 1st row. ${R_1} \to {R_1} - \dfrac{1}{2}{R_2}$ $$ \Rightarrow A \times \left[ {\begin{array}{*{20}{l}} {\dfrac{1}{2} + \dfrac{7}{2}}&{0 - \dfrac{2}{2}} \\\ { - 7}&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1&{\dfrac{1}{2} - \dfrac{1}{2}} \\\ 0&1 \end{array}} \right]$$ On simplification we get, $$ \Rightarrow A \times \left[ {\begin{array}{*{20}{l}} 4&{ - 1} \\\ { - 7}&2 \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 1&0 \\\ 0&1 \end{array}} \right]$$ Now we have the identity matrix on the RHS. We know that a matrix multiplied with matrix itself will give the identity matrix. $ \Rightarrow A \times {A^{ - 1}} = I$ On comparing the equations, we get $$ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{l}} 4&{ - 1} \\\ { - 7}&2 \end{array}} \right]$$ So, the multiplicative inverse of the given matrix is $$\left[ {\begin{array}{*{20}{l}} 4&{ - 1} \\\ { - 7}&2 \end{array}} \right]$$ **So, the correct answer is option B.** **Note:** Alternate solution to this problem is given by, We know that, for a square matrix of order 2, the inverse is obtained by interchanging the diagonal matrix and changing the sign of the other 2 matrices. The inverse of the matrix $\left[ {\begin{array}{*{20}{l}} a&b; \\\ c&d; \end{array}} \right]$ is given by $\left[ {\begin{array}{*{20}{l}} d&{ - b} \\\ { - c}&a; \end{array}} \right]$ So, the inverse of the matrix $\left[ {\begin{array}{*{20}{l}} 2&1 \\\ 7&4 \end{array}} \right]$ is given by $$\left[ {\begin{array}{*{20}{l}} 4&{ - 1} \\\ { - 7}&2 \end{array}} \right]$$ So, the multiplicative inverse of the given matrix is $$\left[ {\begin{array}{*{20}{l}} 4&{ - 1} \\\ { - 7}&2 \end{array}} \right]$$