Question

There are two possible values of $A$ in the Solution of the Matrix Equation

{2A + 1}&{ - 5} \\\ { - 4}&A; \end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}} {A - 5}&B; \\\ {2A - 2}&C; \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {14}&D; \\\ E&F; \end{array}} \right]$$ Where $A,B,C,D,E,F$ are Real Numbers. The absolute value of the difference between these two solutions is $ a)\dfrac{8}{3} \\\ b)\dfrac{{11}}{3} \\\ c)\dfrac{1}{3} \\\ d)\dfrac{{19}}{3} \\\ $

Answer 1

To solve this type of problem we should know about the Inverse of the matrix. Consider a matrix $A$, The inverse ${A^{ - 1}}$ of the Matrix $A$ can be found out by using the Formula
${A^{ - 1}} = \dfrac{1}{{|A|}}adjA$
Where $|A|$ is the determinant of a matrix $A$
$adjA$ is the adjugate or adjoint of a matrix $A$.
Consider a Square Matrix A = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]
The adjugate of this Matrix $A$ is \left[ {\begin{array}{*{20}{c}} d&{ - b} \\\ { - c}&a; \end{array}} \right]

Complete step by step answer:
In this problem, we are going to find the two possible values of $A$. After finding the Values of $A$, we calculate the difference between the two values of $A$.
Let us find the two Possible values of $A$.
It is given in the problem that
{\left[ {\begin{array}{*{20}{c}} {2A + 1}&{ - 5} \\\ { - 4}&A; \end{array}} \right]^{ - 1}}\left[ {\begin{array}{*{20}{c}} {A - 5}&B; \\\ {2A - 2}&C; \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {14}&D; \\\ E&F; \end{array}} \right]\xrightarrow[{\begin{array}{*{20}{c}} {}&{} \end{array}}]{} (1)
We know the formula for finding the inverse of a matrix $A$ is
${A^{ - 1}} = \dfrac{1}{{|A|}}adjA$
Consider a matrix A = \left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]
We know $|A| = ad - bc$
{\left[ {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right]^{ - 1}} = \dfrac{1}{{ad - bc}}adjA
= \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}} d&{ - b} \\\ { - c}&a; \end{array}} \right]

{2A + 1}&{ - 5} \\\ { - 4}&A; \end{array}} \right]^{ - 1}}$$ Substitute $a = 2A + 1,b = - 5,c = - 4,d = A$ ${\left[ {\begin{array}{*{20}{c}} {2A + 1}&{ - 5} \\\ { - 4}&A; \end{array}} \right]^{ - 1}} = \dfrac{1}{{A(2A + 1) - 20}}\left[ {\begin{array}{*{20}{c}} A&5 \\\ 4&{2A + 1} \end{array}} \right]$ $$ = \dfrac{1}{{2{A^2} + A - 20}}\left[ {\begin{array}{*{20}{c}} A&5 \\\ 4&{2A + 1} \end{array}} \right]$$ Substitute the above values in Equation $(1)$ $$\dfrac{1}{{2{A^2} + A - 20}} \left[ {\begin{array}{*{20}{c}} A&5 \\\ 4&{2A + 1} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {A - 5}&B; \\\ {2A - 2}&C; \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {14}&D; \\\ E&F; \end{array}} \right]$$ Multiply the above two matrices on the left-hand side. $\dfrac{1}{{2{A^2} + A - 20}} \left[ {\begin{array}{*{20}{c}} {A(A - 5) + 5(2A - 2)}&{AB + 5C} \\\ {4(A - 5) + (2A + 1)(2A - 2)}&{4B + (2A + 1)C} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {14}&D; \\\ E&F; \end{array}} \right]$ Form an Equation by taking the first element of Matrix on both Sides $$\dfrac{1}{{2{A^2} + A - 20}}(A(A - 5) + 5(2A - 2)) = 14$$ Simplify the above equation $$A(A - 5) + 5(2A - 2) = 14(2{A^2} + A - 20)$$ We can simplify the above equation again $ {A^2} - 5A + 10A - 10 = 28{A^2} + 14A - 280 \\\ 28{A^2} - {A^2} + 14A - 5A - 280 + 10 = 0 \\\ 27{A^2} + 9A - 270 = 0 \\\ 9(3{A^2} + A - 30) = 0 \\\ 3{A^2} + A - 30 = 0 \\\ $ Simplify the above quadratic equation $ 3{A^2} - 9A + 10A - 30 = 0 \\\ 3A(A - 3) + 10(A - 3) = 0 \\\ (A - 3)(3A + 10) = 0 \\\ (A - 3) = 0,(3A + 10) = 0 \\\ A = 3,A = \dfrac{{ - 10}}{3} \\\ $ The possible values of $A$ are $3$ and $\dfrac{{ - 10}}{3}$ The difference between these two solutions $ = 3 - (\dfrac{{ - 10}}{3})$ $ = 3 + \dfrac{{10}}{3} \\\ = \dfrac{{9 + 10}}{3} = \dfrac{{19}}{3} \\\ $ Its absolute value is $\dfrac{{19}}{3}$ **So, the correct answer is “Option d”.** **Note:** We should find the adjugate of the matrix correctly by interchanging the elements in the diagonal and change the Sign of the remaining elements. If we made any mistakes in the change of sign, the answer will be wrong. To solve this problem easily, we should know the Inverse of the Matrix Formula $${A^{ - 1}} = \dfrac{1}{{|A|}}adjA$$ Where $|A|$ is the determinant of a matrix $A$, $adjA$ is the adjugate or adjoint of a matrix $A$.