Question

If A = \left[ {\begin{array}{*{20}{l}}1&2&3\\\\{ - 1}&1&2\\\1&2&4\end{array}} \right], then $\left( {{A^2} - 5A} \right){A^{ - 1}} =$
A.\left[ {\begin{array}{*{20}{l}}4&2&3\\\\{ - 1}&4&2\\\1&2&1\end{array}} \right]
B.\left[ {\begin{array}{*{20}{l}}{ - 4}&2&3\\\\{ - 1}&{ - 4}&2\\\1&2&{ - 1}\end{array}} \right]
C.\left[ {\begin{array}{*{20}{l}}{ - 4}&{ - 1}&1\\\2&{ - 4}&2\\\3&2&{ - 1}\end{array}} \right]
D.\left[ {\begin{array}{*{20}{l}}{ - 1}&{ - 2}&1\\\4&{ - 2}&{ - 3}\\\1&4&{ - 2}\end{array}} \right]

Answer 1

We will find ${A^2}$, $5A$ and ${A^{ - 1}}$ using matrix operations. We will substitute the values we will obtain in the expression given in the question and solve accordingly. Then we will compare the result with options that are given and choose the correct option.

Formulas used: We will use the following formulas:
1.Matrix multiplication:\left[ {\begin{array}{*{20}{l}}a&b;&c;\\\d&e;&f;\\\g&h;&i;\end{array}} \right]\left[ {\begin{array}{*{20}{l}}j&k;&l;\\\m&n;&o;\\\p&q;&r;\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}{aj + bm + cp}&{ak + bn + cq}&{al + bo + cr}\\\\{dj + em + fp}&{dk + en + fq}&{dl + eo + fr}\\\\{gj + hm + ip}&{gk + hn + iq}&{gl + ho + ir}\end{array}} \right]
2.Multiplication of a scalar $n$ with a matrix:
n\left[ {\begin{array}{*{20}{l}}a&b;&c;\\\d&e;&f;\\\g&h;&i;\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}{an}&{bn}&{cn}\\\\{dn}&{en}&{fn}\\\\{gn}&{hn}&{in}\end{array}} \right]
3.The determinant of a matrix \left[ {\begin{array}{*{20}{l}}a&b;&c;\\\d&e;&f;\\\g&h;&i;\end{array}} \right]is $a\left( {ei - fh} \right) - b\left( {di - fg} \right) + c\left( {dh - eg} \right)$

Complete step-by-step answer:
We will first find ${A^2}$. We will substitute 1 for $a$ and $j$, 2 for $b$ and $k$, 3 for $c$ and $l$, -1 for $d$ and $m$, 1 for $e$ and $n$, 2 for $f$ and $o$, 1 for $g$ and $p$, 2 for $h$ and $q$ and 4 for $i$ and $r$ in the formula \left[ {\begin{array}{*{20}{l}}a&b;&c;\\\d&e;&f;\\\g&h;&i;\end{array}} \right]\left[ {\begin{array}{*{20}{l}}j&k;&l;\\\m&n;&o;\\\p&q;&r;\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}{aj + bm + cp}&{ak + bn + cq}&{al + bo + cr}\\\\{dj + em + fp}&{dk + en + fq}&{dl + eo + fr}\\\\{gj + hm + ip}&{gk + hn + iq}&{gl + ho + ir}\end{array}} \right]:
\begin{array}{l}{A^2} = \left[ {\begin{array}{*{20}{l}}1&2&3\\\\{ - 1}&1&2\\\1&2&4\end{array}} \right]\left[ {\begin{array}{*{20}{l}}1&2&3\\\\{ - 1}&1&2\\\1&2&4\end{array}} \right]\\\\{A^2} = \left[ {\begin{array}{*{20}{l}}{1 - 2 + 3}&{2 + 2 + 6}&{3 + 4 + 12}\\\\{ - 1 - 1 + 2}&{ - 2 + 1 + 4}&{ - 3 + 2 + 8}\\\\{1 - 2 + 4}&{2 + 2 + 8}&{3 + 4 + 16}\end{array}} \right]\\\\{A^2} = \left[ {\begin{array}{*{20}{l}}2&{10}&{19}\\\0&3&7\\\3&{12}&{23}\end{array}} \right]\end{array}
We will now find $- 5A$. We will substitute $- 5$ for $n$, 1 for $a$, 2 for $b$, 3 for $c$, -1 for $d$, 1 for $e$, 2 for $f$, 1 for $g$, 2 for $h$ and 4 for $i$ in the formula n\left[ {\begin{array}{*{20}{l}}a&b;&c;\\\d&e;&f;\\\g&h;&i;\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}{an}&{bn}&{cn}\\\\{dn}&{en}&{fn}\\\\{gn}&{hn}&{in}\end{array}} \right]:
\begin{array}{l} - 5A = - 5\left[ {\begin{array}{*{20}{l}}1&2&3\\\\{ - 1}&1&2\\\1&2&4\end{array}} \right]\\\ - 5A = \left[ {\begin{array}{*{20}{l}}{ - 5}&{ - 10}&{ - 15}\\\5&{ - 5}&{ - 10}\\\\{ - 5}&{ - 10}&{ - 20}\end{array}} \right]\end{array}
We will find ${A^{ - 1}}$ using elementary row operations:
\begin{array}{l}A = AI\\\\\left[ {\begin{array}{*{20}{l}}1&2&3\\\\{ - 1}&1&2\\\1&2&4\end{array}} \right] = A\left[ {\begin{array}{*{20}{l}}1&0&0\\\0&1&0\\\0&0&1\end{array}} \right]\end{array}
We will perform the following operation on the above matrix.
$\begin{array}{l}{R_2} \to {R_2} + {R_1}\\\\{R_3} \to {R_3} - {R_1}\end{array}$
\left[ {\begin{array}{*{20}{l}}1&2&3\\\0&3&5\\\0&0&1\end{array}} \right] = A\left[ {\begin{array}{*{20}{l}}1&0&0\\\1&1&0\\\\{ - 1}&0&1\end{array}} \right]
We will perform the following operation on the above matrix.
$\begin{array}{l}{R_2} \to \dfrac{{{R_2}}}{3}\\\\{R_1} \to {R_1} - 2{R_2}\end{array}$
\left[ {\begin{array}{*{20}{l}}1&0&{ - \dfrac{1}{3}}\\\0&1&{\dfrac{5}{3}}\\\0&0&1\end{array}} \right] = A\left[ {\begin{array}{*{20}{l}}{\dfrac{1}{3}}&{ - \dfrac{2}{3}}&0\\\\{\dfrac{1}{3}}&{\dfrac{1}{3}}&0\\\\{ - 1}&0&1\end{array}} \right]
Again we will perform the following operation on the above matrix.
$\begin{array}{l}{R_1} \to {R_1} + \dfrac{{{R_3}}}{3}\\\\{R_2} \to {R_2} - \dfrac{5}{3}{R_3}\end{array}$
\left[ {\begin{array}{*{20}{l}}1&0&0\\\0&1&0\\\0&0&1\end{array}} \right] = A\left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right]
We know that $I = A{A^{ - 1}}$. We will compare this equation with the equation obtained above.
From the comparison we can conclude that {A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right].
We will substitute \left[ {\begin{array}{*{20}{l}}2&{10}&{19}\\\0&3&7\\\3&{12}&{23}\end{array}} \right] for ${A^2}$, \left[ {\begin{array}{*{20}{l}}{ - 5}&{ - 10}&{ - 15}\\\5&{ - 5}&{ - 10}\\\\{ - 5}&{ - 10}&{ - 20}\end{array}} \right] for $- 5A$ and \left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right] for ${A^{ - 1}}$ in the expression given in the question:
\begin{array}{l}\left( {{A^2} - 5A} \right){A^{ - 1}} = \left( {\left[ {\begin{array}{*{20}{l}}2&{10}&{19}\\\0&3&7\\\3&{12}&{23}\end{array}} \right] + \left[ {\begin{array}{*{20}{l}}{ - 5}&{ - 10}&{ - 15}\\\5&{ - 5}&{ - 10}\\\\{ - 5}&{ - 10}&{ - 20}\end{array}} \right]} \right)\left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right]\\\\\left( {{A^2} - 5A} \right){A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{ - 3}&0&4\\\5&{ - 2}&{ - 3}\\\\{ - 2}&2&3\end{array}} \right]\left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right]\end{array}
We will simplify this using the 1st formula. We will substitute $- 3$ for $a$, 0 for $b$, 4 for $c$, 5 for $d$, $- 2$ for $e$, $- 3$ for $f$, $- 2$ for $g$, 2 for $h$, 3 for $i$, 0 for $j$, $- \dfrac{2}{3}$ for $k$, $\dfrac{1}{3}$ for $l$, 2 for $m$, $\dfrac{1}{3}$for $n$, $- \dfrac{5}{3}$ for $o$, $- 1$ for and $p$, 0 for $q$ and 1 for and $r$ in the 1st formula:
\begin{array}{l}\left( {{A^2} - 5A} \right){A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{0 + 0 - 4}&{2 + 0 + 0}&{ - 1 + 0 + 4}\\\\{0 - 4 + 3}&{ - \dfrac{{10}}{3} - \dfrac{2}{3} + 0}&{\dfrac{5}{3} + \dfrac{{10}}{3} - 3}\\\\{0 + 4 - 3}&{\dfrac{4}{3} + \dfrac{2}{3} + 0}&{ - \dfrac{2}{3} - \dfrac{{10}}{3} + 3}\end{array}} \right]\\\\\left( {{A^2} - 5A} \right){A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{ - 4}&2&3\\\\{ - 1}&{ - 4}&2\\\1&2&{ - 1}\end{array}} \right]\end{array}
$\therefore$ Option B is the correct option.

Note: We can also find the ${A^{ - 1}}$ using minors and cofactors.
First, we will find the minors of the matrix. To find the minor of an element, we will ignore the values of its row and column and find the determinant of the remaining values.
\begin{array}{l}{M_{11}} = \left[ {\begin{array}{*{20}{l}}1&2\\\2&4\end{array}} \right]\\\\{M_{11}} = 4 - 4\\\\{M_{11}} = 0\end{array} \begin{array}{l}{M_{12}} = \left[ {\begin{array}{*{20}{l}}{ - 1}&2\\\1&4\end{array}} \right]\\\\{M_{12}} = - 4 - 2\\\\{M_{12}} = - 6\end{array} \begin{array}{l}{M_{13}} = \left[ {\begin{array}{*{20}{l}}{ - 1}&1\\\1&2\end{array}} \right]\\\\{M_{13}} = - 2 - 1\\\\{M_{13}} = - 3\end{array}
\begin{array}{l}{M_{21}} = \left[ {\begin{array}{*{20}{l}}2&3\\\2&4\end{array}} \right]\\\\{M_{21}} = 8 - 6\\\\{M_{21}} = 2\end{array} \begin{array}{l}{M_{22}} = \left[ {\begin{array}{*{20}{l}}1&3\\\1&4\end{array}} \right]\\\\{M_{22}} = 4 - 3\\\\{M_{22}} = 1\end{array} \begin{array}{l}{M_{23}} = \left[ {\begin{array}{*{20}{l}}1&2\\\1&2\end{array}} \right]\\\\{M_{23}} = 2 - 2\\\\{M_{23}} = 0\end{array}
\begin{array}{l}{M_{31}} = \left[ {\begin{array}{*{20}{l}}2&3\\\1&2\end{array}} \right]\\\\{M_{31}} = 4 - 3\\\\{M_{31}} = 1\end{array} \begin{array}{l}{M_{32}} = \left[ {\begin{array}{*{20}{l}}1&3\\\\{ - 1}&2\end{array}} \right]\\\\{M_{32}} = 2 - \left( { - 3} \right)\\\\{M_{32}} = 5\end{array} \begin{array}{l}{M_{33}} = \left[ {\begin{array}{*{20}{l}}1&2\\\\{ - 1}&1\end{array}} \right]\\\\{M_{33}} = 1 - \left( { - 2} \right)\\\\{M_{33}} = 3\end{array}
We will create a matrix of minors:
\left[ {\begin{array}{*{20}{l}}{{M_{11}}}&{{M_{12}}}&{{M_{13}}}\\\\{{M_{21}}}&{{M_{22}}}&{{M_{23}}}\\\\{{M_{31}}}&{{M_{32}}}&{{M_{33}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}0&{ - 6}&{ - 3}\\\2&1&0\\\1&5&3\end{array}} \right]
We will find the matrix of cofactors. For this, we will change the signs of alternate elements of the minor matrix. The cofactor matrix is given by:
C = \left[ {\begin{array}{*{20}{l}}0&6&{ - 3}\\\\{ - 2}&1&0\\\1&{ - 5}&3\end{array}} \right]
We will find the transpose of the matrix:
C' = \left[ {\begin{array}{*{20}{l}}0&{ - 2}&1\\\6&1&{ - 5}\\\\{ - 3}&0&3\end{array}} \right]
We will find the determinant of $A$ . We will substitute 1 for $a$, 2 for $b$, 3 for $c$, -1 for $d$, 1 for $e$, 2 for $f$, 1 for $g$, 2 for $h$ and 4 for $i$ in the 3rd formula:
$\begin{array}{l}\left| A \right| = 1\left( {4 - 4} \right) - 2\left( { - 4 - 2} \right) + 3\left( { - 2 - 1} \right)\\\\\left| A \right| = 0 - 2\left( { - 6} \right) + 3\left( { - 3} \right)\\\\\left| A \right| = 3\end{array}$
The inverse of the matrix will be $\dfrac{1}{{\left| A \right|}}C'$ :
\begin{array}{l}{A^{ - 1}} = \dfrac{1}{3}\left[ {\begin{array}{*{20}{l}}0&{ - 2}&1\\\6&1&{ - 5}\\\\{ - 3}&0&3\end{array}} \right]\\\\{A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right]\end{array}
We have calculated the inverse of the matrix using an alternate method. Now we will proceed in the same way.
We will substitute \left[ {\begin{array}{*{20}{l}}2&{10}&{19}\\\0&3&7\\\3&{12}&{23}\end{array}} \right] for ${A^2}$, \left[ {\begin{array}{*{20}{l}}{ - 5}&{ - 10}&{ - 15}\\\5&{ - 5}&{ - 10}\\\\{ - 5}&{ - 10}&{ - 20}\end{array}} \right] for $- 5A$ and \left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right] for ${A^{ - 1}}$ in the expression given in the question:
\begin{array}{l}\left( {{A^2} - 5A} \right){A^{ - 1}}\\\ = \left( {\left[ {\begin{array}{*{20}{l}}2&{10}&{19}\\\0&3&7\\\3&{12}&{23}\end{array}} \right] + \left[ {\begin{array}{*{20}{l}}{ - 5}&{ - 10}&{ - 15}\\\5&{ - 5}&{ - 10}\\\\{ - 5}&{ - 10}&{ - 20}\end{array}} \right]} \right)\left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right]\\\ = \left[ {\begin{array}{*{20}{l}}{ - 3}&0&4\\\5&{ - 2}&{ - 3}\\\\{ - 2}&2&3\end{array}} \right]\left[ {\begin{array}{*{20}{l}}0&{ - \dfrac{2}{3}}&{\dfrac{1}{3}}\\\2&{\dfrac{1}{3}}&{ - \dfrac{5}{3}}\\\\{ - 1}&0&1\end{array}} \right]\end{array}
We will simplify this using the 1st formula. We will substitute $- 3$ for $a$, 0 for $b$, 4 for $c$, 5 for $d$, $- 2$ for $e$, $- 3$ for $f$, $- 2$ for $g$, 2 for $h$, 3 for $i$, 0 for $j$, $- \dfrac{2}{3}$ for $k$, $\dfrac{1}{3}$ for $l$, 2 for $m$, $\dfrac{1}{3}$for $n$, $- \dfrac{5}{3}$ for $o$, $- 1$ for and $p$, 0 for $q$ and 1 for and $r$ in the 1st formula:
\begin{array}{l}\left( {{A^2} - 5A} \right){A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{0 + 0 - 4}&{2 + 0 + 0}&{ - 1 + 0 + 4}\\\\{0 - 4 + 3}&{ - \dfrac{{10}}{3} - \dfrac{2}{3} + 0}&{\dfrac{5}{3} + \dfrac{{10}}{3} - 3}\\\\{0 + 4 - 3}&{\dfrac{4}{3} + \dfrac{2}{3} + 0}&{ - \dfrac{2}{3} - \dfrac{{10}}{3} + 3}\end{array}} \right]\\\\\Rightarrow \left( {{A^2} - 5A} \right){A^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{ - 4}&2&3\\\\{ - 1}&{ - 4}&2\\\1&2&{ - 1}\end{array}} \right]\end{array}
$\therefore$ Option B is the correct option.