Question

Let $B = {A^3} - 2{A^2} + 3A - I$ where $I$ is a unit matrix and $A = \left[ {\begin{array}{*{20}{c}}
  1&3&2 \\
  2&0&3 \\
  1&{ - 1}&1
\end{array}} \right]$ then the transpose of matrix $B$ is equal to
A.$\left[ {\begin{array}{*{20}{c}}
8&{14}&7 \\
{21}&1&{ - 7} \\
{14}&{21}&8
\end{array}} \right]$
B.$\left[ {\begin{array}{*{20}{c}}
2&{21}&{14} \\
{14}&1&{21} \\
7&{ - 7}&8
\end{array}} \right]$
C.$\left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right]$
D.$\left[ {\begin{array}{*{20}{c}}
3&1&0 \\
1&1&0 \\
3&1&0
\end{array}} \right]$

Answer 1

Hint: We have been given a $3 \times 3$ matrix $A$ using which we have to find the matrix $B$ as given in the equation $B = {A^3} - 2{A^2} + 3A - I$. $I$ is a unit matrix, i.e. $I = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$. We can evaluate ${A^3}$ and ${A^2}$ by multiplication of the matrix. The transpose of a matrix is a matrix where the rows and columns of a given matrix are interchanged with each other.

Complete step-by-step answer:
We have been given a matrix $A = \left[ {\begin{array}{*{20}{c}}
  1&3&2 \\
  2&0&3 \\
  1&{ - 1}&1
\end{array}} \right]$. Using this matrix we have to find the matrix $B$ using the equation $B = {A^3} - 2{A^2} + 3A - I$, where $I$ is a unit matrix.
Since $I$ is a unit matrix, $I$ in this case will be of the order same as that of matrix $A$, i.e. $3 \times 3$.
Thus, $I = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
We can find ${A^2}$ as multiplication of matrix $A$ with itself, i.e. ${A^2} = A.A$
We can perform the multiplication of the matrix to find the square as follows,
${A^2} = A.A$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  1&3&2 \\
  2&0&3 \\
  1&{ - 1}&1
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
  1&3&2 \\
  2&0&3 \\
  1&{ - 1}&1
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {(1 \times 1) + (3 \times 2) + (2 \times 1)}&{(1 \times 3) + (3 \times 0) + (2 \times - 1)}&{(1 \times 2) + (3 \times 3) + (2 \times 1)} \\
  {(2 \times 1) + (0 \times 2) + (3 \times 1)}&{(2 \times 3) + (0 \times 0) + (3 \times - 1)}&{(2 \times 2) + (0 \times 3) + (3 \times 1)} \\
  {(1 \times 1) + ( - 1 \times 2) + (1 \times 1)}&{(1 \times 3) + ( - 1 \times 0) + (1 \times - 1)}&{(1 \times 2) + ( - 1 \times 3) + (1 \times 1)}
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  {1 + 6 + 2}&{3 + 0 - 2}&{2 + 9 + 2} \\
  {2 + 0 + 3}&{6 + 0 - 3}&{4 + 0 + 3} \\
  {1 - 2 + 1}&{3 + 0 - 1}&{2 - 3 + 1}
\end{array}} \right]$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
  9&1&{13} \\
  5&3&7 \\
  0&2&0
\end{array}} \right]$
Thus, we get ${A^2} = \left[ {\begin{array}{*{20}{c}}
  9&1&{13} \\
  5&3&7 \\
  0&2&0
\end{array}} \right]$
Similarly, we can find ${A^3}$ as multiplication of matrix $A$ with ${A^2}$, i.e. ${A^3} = {A^2}.A$
We can perform the multiplication of the matrix to find the cube as follows,
${A^3} = {A^2}.A$
$ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  9&1&{13} \\
  5&3&7 \\
  0&2&0
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
  1&3&2 \\
  2&0&3 \\
  1&{ - 1}&1
\end{array}} \right]$
$ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  {(9 \times 1) + (1 \times 2) + (13 \times 1)}&{(9 \times 3) + (1 \times 0) + (13 \times - 1)}&{(9 \times 2) + (1 \times 3) + (13 \times 1)} \\
  {(5 \times 1) + (3 \times 2) + (7 \times 1)}&{(5 \times 3) + (3 \times 0) + (7 \times - 1)}&{(5 \times 2) + (3 \times 3) + (7 \times 1)} \\
  {(0 \times 1) + (2 \times 2) + (0 \times 1)}&{(0 \times 3) + (2 \times 0) + (0 \times - 1)}&{(0 \times 2) + (2 \times 3) + (0 \times 1)}
\end{array}} \right]$
$ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  {9 + 2 + 13}&{27 + 0 - 13}&{18 + 3 + 13} \\
  {5 + 6 + 7}&{15 + 0 - 7}&{10 + 9 + 7} \\
  {0 + 4 + 0}&{0 + 0 - 0}&{0 + 6 + 0}
\end{array}} \right]$
$ \Rightarrow {A^3} = \left[ {\begin{array}{*{20}{c}}
  {24}&{14}&{34} \\
  {18}&8&{26} \\
  4&0&6
\end{array}} \right]$
Thus, we get ${A^3} = \left[ {\begin{array}{*{20}{c}}
  {24}&{14}&{34} \\
  {18}&8&{26} \\
  4&0&6
\end{array}} \right]$
Now we will use the given equation $B = {A^3} - 2{A^2} + 3A - I$ to find the matrix $B$.
We can use the matrix ${A^2}$ and ${A^3}$ in the given equation to find the matrix $B$.
Here, ${A^3} = \left[ {\begin{array}{*{20}{c}}
  {24}&{14}&{34} \\
  {18}&8&{26} \\
  4&0&6
\end{array}} \right]$
     ${A^2} = \left[ {\begin{array}{*{20}{c}}
  9&1&{13} \\
  5&3&7 \\
  0&2&0
\end{array}} \right]$
     $A = \left[ {\begin{array}{*{20}{c}}
  1&3&2 \\
  2&0&3 \\
  1&{ - 1}&1
\end{array}} \right]$
$I = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]$
Thus, we can write,
\[B = {A^3} - 2{A^2} + 3A - I\]
\[ \Rightarrow B = \left[ {\begin{array}{*{20}{c}}
  {24}&{14}&{34} \\
  {18}&8&{26} \\
  4&0&6
\end{array}} \right] - 2.\left[ {\begin{array}{*{20}{c}}
  9&1&{13} \\
  5&3&7 \\
  0&2&0
\end{array}} \right] + 3.\left[ {\begin{array}{*{20}{c}}
  1&3&2 \\
  2&0&3 \\
  1&{ - 1}&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]\]
\[ \Rightarrow B = \left[ {\begin{array}{*{20}{c}}
  {24}&{14}&{34} \\
  {18}&8&{26} \\
  4&0&6
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {9 \times 2}&{1 \times 2}&{13 \times 2} \\
  {5 \times 2}&{3 \times 2}&{7 \times 2} \\
  {0 \times 2}&{2 \times 2}&{0 \times 2}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  {1 \times 3}&{3 \times 3}&{2 \times 3} \\
  {2 \times 3}&{0 \times 3}&{3 \times 3} \\
  {1 \times 3}&{ - 1 \times 3}&{1 \times 3}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]\]
\[ \Rightarrow B = \left[ {\begin{array}{*{20}{c}}
  {24}&{14}&{34} \\
  {18}&8&{26} \\
  4&0&6
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {18}&2&{26} \\
  {10}&6&{14} \\
  0&4&0
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  3&9&6 \\
  6&0&9 \\
  3&{ - 3}&3
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]\]
\[ \Rightarrow B = \left[ {\begin{array}{*{20}{c}}
  {24 - 18 + 3 - 1}&{14 - 2 + 9 - 0}&{34 - 26 + 6 - 0} \\
  {18 - 10 + 6 - 0}&{8 - 6 + 0 - 1}&{26 - 14 + 9 - 0} \\
  {4 - 0 + 3 - 0}&{0 - 4 - 3 - 0}&{6 - 0 + 3 - 1}
\end{array}} \right]\]
\[ \Rightarrow B = \left[ {\begin{array}{*{20}{c}}
  8&{21}&{14} \\
  {14}&1&{21} \\
  7&{ - 7}&8
\end{array}} \right]\]
We got the matrix \[B = \left[ {\begin{array}{*{20}{c}}
  8&{21}&{14} \\
  {14}&1&{21} \\
  7&{ - 7}&8
\end{array}} \right]\]
We can find the transpose of the matrix $B$ by interchanging the rows and columns as follows,
${B^T} = {\left[ {\begin{array}{*{20}{c}}
  8&{21}&{14} \\
  {14}&1&{21} \\
  7&{ - 7}&8
\end{array}} \right]^T} = \left[ {\begin{array}{*{20}{c}}
  8&{14}&7 \\
  {21}&1&{ - 7} \\
  {14}&{21}&8
\end{array}} \right]$
Thus, we get the resultant matrix ${B^T} = \left[ {\begin{array}{*{20}{c}}
  8&{14}&7 \\
  {21}&1&{ - 7} \\
  {14}&{21}&8
\end{array}} \right]$
Hence, option A is correct in the given question.
So, the correct answer is “Option A”.

Note: The order of the square or cube of a square matrix is same as that of the given matrix. To find the cube of a matrix, we can either use ${A^3} = {A^2}.A$ or ${A^3} = A.{A^2}$, both will yield the same result. We take the order of the unit matrix same as that of the given matrix, i.e. $3 \times 3$ in this case. The addition and subtraction of the terms containing matrices are simply done with corresponding elements.