Question

If \[A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 4}&1
\end{array}} \right]\] , then \[adj(3{A^2} + 12A)\] is equal to
A. \[\left[ {\begin{array}{*{20}{c}}
 {72}&{ - 84} \\
 { - 63}&{51}
\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}
{51}&{63} \\
{84}&{72}
\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}
{51}&{84} \\
{63}&{72}
\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}
{72}&{ - 63} \\
{ - 84}&{51}
\end{array}} \right]\]

Answer 1

Hint: As you can see, this question is based on matrices. You must be familiar with the concept of matrices. Matric is defined as a set of numbers arranged in rows and columns so as to form a rectangular array. You are given a square matrix A and you need to find an adjoint of \[(3{A^2} + 12A)\] . The adjoint of a square matrix \[A = {\left[ {{a_{ij}}} \right]_{n \times n}}\] is defined as the transpose of the matrix \[{\left[ {{a_{ij}}} \right]_{n \times n}}\] , where \[{a_{ij}}\] is the cofactor of the element \[{a_{ij}}\] and adjoint of matrix A is denoted as adj A.

Complete step by step solution
Given: The matrix A is given as \[A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 4}&1
\end{array}} \right]\]and we need to choose the value of \[adj(3{A^2} + 12A)\] from the option mentioned in the question.
To find out the matrix \[3{A^2}\] using the matrix A.
Hence, we have,
\[A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 4}&1
\end{array}} \right]\]
To find out the matrix \[3{A^2}\], we need to multiply the matrix A with itself and then multiply it again with 3.
So,
\[
  3{A^2} = 3\left( {A \cdot A} \right) \\
   \Rightarrow 3{A^2} = 3\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 4}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 4}&1
\end{array}} \right] \\
   \Rightarrow 3{A^2} = 3\left[ {\begin{array}{*{20}{c}}
  {16}&{ - 9} \\
  { - 12}&{13}
\end{array}} \right] (Using{\text{ }}the{\text{ }}law{\text{ }}of{\text{ }}product{\text{ }}of{\text{ }}two{\text{ }}matrices) \\
   \Rightarrow 3{A^2} = \left[ {\begin{array}{*{20}{c}}
  {48}&{ - 27} \\
  { - 36}&{39}
\end{array}} \right] \\
 \]
To find out the matrix 12A
We are given, \[A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 4}&1
\end{array}} \right]\]
So,
\[
   \Rightarrow 12A = 12\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 4}&1
\end{array}} \right] \\
   \Rightarrow 12A = 12\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 4}&1
\end{array}} \right] \\
   \Rightarrow 12A = \left[ {\begin{array}{*{20}{c}}
  {24}&{ - 36} \\
  { - 48}&{12}
\end{array}} \right] \\
 \]
To find out the matrix \[3{A^2} + 12A\]
We have,
\[
   \Rightarrow 3{A^2} = \left[ {\begin{array}{*{20}{c}}
  {48}&{ - 27} \\
  { - 36}&{39}
\end{array}} \right] \,and \\
   \Rightarrow 12A = \left[ {\begin{array}{*{20}{c}}
  {24}&{ - 36} \\
  { - 48}&{12}
\end{array}} \right] \\
 \]
So, we get
\[
   \Rightarrow 3{A^2} + 12A = \left[ {\begin{array}{*{20}{c}}
  {48}&{ - 27} \\
  { - 36}&{39}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  {24}&{ - 36} \\
  { - 48}&{12}
\end{array}} \right] \\
   \Rightarrow 3{A^2} + 12A = \left[ {\begin{array}{*{20}{c}}
  {72}&{ - 63} \\
  { - 84}&{51}
\end{array}} \right] \\
 \]
To find out \[adj(3{A^2} + 12A)\]
We know that adjoint of a matrix \[P = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right]\] is given by
\[
  Adj{\text{ }}P{\text{ }} = {\text{ }}Transpose{\text{ }}of\;\left[ {\begin{array}{*{20}{c}}
  {cofacto{r_{11}}}&{cofacto{r_{12}}} \\
  {cofacto{r_{21}}}&{cofacto{r_{22}}}
\end{array}} \right] \\
  For,\,\,3{A^2} + 12A = \left[ {\begin{array}{*{20}{c}}
  {72}&{ - 63} \\
  { - 84}&{51}
\end{array}} \right] \\
 \]
We have,
\[
  72:cofacto{r_{11}} = {( - 1)^{1 + 1}} \times minor \\
     = {( - 1)^2} \times 51 \\
     = 51 \\
   - 82:cofacto{r_{21}} = {( - 1)^{2 + 1}} \times minor \\
     = {( - 1)^3} \times ( - 63) \\
     = 63 \\
   - 63:cofacto{r_{12}} = {( - 1)^{1 + 2}} \times minor \\
     = {( - 1)^3} \times ( - 84) \\
     = 84 \\
  51:cofacto{r_{22}} = {( - 1)^{2 + 2}} \times minor \\
     = {( - 1)^3} \times (72) \\
     = 72 \\
 \]
After putting the values of the cofactors in the transpose matrix, we obtain,
\[
  adj(3{A^2} + 12A) = {\text{ }}Transpose{\text{ }}of\;\left[ {\begin{array}{*{20}{c}}
  {51}&{84} \\
  {63}&{72}
\end{array}} \right] \\
  So, \\
  adj(3{A^2} + 12A) = \left[ {\begin{array}{*{20}{c}}
  {51}&{84} \\
  {63}&{72}
\end{array}} \right] \\
 \]
Hence, the correct option is B.

Note: Students often forget to find the transpose of the matrix and make mistakes. You should always be careful and find the transpose and then only, it will give you the correct value of the adjoint of that matrix. Also, not to make mistakes in multiplication of two matrices.