Question

If A = $\left[ \begin{matrix}
   4 & 6 & -1 \\
   3 & 0 & 2 \\
   1 & -2 & 5 \\
\end{matrix} \right]$ , B = $\left[ \begin{matrix}
   2 \\
   0 \\
   -1 \\
\end{matrix}\,\,\,\,\begin{matrix}
   4 \\
   1 \\
   2 \\
\end{matrix} \right]$ , C = $\left[ \begin{matrix}
   3 \\
   1 \\
   2 \\
\end{matrix}\, \right]$ then the expression which is not defined is
A . ${{A}^{2}}+2B-2A$
B . $CC'$
C . $B'C$
D . $AB$

Answer 1

Hint: in this question, we have to find the option which is not defined. For this , we solve all the given options and find out the option which is not defined and choose the correct option.
Complete Step- by- Step Solution:
We have given the matrices A = $\left[ \begin{matrix}
   4 & 6 & -1 \\
   3 & 0 & 2 \\
   1 & -2 & 5 \\
\end{matrix} \right]$ , B = $\left[ \begin{matrix}
   2 \\
   0 \\
   -1 \\
\end{matrix}\,\,\,\,\begin{matrix}
   4 \\
   1 \\
   2 \\
\end{matrix} \right]$ and C = $\left[ \begin{matrix}
   3 \\
   1 \\
   2 \\
\end{matrix}\, \right]$
Which are of different order.
To find out the correct option, we will check all the options
First option is ${{A}^{2}}+2B-2A$
${{A}^{2}}$= $\left[ \begin{matrix}
   4 & 6 & -1 \\
   3 & 0 & 2 \\
   1 & -2 & 5 \\
\end{matrix} \right]$ $\left[ \begin{matrix}
   4 & 6 & -1 \\
   3 & 0 & 2 \\
   1 & -2 & 5 \\
\end{matrix} \right]$
Multiply both the matrix, we get
${{A}^{2}}$= $\left[ \begin{matrix}
   16+18-1 & 24+0+2 & -4+12-5 \\
   12+0+2 & 18+0-4 & -3+0+10 \\
   4-6+5 & 6+0-10 & -1-4+25 \\
\end{matrix} \right]$
Simplify further, we get
${{A}^{2}}$= $\left[ \begin{matrix}
   33 & 26 & 3 \\
   14 & 14 & 7 \\
   3 & -4 & 20 \\
\end{matrix} \right]$
Now we find ${{A}^{2}}+2B-2A$
${{A}^{2}}+2B-2A$= $\left[ \begin{matrix}
   33 & 26 & 3 \\
   14 & 14 & 7 \\
   3 & -4 & 20 \\
\end{matrix} \right]$+ 2 $\left[ \begin{matrix}
   2 \\
   0 \\
   -1 \\
\end{matrix}\,\,\,\,\begin{matrix}
   4 \\
   1 \\
   2 \\
\end{matrix} \right]$ - 2 $\left[ \begin{matrix}
   4 & 6 & -1 \\
   3 & 0 & 2 \\
   1 & -2 & 5 \\
\end{matrix} \right]$
We cannot add the above matrices as they are not of same order.
So ${{A}^{2}}+2B-2A$ is not defined.
Now we solve Option [ B ]
We have to find CC’
C = $\left[ \begin{matrix}
   3 \\
   1 \\
   2 \\
\end{matrix}\, \right]$ and then C’ = $\left[ \begin{matrix}
   3 & 1 & 2 \\
\end{matrix} \right]$
Multiply both matrix, we get
C'C = $\left[9+1+4\right]=\left[14\right]$
Now we solve Option [ C ]
Given B = $\left[ \begin{matrix}
   2 \\
   0 \\
   -1 \\
\end{matrix}\,\,\,\,\begin{matrix}
   4 \\
   1 \\
   2 \\
\end{matrix} \right]$ then B’ = $\left[ \begin{align}
  & \begin{matrix}
   2 & 0 & -1 \\
\end{matrix} \\
 & \begin{matrix}
   4 & 1 & 2 \\
\end{matrix} \\
\end{align} \right]$
And C = $\left[ \begin{matrix}
   3 \\
   1 \\
   2 \\
\end{matrix}\, \right]$
Then B’C = $\left[ \begin{align}
  & \begin{matrix}
   2 & 0 & -1 \\
\end{matrix} \\
 & \begin{matrix}
   4 & 1 & 2 \\
\end{matrix} \\
\end{align} \right]$$\left[ \begin{matrix}
   3 \\
   1 \\
   2 \\
\end{matrix}\, \right]$
B’C = $\left[ \begin{align}
  & 4 \\
 & 17 \\
\end{align} \right]$
Now we solve Option [ D ]
AB = $\left[ \begin{matrix}
   4 & 6 & -1 \\
   3 & 0 & 2 \\
   1 & -2 & 5 \\
\end{matrix} \right]$$\left[ \begin{matrix}
   2 \\
   0 \\
   -1 \\
\end{matrix}\,\,\,\,\begin{matrix}
   4 \\
   1 \\
   2 \\
\end{matrix} \right]$
AB = $\left[ \begin{matrix}
   9 \\
   4 \\
   -3 \\
\end{matrix}\,\,\,\,\begin{matrix}
   20 \\
   16 \\
   12 \\
\end{matrix} \right]$
Hence we see that all the options are defined except Option [ A ]

Thus, Option ( A) is correct.

Note: In these types of questions, Students made mistakes in multiplying the matrices. Matrices are the set of numbers which are arranged in rows and columns to make a rectangular array. In Multiplication of matrices, remember that the number of column of first matrix match the number of rows of second matrix. In multiplication of matrices, if we take $A[4\times 3]$and $B[3\times 4]$multiplication of both the matrix is possible but If we take $A[4\times 3]$and $B[4\times 3]$multiplication of matrix is not possible.