Question
Let $\text{A}=\begin{bmatrix}1&0\\2&1\end{bmatrix},$ and $U_1, U_2$ are e first and second columns respectively of a $2 \times 2$ matrix $U$. Also, let the column matrices $U_1$ and $U_2$ satisfying $\text{AU}_1=\begin{bmatrix}1\\0\end{bmatrix}$ and $\text{AU}_2=\begin{bmatrix}2\\3\end{bmatrix}.$
Based on the above information, answer the following questions.
  1. The matrix $U_1 + U_2$ is equal to:
  1. $\begin{bmatrix}1\\-1\end{bmatrix}$
  2. $\begin{bmatrix}2\\-2\end{bmatrix}$
  3. $\begin{bmatrix}3\\-3\end{bmatrix}$
  4. $\begin{bmatrix}4\\-4\end{bmatrix}$
  1. The value of $|U|$ is:
  1. $2$
  2. $-2$
  3. $3$
  4. $-3$
  1. If $\text{X}=\begin{bmatrix}3&2\end{bmatrix}\text{U}\begin{bmatrix}3\\2\end{bmatrix},$ then the value of |X| =
  1. $3$
  2. $-3$
  3. $-5$
  4. $5$
  1. The minor of element at the position $a_{22}$ in $U$ is:
  1. $1$
  2. $2$
  3. $-2$
  4. $-1$
  1. If $\text{U}=[\text{a}_\text{ij}]_{2\times2},$ then the value of $a_{11}A_{11}+ a_{12}A_{12},$ where $A_{ij} $ denotes the cofactor of $a_{ij},$ is:
  1. $1$
  2. $2$
  3. $-3$
  4. $3$

Answer

  1. (c) $\begin{bmatrix}3\\-3\end{bmatrix}$
Solution:
We have, $\text{A}=\begin{bmatrix}1&0\\2&1\end{bmatrix}$
Let $\text{U}_1=\begin{bmatrix}\text{a}\\\text{b}\end{bmatrix}$ and $\text{AU}_1=\begin{bmatrix}1\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}\text{a}\\\text{b}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}\Rightarrow\begin{bmatrix}\text{a}\\2\text{a}+\text{b}\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$
$⇒ a = 1$ and $2a + b = 0 ⇒ a = 1$ and $b = -2.$
Let $\text{U}_2=\begin{bmatrix}\text{c}\\\text{d}\end{bmatrix}$ then $\text{AU}_2=\begin{bmatrix}2\\3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1&0\\2&1\end{bmatrix}\begin{bmatrix}\text{c}\\\text{d}\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}\Rightarrow\begin{bmatrix}\text{c}\\2\text{c}+\text{d}\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}$
$⇒ c = 2$ and $2c + d = 3$
$⇒ c = 2$ and $d = 3 - 4= -1$
Thus, $\text{U}_1+\text{U}_2=\begin{bmatrix}1\\-2\end{bmatrix}+\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}3\\-3\end{bmatrix}$
  1. (c) $3$
Solution:
Clearly, $\text{U}=\begin{bmatrix}1&2\\-2&-1\end{bmatrix}$
$\therefore|\text{U}|=\begin{vmatrix}1&2\\-2&-1\end{vmatrix}=-1+4=3$
  1. (d) $5$
Solution:
We have, $\text{X}=\begin{bmatrix}3&2\end{bmatrix}\begin{bmatrix}1&2\\-2&-1\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}$
$=\begin{bmatrix}3&2\end{bmatrix}\begin{bmatrix}7\\-8\end{bmatrix}=[21-16]=[5]$
$\therefore|\text{X}|=5$
  1. (a) $1$
Solution:
$a_{22}$ in U is $-1$ and its minor is $1.$
  1. (d) $3$
Solution:
Since, the sum of products of elements of any row (or column) with their corresponding cofactors is equal to the value of determinant.
$\therefore a_{11}A_{11} + a_{12}A_{12} = |U| = 3$

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