Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Matrices

Question:

Let $A=\begin{bmatrix}1&0&0\\2&1&0\\3&2&1\end{bmatrix}$ and $U_1,U_2,U_3$ be column matrices satisfying $AU_1=\begin{bmatrix}1\\0\\0\end{bmatrix},AU_2=\begin{bmatrix}2\\3\\6\end{bmatrix},AU_3=\begin{bmatrix}2\\3\\1\end{bmatrix}$. If $U$ is 3 × 3 matrix whose columns are $U_1,U_2,U_3$, then $|U|=$

Options:

3

-3

3/2

2

Correct Answer:

3

Explanation:

Let $U_1=\begin{bmatrix}a\\b\\c\end{bmatrix},U_2=\begin{bmatrix}p\\q\\r\end{bmatrix}$ and $U_3=\begin{bmatrix}x\\y\\z\end{bmatrix}$. Then,

$AU_1=\begin{bmatrix}1\\0\\0\end{bmatrix}⇒\begin{bmatrix}a\\2a+b\\3a+2b+c\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix}⇒a=1,b=-2, c=1$

$AU_2=\begin{bmatrix}2\\3\\6\end{bmatrix}⇒\begin{bmatrix}p\\2p+q\\3p+2q+r\end{bmatrix}=\begin{bmatrix}2\\3\\6\end{bmatrix}⇒p=2, q=-1, r = -4$

and,

$AU_3=\begin{bmatrix}2\\3\\1\end{bmatrix}⇒\begin{bmatrix}x\\2x+y\\3x+2y+z\end{bmatrix}⇒x=2, y=-1, z=-3$

$∴U\begin{bmatrix}U_1&U_2&U_3\end{bmatrix}=\begin{bmatrix}1&2&2\\-2&-1&-1\\1&-4&-3\end{bmatrix}$

$⇒|U|=\begin{vmatrix}1&2&2\\-2&-1&-1\\1&-4&-3\end{vmatrix}=\begin{vmatrix}1&2&0\\-2&-1&0\\1&-4&1\end{vmatrix}$ Applying $C_3→ C_3-C_2$

$⇒|U|=3$