Let A={pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 {pmatrix}. If u₁ and u₂ are column matrices such that A u₁={pmatrix} 1 \\ 0 \\ 0 {pmatrix} and A u₂={pmatrix} 0 \\ 1 \\ 0 {pmatrix}, then u₁+u₂ is equal to : [AIEEE 2012]

Correct Answer is : (1−1−1){pmatrix} 1 \\ -1 \\ -1 {pmatrix}1−1−1

{pmatrix} -1 \\ 1 \\ 0 {pmatrix}

{pmatrix} -1 \\ 1 \\ -1 {pmatrix}

{pmatrix} -1 \\ -1 \\ 0 {pmatrix}

{pmatrix} 1 \\ -1 \\ -1 {pmatrix}

Correct Answer is : (1−1−1){pmatrix} 1 \\ -1 \\ -1 {pmatrix}1−1−1

Test Index

AIEEE 2012 Solved Paper

Section: Mathematics

Question : 75 of 89

Marks: +1, -0

Let

A=\begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}

u_1

u_2

A u_1=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}

A u_2=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

u_1+u_2

Solution:

Let

A u_1=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \;\; A u_2=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

Then,

A u_1+A u_2=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}

\Rightarrow A (u_1+u_2) =\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}

.....(1)

Also,

A=\begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{pmatrix}

\Rightarrow |A|=1(1)-0(2)+0(4-3)=1

We know,

\;A^{-1}=\;\frac{1}{|A|} \text{adj} A \Rightarrow A^{-1}=\text{adj}(A)

\;(\;\text{ as }\;|A|=1 \;\text{ ) }\;

Now, from equation (1), we have

u_1+u_2=A^{-1}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}

=\begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}

=\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}

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