Register

GO Classes DPP | GATE CS, DA | Linear Algebra | Matrix

398 views
7 7 votes

As we say goodbye to $2026$ batch at GoClasses and begin with $2027$, our matrices $\mathrm{A}$ and $\mathrm{B}$ are throwing a big New Batch's party like we do at the felicitation. They're doing something special by calculating the expression $\mathrm{A}^{2027} \mathrm{~B}^{2026} v$.

Let's meet the main characters:

$$
\text { Let } \mathbf{A}=\left[\begin{array}{ll}
1 & 4 \\
0 & 2
\end{array}\right], \mathbf{B}=\left[\begin{array}{ll}
1 & -2 \\
0 & 0.5
\end{array}\right], v=\left[\begin{array}{l}
4 \\
1
\end{array}\right]
$$

What is the resultant vector of $\mathrm{A}^{2027} \mathrm{~B}^{2026} v?$
 

  1. $\left[\begin{array}{l}4 \\ 1\end{array}\right]$
  2. $\left[\begin{array}{l}8 \\ 1\end{array}\right]$
  3. $\left[\begin{array}{l}8 \\ 2\end{array}\right]$
  4. $\left[\begin{array}{l}4 \\ 2\end{array}\right]$

 

 

4 Answers

3 3 votes

As A and B are UTM , so

$A \to \lambda \to 1, 2$

$B \to \lambda' \to 1, 0.5$

Now,

$B\mathbf{v}$

$= \begin{bmatrix} 1 & -2 \\ 0 & 0.5 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \end{bmatrix}$

$= \begin{bmatrix} 4 - 2 \\ 0.5 \end{bmatrix} = \begin{bmatrix} 2 \\ 0.5 \end{bmatrix}$

$= \frac{1}{2} \begin{bmatrix} 4 \\ 1 \end{bmatrix}$

$= 0.5 \mathbf{v}$

ie, 

$B \mathbf{v} = 0.5 \mathbf{v}$

$B^{2026} \mathbf{v} = (0.5)^{2026} \mathbf{v}$ ------eq(1)

and also,

$A \mathbf{v} =$

$\begin{bmatrix} 1 & 4 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \end{bmatrix}$

$= \begin{bmatrix} 4 + 4 \\ 0 + 2 \end{bmatrix}$

$= 2 \begin{bmatrix} 4 \\ 1 \end{bmatrix}$

$A \mathbf{v} = 2 \mathbf{v}$

$A^{2027}v = 2^{2027} \mathbf{v}$ -------eq(2)

 

Now $A^{2027} (B^{2026} \mathbf{v})$

$= A^{2027} (0.5)^{2026} \mathbf{v}$ $\quad$ [using eq (1)]

$= (\frac{1}{2})^{2026} (A^{2027} \mathbf{v})$

$= 2^{-2026} \times 2^{2027} \mathbf{v}$ $\quad$ [using eq (2)]

$= 2 \mathbf{v}$

$= \begin{bmatrix} 8 \\ 2 \end{bmatrix}$

 

ie option C

1 1 vote
We first check the product $AB$:

$AB
=\begin{pmatrix}1&4\\0&2\end{pmatrix}
\begin{pmatrix}1&-2\\0&\tfrac12\end{pmatrix}
=\begin{pmatrix}
1\cdot 1+4\cdot 0 & 1\cdot(-2)+4\cdot\tfrac12\\
0\cdot 1+2\cdot 0 & 0\cdot(-2)+2\cdot\tfrac12
\end{pmatrix}
=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I.$

So $B=A^{-1}$. Therefore,

$A^{2027}B^{2026}
=A^{2027}(A^{-1})^{2026}
=A^{2027}A^{-2026}
=A.$

Hence,

$A^{2027}B^{2026}v=Av.$

Now compute $Av$:

$Av
=\begin{pmatrix}1&4\\0&2\end{pmatrix}
\begin{pmatrix}4\\1\end{pmatrix}
=\begin{pmatrix}1\cdot 4+4\cdot 1\\0\cdot 4+2\cdot 1\end{pmatrix}
=\begin{pmatrix}8\\2\end{pmatrix}.$

Therefore,
$A^{2027}B^{2026}v=\begin{pmatrix}8\\2\end{pmatrix}.$
Voting & Accepting an answer helps improve the community
Answer:
← Previous · GO Classes DPP | GAT...Next → · GO Classes DPP | GAT...
← Previous in category · GO Classes DPP | GAT...Next in category → · GO Classes DPP | GAT...
Position:
Show:

Related questions

5 5 votes
4 4 answers
449
449 views
GO Classes asked Mar 23
449 views
GO Classes DPP | GATE CS, DA | Linear Algebra | Adjoint of a Matrix
If $I$ is the unit matrix of order $n$, where $k \ne 0$ is a constant, then $\operatorname{adj}(kI)$ is$k^n I$ $k^{n-1} I$ $kI$ $I$
5 5 votes
3 3 answers
450
450 views
GO Classes asked Mar 23
450 views
GO Classes DPP | GATE CS, DA | Linear Algebra | Linear Dependence & Independence
Let $A$ be an $m \times n$-matrix. Consider the system of linear equations $A x=b$, where $x$ and $b$ are the column vectors with dimension $n \times 1$ and $m \times 1$ ...
4 4 votes
3 3 answers
430
430 views
GO Classes asked Mar 23
430 views
GO Classes DPP | GATE CS, DA | Linear Algebra | Linear independence & Orthogonality of Vectors
Consider the following three vectors:$(1,1,1)$, $(1,-2,1)$, $(3,-3,0)$ $\text{S1}:$ Given vectors are linearly independent.$\text{S2}:$ Given vectors are pairwise orthogo...
7 7 votes
3 3 answers
430
430 views
GO Classes asked Mar 23
430 views
GO Classes DPP | GATE CS, DA | Linear Algebra | Determinant
If a matrix $A$ is given by $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+....+a_{n-1}x^{n-1}+a_{n}x^{n},$then the determinant of $A$ is?$\frac{a_{0}}{a_{n}}$ $\frac{a_{n...