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Let $A$ be am $n\times n$ invertible matrix with real entries whose column sums are all equal to $1$. Consider the following statements:

  1. Every column in the matrix $A^{2}$ sums to $2$
  2. Every column in the matrix $A^{3}$ sums to $3$
  3. Every column in the matrix $A^{-1}$ sums to $1$

Which of the following is TRUE?

  1. none of the statements $(1),(2),(3)$ is correct
  2. statement $(1)$ is correct but not statements $(2)$ or $(3)$
  3. statement $(2)$ is correct but not statements $(1)$ or $(3)$
  4. statement $(3)$ is correct but not statements $(1)$ or $(2)$
  5. all the $3$ statements $(1),(2),$ and $(3)$ are correct
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1 Answer

4 votes
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Let $A=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$

$A^{2} = A \cdot A = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}\cdot\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$

  • Every column in the matrix $A^{2}$ sums to $2.\implies \text{False}$

 $A^{3} = A^{2} \cdot A = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}\cdot\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$

  • Every column in the matrix $A^{3}$ sums to $3.\implies \text{False}$

$A^{-1} = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$ 

Every column in the matrix $A^{-1}$ sums to $1.\implies \text{True}$

Lets take another example for statement $3:$

$A = \begin{bmatrix} 3&-6 \\ -2 &7 \end{bmatrix}$ 

$A = \begin{bmatrix} a &b \\ c &d \end{bmatrix}^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix} d &-b \\ -c &a \end{bmatrix}$

In other words: swap the positions of $a$ and $d,$ put negatives in front of $b$ and $c,$ and divide everything by the determinant $(ad-bc).$

$A^{-1} = \dfrac{1}{9}\begin{bmatrix}7 & 6\\ 2 &3 \end{bmatrix}$

  • Every column in the matrix $A^{-1}$ sums to $1.\implies \text{True}$

So, the correct answer is $(D).$

edited by

4 Comments

3rd statement can be proved like :-

If every column in $A$ sums to 1

Then $A^tx=x$ is always true. Here, $x$ is a vector

$\Rightarrow A^{t^{-1}}A^t x = A^{t^{-1}}x$

$\Rightarrow  x= A^{t^{-1}}x = A^{-1^{t}} x$

Which means $A^tx = (A^{-1})^tx = x$

So, if columns of matrix  $A$ sums to 1 then columns of matrix $A^{-1}$ also sums to 1
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@Lakshman Patel RJIT, you need to make a correction in your second array, $A^{-1}$ makes the column sum to 1, So answer is D not A.
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@Lakshman Bhaiya you have written false in the last second line. It should be true. Just a minor mistake I guess.

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@psnt_rwt

Corrected now, thanks

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