# TIFR-2015-Maths-A-1

324 views

Let $A$ be an invertible $10 \times 10$ matrix with real entries such that the sum of each row is $1$. Then

1. The sum of the entries of each row of the inverse of $A$ is $1$.
2. The sum of the entries of each column of the inverse of $A$ is $1$.
3. The trace of the inverse of $A$ is non-zero.
4. None of the above.
0
I guess answer is option "C"

is it right?

Ans: 1. The sum of the entries of each row of the inverse of $A$ is $1$.

Explanation:

Let me first define a vector $e$, which is a vector of all $1$s. (Vector is nothing but a column matrix)

so,

$e = \begin{bmatrix} 1\\ 1\\ ..\\ 1 \end{bmatrix}_{10 \times 1}$

Now, what will be $Ae =\ ?$

It turns out, that since the sum of each row of $A$ is $1$, $Ae$ will be equal to $e$. This is an important point.

so,

$Ae = e$

Multiply both sides by $A^{-1}$

$\implies A^{-1}Ae = A^{-1}e$

$\implies e = A^{-1}e$

$\implies A^{-1}e = e$

The above equation means that when we multiply $A^{-1}$ with $e$, we get $e$. What this implies?

This means that the sum of each row in $A^{-1}$ is $1$.

selected by
0
Wow.... simply amazing and extraordinary answer....
0
is there any simple method than this.

It will be too difficult to guess in exam

## Related questions

1
186 views
Let $A$ be the $2 \times 2$ matrix $\begin{pmatrix} \sin\frac{\pi}{18}&-\sin \frac{4\pi}{9} \\ \sin \frac{4\pi}{9}&\sin \frac {\pi}{18} \end{pmatrix}$. Then the smallest number $n \in \mathbb{N}$ such that $A^{n}=1$ is. $3$ $9$ $18$ $27$
1 vote
Let $A$ be a $10 \times 10$ matrix with complex entries such that all its eigenvalues are non-negative real numbers, and at least one eigenvalue is positive. Which of the following statements is always false ? There exists a matrix $B$ such that $AB-BA = B$. There exists a matrix $B$ such ... $AB-BA = A$. There exists a matrix $B$ such that $AB+BA=A$. There exists a matrix $B$ such that $AB+BA=B$.
Let $n \geq 1$ and let $A$ be an $n \times n$ matrix with real entries such that $A^{k}=0$, for some $k \geq 1$. Let $I$ be the identity $n \times n$ matrix. Then. $I+A$ need not be invertible. Det $(I+A)$ can be any non-zero real number. Det $(I+A) = 1$ $A^{n}$ is a non-zero matrix.
Let $X=\left\{(x, y) \in \mathbb{R}^{2}: 2x^{2}+3y^{2}= 1\right\}$. Endow $\mathbb{R}^{2}$ with the discrete topology, and $X$ with the subspace topology. Then. $X$ is a compact subset of $\mathbb{R}^{2}$ in this topology. $X$ is a connected subset of $\mathbb{R}^{2}$ in this topology. $X$ is an open subset of $\mathbb{R}^{2}$ in this topology. None of the above.