Question

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{a. If A is an indempotent matrix and I is an}& \\ \text{identity matrix of the same order, then the}& \\ \text{value of}n,\text{such that}(A+I{)}^n=I+127A\text{is}& p.9\\ \text{b. If}(I-A{)}^{-1}=I+A+A^2+....+A^7,& \\ \text{then}{A}^n=0\text{where}n\text{is}& \text{q.}10\\ \text{c. If A is matrix such that}{a}_{ij}=(i+j)(i-j),& \\ \text{then A is singular if order of marix is}& r.7\\ \text{d. If a non-singular matrix A is symmetric,}& \\ \text{show that}{A}^{-1}\text{is also symmetric, then order}& \\ \text{of A can be}& s.8\end{array}$
Which of the following options is/are correct?

• A. $a\to r;b\to s;c\to ,r;d\to p$
• B. $a\to r;b\to p,q,s;c\to ,p,r;d\to p,q,r,s.$
• C. $a\to r;b\to p,,s;c\to ,p,r;d\to r,s.$
• D. $a\to s;b\to p,q,s;c\to ,p,r;d\to q,r,s.$

Answer 1

The correct option is B $a\to r;b\to p,q,s;c\to ,p,r;d\to p,q,r,s.$

$a.$
Since, $A$ is idempotent,
$A^2=A^3=A^4=....=A.$
Now,
$(A+I{)}^n=I{+}^n{C}_{1}A{+}^n{C}_2{A}^2+...{+}^n{C}_n{A}^n$
$=I{+}^n{C}_{1}A{+}^n{C}_{2}A+...{+}^n{C}_{n}A$
$=I{+}^n{C}_{1}A{+}^n{C}_{2}A+...{+}^n{C}_{n}A$
$=I+{(}^n{C}_1{+}^n{C}_2+...{+}^n{C}_n)A$
$=I+(2^n-1)A$
$\Rightarrow 2^n-1=127$
$\Rightarrow n=7$

$b.$
We have,
$(I-A)(I+A+A^2+...+A^7)$
$=I+A+A^2+...+A^7+(-A-A^2-A^3-A^4....-A^8)$
$=I-A^8$
$=I(\text{if}{A}^8=0)$

$c.$
Here matrix $A$ is skew-symmetric and since $|A|=|A^T|$
$\Rightarrow |A|=(-1{)}^n|A|$
$\Rightarrow |A|(1-(-1{)}^n)=0$
As $n$ is odd, hence $|A|=0.$
Hence, $A$ is singular.

$d.$
If $A$ is symmetric, $A^{-1}$ is also symmetric for matrix of any order.