Question

Select all statements below which are true for all invertible n×nn×n matrices aa and bb

Answer 1

Answer:

The statements that are correct are:

c) $(in+a)(in+a^{-1})=2in+a+a^{-1}$

d) $a^2b^7$ is invertible.

e) $a+b$ is invertible.

Step-by-step explanation:

We are given that:

a and b are invertible n×n matrices.

We have to tell which of the following statements are true.

a)

$(ab)^{-1}=a^{-1}b^{-1}$

This statement is false.

Since:

$(ab)^{-1}=b^{-1}a^{-1}$ and it may not be equal to the term $a^{-1}b^{-1}$

b)

$aba{-1}=b$

This expression could also be written as:

$ab=ba$

Since on Post multiplying by a on both the sides.

But here we don't know whether the matrices are commutative or not.

Hence, the statement is false.

c)

$(in+a)(in+a^{-1})=2in+a+a^{-1}$

This statement is true.

since,

$(in+a)(in+a^{-1})=in(in+a^{-1})+a(in+a^{-1})\\\\=in^2+in.a^{-1}+a.in+aa^{-1}\\\\=in+a^{-1}+a+in\\\\=in+a^{-1}+a$

where in denote the identity matrix.

and we know that:

$in^2=in$

d)

$a^2b^7$ is invertible.

This statement is true.

Since we know that prodct of invertible matrices is also invertible.

As $a$ is invertible so is $a^2$.

Also $b$ is invertible so is $b^7$.

Hence Product of  $a^2$ and  $b^7$ is also invertible.

i.e. $a^2b^7$ is invertible.

e)

$a+b$ is invertible.

This statement is true as sum of two invertible matrices is invertible.

f)

$(a+b).(a-b)=a^2-b^2$

This statement is false.

Since,

$(a+b).(a-b)=a(a-b)+b(a-b)\\\\=a.a-a.b+b.a-b.b\\\\=a^2-ab+ba-b^2$

Now as we are not given that:

$ab=ba$

Hence, we could not say that:

$(a+b).(a-b)=a^2-b^2$

Answer 2

The correct statements which are true for all invertible $\left({n\times n}\right)\cdot\left({n\times n}\right)$ are:

(c). $\left({in+a}\right)\left({in+{a^{-1}}}\right)=2in+a+{a^{-1}}$

(d). ${a^2}{b^7}$ is invertible.

(e). $a+b$ is invertible.

Further Explanation:

Given:

The matrix $a$ and $b$ are invertible $\left({n\times n}\right)$ matrices.

Calculation:

(a)

The statement is ${\left({ab}\right)^{-1}}={a^{-1}}{b^{-1}}$ false.

$\begin{aligned}{\left({ab}\right)^{-1}}&={a^{-1}}{b^{-1}}\\\left({ab}\right){\left({ab}\right)^{-1}}&=\left({ab}\right){a^{-1}}{b^{-1}}\\I&\ne ab{a^{-1}}{b^{-1}}\\\end{aligned}$

The statement is ${\left({ab}\right)^{-1}}={a^{-1}}{b^{-1}}$ false.

(b)

The statement is $ab{a^{-1}}=b$.

Now multiply by a both the side.

$\begin{aligned}ab{a^{-1}}a&=ba\\ab&\ne ba\\\end{aligned}$

The statement is $ab{a^{-1}}=b$ is false.

(c)

The statement is $\left({in+a}\right)\left({in+{a^{-1}}}\right)=2in+a+{a^{-1}}$.

Solve the above equation to check whether it is invertible.

$\begin{aligned}\left({in+a}\right)\left({in+{a^{-1}}}\right)&=in\left({in+{a^{-1}}}\right)+a\left({in+{a^{-1}}}\right)\\&=i{n^2}+in\cdot{a^{-1}}+a\cdot in+a{a^{-1}}\\&=in+{a^{-1}}+a+in\\&=in+{a^{-1}}+a\\\end{aligned}$

The statement is true.

(d)

The statement is ${a^2}{b^7}$.

The product of invertible matrices is always invertible.

As $a$ is invertible so ${a^2}$ is also invertible.

As $b$ is invertible so ${b^7}$ is also invertible.

Hence, the product of ${a^2}$ and ${b^7}$ is also invertible.

The statement is true.

(e)

The statement $a+b$ is true as $a+b$ is always invertible.

(f)

The statement is $\left({a+b}\right)\cdot\left({a-b}\right)={a^2}-{b^2}$.

Solve the equation to check the inevitability.

$\left({a+b}\right)\times\left({a-b}\right)={a^2}-ab+ba-{b^2}$

The statement is not true as $ab=ba$.

Hence, the correct statements which are true for all invertible $\left({n\times n}\right)\cdot\left({n\times n}\right)$ are:

(c). $\left({in+a}\right)\left({in+{a^{-1}}}\right)=2in+a+{a^{-1}}$

(d). ${a^2}{b^7}$ is invertible.

(e). $a+b$ is invertible.

Learn more:

1. Learn more about unit conversion brainly.com/question/4837736

2. Learn more about non-collinear brainly.com/question/4165000

3. Learn more about binomial and trinomial brainly.com/question/1394854

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Linear equation

Keywords: Invertible, matrices, matrix, statement, function, true, determinants, elements, inverse.