a) If $$ A^{3}=0 $$ then $$ I-A $$ is invertible. b) If $$ A=A^{-1} $$, then $$ A=I $$. c) The vectors $$ v_{1}=(1,2,-1), v_{2}=(0,4,-3) $$ and $$ v_{3}=(2,0,-1) $$ are independent. d) $$ A B A^{T} $$ is symmetric when $$ A $$ and $$ B $$ are square symmetric matrices. e) Let $$ A $$ and $$ B $$ are $$ 2 imes 2 $$ matrices where $$ B
eq 0 $$. If $$ A B=0 $$ then, $$ A=0 $$.

Question

a) If $$ A^{3}=0 $$ then $$ I-A $$ is invertible. b) If $$ A=A^{-1} $$, then $$ A=I $$. c) The vectors $$ v_{1}=(1,2,-1), v_{2}=(0,4,-3) $$ and $$ v_{3}=(2,0,-1) $$ are independent. d) $$ A B A^{T} $$ is symmetric when $$ A $$ and $$ B $$ are square symmetric matrices. e) Let $$ A $$ and $$ B $$ are $$ 2 	imes 2 $$ matrices where $$ B 
eq 0 $$. If $$ A B=0 $$ then, $$ A=0 $$.

UpStudy AI · Accepted Answer

**(a)** We are given that $$ A^3 = 0 $$. This means that $$ A $$ is nilpotent, so there exists a positive integer (here, 3) such that $$ A^3 = 0 $$. To show that $$ I-A $$ is invertible, we can use the finite geometric series:

$$
(I-A)(I+A+A^2) = I - A^3 = I-0 = I.
$$

Thus, the inverse of $$ I-A $$ is given by

$$
(I-A)^{-1} = I+A+A^2,
$$

which proves that $$ I-A $$ is invertible.

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**(b)** We are given $$ A = A^{-1} $$. Multiplying both sides by $$ A $$ leads to

$$
A^2 = I.
$$

The equation $$ A^2 = I $$ does not imply that $$ A = I $$ because it also has the solution $$ A = -I $$ (and in general other matrices may satisfy $$ A^2=I $$ without being the identity). For example, if

$$
A = -I,
$$

then $$ A = A^{-1} $$ holds because

$$
(-I)^{-1} = -I.
$$

Thus, the statement is false.

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**(c)** We are given the vectors $$ v_{1}=(1,2,-1) $$, $$ v_{2}=(0,4,-3) $$, and $$ v_{3}=(2,0,-1) $$ in $$ \mathbb{R}^3 $$. To check their independence, consider a linear combination that equals the zero vector:

$$
c_1 v_1 + c_2 v_2 + c_3 v_3 = (0,0,0).
$$

This gives the system of equations:

$$
\begin{cases}
c_1 + 2c_3 = 0, \
2c_1 + 4c_2 = 0, \
- c_1 - 3c_2 - c_3 = 0.
\end{cases}
$$

From the first equation, we have

$$
c_1 = -2c_3.
$$

Substitute into the second equation:

$$
2(-2c_3) + 4c_2 = -4c_3 + 4c_2 = 0 \quad \Longrightarrow \quad c_2 = c_3.
$$

Now substitute $$ c_1 = -2c_3 $$ and $$ c_2 = c_3 $$ into the third equation:

$$
-(-2c_3) - 3c_3 - c_3 = 2c_3 - 4c_3 = -2c_3 = 0.
$$

Thus, $$ c_3 = 0 $$, which then implies $$ c_1 = 0 $$ and $$ c_2 = 0 $$. Since the only solution is the trivial solution, the vectors are linearly independent.

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**(d)** We are to check whether $$ A B A^{T} $$ is symmetric when $$ A $$ and $$ B $$ are symmetric matrices. Since $$ B $$ is symmetric, $$ B = B^{T} $$. Also, because $$ A $$ is symmetric, $$ A^{T} = A $$. Compute the transpose:

$$
(ABA^{T})^{T} = (A^{T})^{T}B^{T}A^{T} = A B A^{T}.
$$

Thus, $$ ABA^{T} $$ is symmetric.

---

**(e)** Let $$ A $$ and $$ B $$ be $$ 2 \times 2 $$ matrices, with $$ B \neq 0 $$. The statement says if $$ AB = 0 $$, then $$ A = 0 $$. This is not necessarily true. For example, if $$ B $$ is a matrix of rank 1, it is possible for a nonzero matrix $$ A $$ to have a kernel that contains the image of $$ B $$ so that $$ AB = 0 $$. Therefore, $$ A $$ need not be zero, and the statement is false.
**(a)** If $$ A^3 = 0 $$, then $$ I - A $$ is invertible. **(b)** If $$ A = A^{-1} $$, then $$ A $$ does not necessarily equal the identity matrix. **(c)** The vectors $$ v_1 = (1,2,-1) $$, $$ v_2 = (0,4,-3) $$, and $$ v_3 = (2,0,-1) $$ are linearly independent. **(d)** If $$ A $$ and $$ B $$ are symmetric matrices, then $$ A B A^{T} $$ is symmetric. **(e)** If $$ A $$ and $$ B $$ are $$ 2 \times 2 $$ matrices with $$ B \neq 0 $$ and $$ A B = 0 $$, it does not necessarily follow that $$ A = 0 $$.

a) If then is invertible.
b) If , then .
c) The vectors and are independent.
d) is symmetric when and are square symmetric matrices.
e) Let and are matrices where . If then, .

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