The following problems are True or False.

Let $A$ and $B$ be $n\times n$ matrices.

(a) If $AB=B$, then $B$ is the identity matrix.
(b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.
(c) If $A$ is invertible, then $ABA^{-1}=B$.
(d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.
(e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Solution.

(a) True or False: if $AB=B$, then $B$ is the identity matrix.

False. For example, if $B$ is the zero matrix, then of course we have $AB=B$ as both sides are the zero matrix.

(b) True or False: if the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions.

False. If the coefficient matrix $A$ is invertible, the system has a unique solution $\mathbf{x}=A^{-1}\mathbf{b}$.

(c) True or False: if $A$ is invertible, then $ABA^{-1}=B$.

False. The given equality is equivalent to $AB=BA$. Even $A$ is invertible, matrix multiplication is not commutative. As a counterexample, consider
\[A=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}.\] Note that the determinant of $A$ is $\det(A)=1\neq 0$. Hence $A$ is invertible.
Yet, we have
\[AB=\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}=\begin{bmatrix}
0 & 1\\
-1& 1
\end{bmatrix}\] and
\[BA=\begin{bmatrix}
1 & 0\\
-1& 1
\end{bmatrix}\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}=\begin{bmatrix}
1 & 1\\
-1& 0
\end{bmatrix},\] and hence $AB\neq BA$.

(d) True or False: if $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix.

True. Since $A$ is idempotent, we have $A^2=A$. As $A$ is nonsingular, it is invertible. Thus, the inverse matrix $A^{-1}$ exists. Then we have
\[I=A^{-1}A=A^{-1}A^2=IA=A.\] Hence, such a matrix $A$ must be the identity matrix $I$.

(e) True or False: if $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.

Note that any homogeneous system has the zero solution. In addition to the zero solution, this system has a solution $x_1=0, x_2=0, x_3=1$. So, it has at least two solutions. The only possibilities for the number of solutions of a system of linear equations are zero, one, or infinitely many.
So, we conclude that the system must have infinitely many solutions.

The post True or False Problems on Midterm Exam 1 at OSU Spring 2018 first appeared on Problems in Mathematics.

Find the vector form solution $\mathbf{x}$ of the equation $A\mathbf{x}=\mathbf{0}$, where $A=\begin{bmatrix}
1 & 1 & 1 & 1 &2 \\
1 & 2 & 4 & 0 & 5 \\
3 & 2 & 0 & 5 & 2 \\
\end{bmatrix}$. Also, find two linearly independent vectors $\mathbf{x}$ satisfying $A\mathbf{x}=\mathbf{0}$.

Solution.

Find the vector form solution $\mathbf{x}$ of the equation $A\mathbf{x}=\mathbf{0}$

We reduce the augmented matrix as follows:
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrrr|r}
1 & 1 & 1 & 1 &2 & 0 \\
1 & 2 & 4 & 0 & 5 & 0 \\
3 & 2 & 0 & 5 & 2 & 0 \\
\end{array} \right] \xrightarrow[R_3-3R_1]{R_2-R_1}
\left[\begin{array}{rrrrr|r}
1 & 1 & 1 & 1 &2 & 0 \\
0 & 1 & 3 & -1 & 3 & 0 \\
0 & -1 & -3 & 2 & -4 & 0 \\
\end{array} \right] \\[6pt] \xrightarrow[R_3+R_2]{R_1-R_2}
\left[\begin{array}{rrrrr|r}
1 & 0 & -2 & 2 &-1 & 0 \\
0 & 1 & 3 & -1 & 3 & 0 \\
0 & 0 & 0 & 1 & -1 & 0 \\
\end{array} \right] \xrightarrow[R_2+R_3]{R_1-2R_3}
\left[\begin{array}{rrrrr|r}
1 & 0 & -2 & 0 &1 & 0 \\
0 & 1 & 3 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 & -1 & 0 \\
\end{array} \right].
\end{align*}
Hence, the solution of the system is
\begin{align*}
x_1&=2x_3-x_5\\
x_2&=-3x_3-2x_5\\
x_4&=x_5,
\end{align*}
where $x_3, x_5$ are free variables.

The vector form of the general solution is
\begin{align*}
\mathbf{x}&=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}=\begin{bmatrix}
2x_3-x_5 \\
-3x_3-2x_5 \\
x_3 \\
x_5 \\
x_5
\end{bmatrix}\\[6pt] &=x_3\begin{bmatrix}
2 \\
-3 \\
1 \\
0 \\
0
\end{bmatrix}+x_5\begin{bmatrix}
-1 \\
-2 \\
0 \\
1 \\
1
\end{bmatrix}.
\end{align*}

Find two linearly independent vectors $\mathbf{x}$ satisfying $A\mathbf{x}=\mathbf{0}$

For example, setting $x_3=1, x_5=0$, we see that $\begin{bmatrix}
2 \\
-3 \\
1 \\
0 \\
0
\end{bmatrix}$ is a solution. Similarly, setting $x_3=0, x_5=1$, we see that $\begin{bmatrix}
-1 \\
-2 \\
0 \\
1 \\
1
\end{bmatrix}$ is another solution.
It is straightforward to check that these two vectors are linearly independent.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

For the second part, some students chose the zero vector.
But note that the zero vector and another nonzero vector are always linearly dependent.

The post Find the Vector Form Solution to the Matrix Equation $A\mathbf{x}=\mathbf{0}$ first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ nonsingular matrix. Let $\mathbf{v}, \mathbf{w}$ be linearly independent vectors in $\R^n$. Prove that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

Proof.

Suppose that we have a linear combination
\[c_1(A\mathbf{v})+c_2(A\mathbf{w})=\mathbf{0},\] where $c_1, c_2$ are scalars.
Out goal is to show that $c_1=c_2=0$.
Factoring out $A$, we have
\[A(c_1\mathbf{v}+c_2\mathbf{w})=\mathbf{0}.\]

Note that since $A$ is a nonsingular matrix, the equation $A\mathbf{x}=\mathbf{0}$ has only the zero solution $\mathbf{x}=\mathbf{0}$.
The above equality yields that $c_1\mathbf{v}+c_2\mathbf{w}$ is a solution to $A\mathbf{x}=\mathbf{0}$.
Hence, we have
\[c_1\mathbf{v}+c_2\mathbf{w}=\mathbf{0}.\]

By assumption, the vectors $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, and this implies that $c_1=c_2=0$.

We have shown that whenever we have $c_1(A\mathbf{v})+c_2(A\mathbf{w})=\mathbf{0}$, we must have $c_1=c_2=0$. This yields that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is to ignore the logic and write down whatever you know.
For example, some students started with $c_1\mathbf{v}+c_2\mathbf{w}=\mathbf{0}$ and since $\mathbf{v}$ and $\mathbf{w}$ are linearly independent, we have $c_1=c_2=0$.
Multiplying by $A$, we have $c_1A\mathbf{v}+c_2A\mathbf{w}=\mathbf{0}$ and $c_1=c_2=0$.

This argument is totally wrong.

The above argument is wrong because it started with different vectors than we want to prove to be linearly independent.
There is nothing wrong about the mathematical operations in the above arguments. However, that argument does not prove that the vectors $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent.

We should first assume that $c_1A\mathbf{v}+c_2A\mathbf{w}=\mathbf{0}$ and prove that $c_1=c_2=0$.

The post If $\mathbf{v}, \mathbf{w}$ are Linearly Independent Vectors and $A$ is Nonsingular, then $A\mathbf{v}, A\mathbf{w}$ are Linearly Independent first appeared on Problems in Mathematics.

(a) Find a $3\times 3$ nonsingular matrix $A$ satisfying $3A=A^2+AB$, where \[B=\begin{bmatrix}
2 & 0 & -1 \\
0 &2 &-1 \\
-1 & 0 & 1
\end{bmatrix}.\]

(b) Find the inverse matrix of $A$.

Solution

(a) Find a $3\times 3$ nonsingular matrix $A$.

Assume that $A$ is nonsingular. Then the inverse matrix $A^{-1}$ exists.
Multiplying the given equality by $A^{-1}$ on the left, we obtain
\[3A^{-1}A=A^{-1}(A^2+AB)=A^{-1}A^2+A^{-1}AB=A+IB=A+B.\] Note that the left most term is equal to $3I$.
Hence, we have $3I=A+B$. Solving for $A$, we have
\[A=3I-B=\begin{bmatrix}
3 & 0 & 0 \\
0 &3 &0 \\
0 & 0 & 3
\end{bmatrix}-\begin{bmatrix}
2 & 0 & -1 \\
0 &2 &-1 \\
-1 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &1 \\
1 & 0 & 2
\end{bmatrix}.\]

As we see in part (b), this matrix is actually invertible.

(b) Find the inverse matrix of $A$.

To find the inverse matrix of $A$, we reduce the augmented matrix $[A\mid I]$ as follows:
\begin{align*}
[A\mid I]= \left[\begin{array}{rrr|rrr}
1 & 0 & 1 & 1 &0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
1 & 0 & 2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_3-R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 1 & 1 &0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & 0 & 1 \\
\end{array} \right]\\[6pt] \xrightarrow[R_2-R_3]{R_1-R_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 2 &0 & -1 \\
0 & 1 & 0 & 1 & 1 & -1 \\
0 & 0 & 1 & -1 & 0 & 1 \\
\end{array} \right].
\end{align*}
The left 3 by 3 part is now the identity matrix.
So the inverse matrix is given by the right 3 by 3 part:
\[A^{-1}=\begin{bmatrix}
2 & 0 & -1 \\
1 &1 &-1 \\
-1 & 0 & 1
\end{bmatrix}.\]

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is: $A=3-B$. Don’t forget the identity matrix $I$. $3$ is a number but $3I$ is a matrix.
Also $3I$ is the 3 by 3 matrix whose diagonal entries are 3 and 0 elsewhere. Note that $3I$ is not the 3 by3 matrix whose entries are all 3.

The post Find a Nonsingular Matrix $A$ satisfying A=A^2+AB$ first appeared on Problems in Mathematics.

Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.
If
\[A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=B\begin{bmatrix}
2 \\
6 \\
10
\end{bmatrix},\] then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.

Solution.

Using the given equality, we have
\begin{align*}
\mathbf{0}&=A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}-B\begin{bmatrix}
2 \\
6 \\
10
\end{bmatrix}\\[6pt] &=A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}-2B\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}\\[6pt] &=(A-2B)\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}\\[6pt] &=C\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}.
\end{align*}
Thus, we obtain
\[C\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=\mathbf{0}.\] This implies that the vector $\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}$ is a nonzero solution to the equation $C\mathbf{x}=\mathbf{0}$.
Therefore, the matrix $C$ is singular.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is that once you have
\[A\begin{bmatrix}
1 \\
3 \\
5
\end{bmatrix}=2B\begin{bmatrix}
1 \\
3\\
5
\end{bmatrix},\] then you conclude that $A=2B$, and hence $C$ is the zero matrix.

This is WRONG!! You cannot cancel matrices, like numbers!

Another common mistake is to state something like “Because $C$ has a nonzero solution…”
There is no “solution” to a matrix. We can talk about the solution of a system, for example.

The post Determine whether the Matrix is Nonsingular from the Given Relation first appeared on Problems in Mathematics.

Find all $2\times 2$ symmetric matrices $A$ satisfying $A\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=
\begin{bmatrix}
2 \\
3
\end{bmatrix}$? Express your solution using free variable(s).

Solution.

Let $A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$ be a $2\times 2$ matrix satisfying the conditions. Then as $A$ is symmetric, we have $A^{\trans}=A$. This yields that $b=c$.
So, we find all matrices $A=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}$ satisfying $A\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=
\begin{bmatrix}
2 \\
3
\end{bmatrix}$.
We have
\begin{align*}
\begin{bmatrix}
2 \\
3
\end{bmatrix}=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}\begin{bmatrix}
1 \\
-1
\end{bmatrix}
=\begin{bmatrix}
a-b \\
b-d
\end{bmatrix}.
\end{align*}
Hence, we need $a-b=2$ and $b-d=3$.
Equivalently, $a=b+2, d=b-3$. So, we have
\begin{align*}
A=\begin{bmatrix}
a & b\\
b& d
\end{bmatrix}=\begin{bmatrix}
b+2 & b\\
b& b-3
\end{bmatrix}=b\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}+\begin{bmatrix}
2 & 0\\
0& -3
\end{bmatrix},
\end{align*}
where $b$ is a free variable.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common mistake is not using the assumption that $A$ is symmetric or using wrongly.
A matrix $A$ is symmetric if $A^{\trans}=A$. For a 2 by 2 matrix, this yields that the off-diagonal entries must be the same.
However, note that the diagonal entries can be distinct. Some students assumed the same diagonal entries and concluded that there are no matrices satisfying the conditions.

The post Find All Symmetric Matrices satisfying the Equation first appeared on Problems in Mathematics.

Let
\[A=\begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix}, \quad \mathbf{u}=\begin{bmatrix}
6 \\
5 \\
-3
\end{bmatrix}, \quad \mathbf{v}=\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}, \quad \text{ and } \mathbf{w}=\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}.\]

(a) Express the vector $\mathbf{u}$ as a linear combination of $\mathbf{v}$ and $\mathbf{w}$.

(b) Compute $A^5\mathbf{v}$.

(c) Compute $A^5\mathbf{w}$.

(d) Compute $A^5\mathbf{u}$.

Solution.

(a) Express the vector $\mathbf{u}$ as a linear combination of $\mathbf{v}$ and $\mathbf{w}$.

Our goal here is to find scalars $c_1, c_2$ such that
\[\mathbf{u}=c_1\mathbf{v}+c_2\mathbf{w}.\] This is the same as the matrix equation
\[\begin{bmatrix}
-2 & -2 \\
0 & -1 \\
1 &1
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}=\begin{bmatrix}
6 \\
5 \\
-3
\end{bmatrix}.\] To solve this, we reduced the augmented matrix as follows:
\begin{align*}
\left[\begin{array}{rr|r}
-2 & -2 & 6 \\
0 &-1 &5 \\
1 & 1 & -3
\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_3}
\left[\begin{array}{rr|r}
1 & 1 & -3 \\
0 &-1 &5 \\
-2 & -2 & 6
\end{array}\right]\\[6pt] \xrightarrow[-R_2]{R_3+2R_1}
\left[\begin{array}{rr|r}
1 & 1 & -3 \\
0 &1 &-5 \\
0 & 0 & 0
\end{array}\right] \xrightarrow{R_1-R_2}
\left[\begin{array}{rr|r}
1 & 0 & 2 \\
0 &1 &-5 \\
0 & 0 & 0
\end{array}\right].
\end{align*}
This yields the solution $c_1=2$ and $c_2=-5$.
Hence, we have the linear combination
\[\mathbf{u}=2\mathbf{v}-5\mathbf{w}.\]

(b) Compute $A^5\mathbf{v}$.

We first compute $A\mathbf{v}$. We have
\[A\mathbf{v}= \begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix}
\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}
=\begin{bmatrix}
-4 \\
0 \\
2
\end{bmatrix}=2\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=2\mathbf{v}.
\] Using this relation $A\mathbf{v}=2\mathbf{v}$, we obtain
\[A^2\mathbf{v}=AA\mathbf{v}=A(2\mathbf{v})=2A\mathbf{v}=2(2\mathbf{v})=2^2\mathbf{v}.\] Next, we have
\[A^3\mathbf{v}=AA^2\mathbf{v}=A(2^2\mathbf{v})=2^2A\mathbf{v}=2^2(a\mathbf{v})=2^3\mathbf{v}.\] Repeating this process, we see that $A^5\mathbf{v}=2^5\mathbf{v}$.
Or, we can find this by computing as follows:
\begin{align*}
A^5\mathbf{v}=A^2A^3\mathbf{v}=A^2(2^3\mathbf{v})=2^3A^2\mathbf{v}=2^3(2^2\mathbf{v})=2^5\mathbf{v}.
\end{align*}
In summary, we have
\[A^5\mathbf{v}=2^5\mathbf{v}=32\begin{bmatrix}
-2 \\
0 \\
1
\end{bmatrix}=\begin{bmatrix}
-64 \\
0 \\
32
\end{bmatrix}.\]

(c) Compute $A^5\mathbf{w}$.

First, we note that
\[A\mathbf{w}= \begin{bmatrix}
-4 & -6 & -12 \\
-2 &-1 &-4 \\
2 & 3 & 6
\end{bmatrix} \begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}=-\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=-\mathbf{w}.\] Using this relation $A\mathbf{w}=-\mathbf{w}$ as in part (a), we obtain
\[A^5\mathbf{w}=(-1)^5\mathbf{w}=-\begin{bmatrix}
-2 \\
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}.\]

(d) Compute $A^5\mathbf{u}$.

Using the linear combination $\mathbf{u}=2\mathbf{v}-5\mathbf{w}$ obtained part (a), we compute
\begin{align*}
A^5\mathbf{w}&=A^5(2\mathbf{v}-5\mathbf{w})\\
&=2A^5\mathbf{v}-5A^5\mathbf{w}\\[6pt] &=2\begin{bmatrix}
-64 \\
0 \\
32
\end{bmatrix}-5\begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix} &&\text{by (b), (c)}\\[6pt] &=\begin{bmatrix}
-138 \\
-5 \\
69
\end{bmatrix}.
\end{align*}

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

One common pitfall is to compute $A^5$, but this is time consuming and it is very likely to make a mistake by hand computaiton.

The post Compute $A^5\mathbf{u}$ Using Linear Combination first appeared on Problems in Mathematics.

Consider the following system of linear equations:
\begin{align*}
ax_1+bx_2 &=c\\
dx_1+ex_2 &=f\\
gx_1+hx_2 &=i.
\end{align*}

(a) Write down the augmented matrix.

(b) Suppose that the augmented matrix is row equivalent to the identity matrix. Is the system consistent? Justify your answer.

Solution.

(a) Write down the augmented matrix.

The augmented matrix of the system is
\[\left[\begin{array}{rr|r}
a & b & c \\
d &e &f \\
g & h & i
\end{array}\right].\]

(b) Suppose that the augmented matrix is row equivalent to the identity matrix. Is the system consistent? Justify your answer.

Since the augmented matrix is row equivalent, we have
\[\left[\begin{array}{rr|r}
a & b & c \\
d &e &f \\
g & h & i
\end{array}\right] \xrightarrow{\text{elementary row operations}}
\left[\begin{array}{rr|r}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{array}\right].\] Thus the third row corresponds to the equation $0x_1+0x_2=1$, equivalently, $0=1$.
This yields that the system is inconsistent.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

For part (b), many students wrote that “the identity matrix is nonsingular, so it is consistent”.
Well, if the coefficient matrix of a system is row equivalent to the identity, then this is ture but in our case, the augmented matrix is row-equivalent to the identity matrix.

The post If the Augmented Matrix is Row-Equivalent to the Identity Matrix, is the System Consistent? first appeared on Problems in Mathematics.

Let $A, B, C$ be $n\times n$ invertible matrices. When you simplify the expression
\[C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2,\] which matrix do you get?
(a) $A$
(b) $C^{-1}A^{-1}BC^{-1}AC^2$
(c) $B$
(d) $C^2$
(e) $C^{-1}BC$
(f) $C$

Solution.

In this problem, we use the following facts about inverse matrices: if $P, Q$ are invertible matrices, then we have
\[(PQ)^{-1}=Q^{-1}P^{-1} \text{ and } (P^{-1})^{-1}=P.\]

Using these, we simplify the given expression as follows:
\begin{align*}
&C^{-1}(AB^{-1})^{-1}(CA^{-1})^{-1}C^2\\
&=C^{-1}(B^{-1})^{-1}A^{-1}(A^{-1})^{-1}C^{-1}C^2\\
&=C^{-1}BA^{-1}AC^{-1}C^2\\
&=C^{-1}BIC\\
&=C^{-1}BC,
\end{align*}
where we used the identity $A^{-1}A=I$, identity matrix, in the third step.
Thus, the answer is (e).

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

Here are two common mistakes.

The first one is to use a wrong identity $(PQ)^{-1}=P^{-1}Q^{-1}$. This is WRONG.
If you use this, then most likely you chose (b).

The seocnd one is to think matrix multiplication is commutative. Even though $PQ\neq QP$ for matrices $P, Q$, some students freely cancel terms.
In this case, (c) was chosen.

Be careful, if you have $A^{-1}BA$, then in general it is not equal to $B$. You cannot cancel $A$ becuase $A$ and $A^{-1}$ are not adjacent each other.

The post Using Properties of Inverse Matrices, Simplify the Expression first appeared on Problems in Mathematics.

Determine whether the function $T:\R^2 \to \R^3$ defined by
\[T\left(\, \begin{bmatrix}
x \\
y
\end{bmatrix} \,\right)
=
\begin{bmatrix}
x_+y \\
x+1 \\
3y
\end{bmatrix}\] is a linear transformation.

Solution.

The function $T:\R^2 \to \R^3$ is a not a linear transformation.

Recall that every linear transformation must map the zero vector to the zero vector.

However, we have
\[T\left(\, \begin{bmatrix}
0 \\
0
\end{bmatrix} \,\right)
=\begin{bmatrix}
0+0 \\
0+1 \\
3\cdot 0
\end{bmatrix}=\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \neq \begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.\] So the function $T$ does not map the zero vector $\begin{bmatrix}
0 \\
0
\end{bmatrix}$ to the zero vector $\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}$.
Thus, $T$ is not a linear transformation.

Another solution

Another way to see this is, for example, as follows.
Let
\[\mathbf{u}=\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } \mathbf{v}=\begin{bmatrix}
0 \\
1
\end{bmatrix}.\] (In fact, you may take any two vectors.)

Then we have
\[T(\mathbf{u})+T(\mathbf{v})=T\left(\, \begin{bmatrix}
1 \\
0
\end{bmatrix} \,\right)+T\left(\, \begin{bmatrix}
0 \\
1
\end{bmatrix} \,\right)
=\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}+\begin{bmatrix}
1 \\
1 \\
3
\end{bmatrix}=\begin{bmatrix}
2 \\
3 \\
3
\end{bmatrix}.\] On the other hand, we have
\[T\left(\, \mathbf{u}+\mathbf{v} \,\right) =T\left(\, \begin{bmatrix}
1 \\
1
\end{bmatrix} \,\right)
=\begin{bmatrix}
2 \\
2 \\
3
\end{bmatrix}.\]

Therefore, we see that
\[T(\mathbf{u})+T(\mathbf{v}) \neq T\left(\, \mathbf{u}+\mathbf{v} \,\right),\] and hence $T$ is not a linear transformation.

The post Is the Following Function $T:\R^2 \to \R^3$ a Linear Transformation? first appeared on Problems in Mathematics.