Given any constants $a,b,c$ where $a\neq 0$, find all values of $x$ such that the matrix $A$ is invertible if
\[
A=
\begin{bmatrix}
1 & 0 & c \\
0 & a & -b \\
-1/a & x & x^{2}
\end{bmatrix}
.
\]

Solution.

We know that $A$ is invertible precisely when $\det(A)\neq 0$. We therefore compute, by expanding along the first row,
\begin{align*}
\det(A)
&=
1
\begin{vmatrix}
a & -b \\ x & x^{2}
\end{vmatrix}
+c
\begin{vmatrix}
0 & a \\ -1/a & x
\end{vmatrix}
=
1(ax^{2}+bx)
+c(0+1)
\\
&=
ax^{2}+bx+c
.
\end{align*}
Thus $\det(A)\neq 0$ when $ax^{2}+bx+c\neq 0$. We know by the quadratic formula that $ax^{2}+bx+c=0$ precisely when
\[
x=
\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2}
.
\] Therefore, $A$ is invertible so long as $x$ satisfies both of the following inequalities:
\[
x\neq
\dfrac{-b+\sqrt{b^{2}-4ac}}{2}
,\quad
x\neq
\dfrac{-b-\sqrt{b^{2}-4ac}}{2}
.
\] Click here if solved 435The post Find All Values of $x$ such that the Matrix is Invertible first appeared on Problems in Mathematics.]]> https://yutsumura.com/find-all-values-of-x-such-that-the-matrix-is-invertible/feed/ 0 7035 https://yutsumura.com/find-all-eigenvalues-and-corresponding-eigenvectors-for-the-3times-3-matrix/ https://yutsumura.com/find-all-eigenvalues-and-corresponding-eigenvectors-for-the-3times-3-matrix/#respond Mon, 16 Apr 2018 03:38:31 +0000 https://yutsumura.com/?p=7012 Find all eigenvalues and corresponding eigenvectors for the matrix $A$ if \[ A= \begin{bmatrix} 2 & -3 & 0 \\ 2 & -5 & 0 \\ 0 & 0 & 3 \end{bmatrix} . \]...

Let $A$ be the matrix given by
\[
A=
\begin{bmatrix}
-2 & 0 & 1 \\
-5 & 3 & a \\
4 & -2 & -1
\end{bmatrix}
\] for some variable $a$. Find all values of $a$ which will guarantee that $A$ has eigenvalues $0$, $3$, and $-3$.

Solution.

Let $p(t)$ be the characteristic polynomial of $A$, i.e. let $p(t)=\det(A-tI)=0$. By expanding along the second column of $A-tI$, we can obtain the equation
\begin{align*}
p(t)
&=
\det\left(
\begin{bmatrix}
-2 & 0 & 1 \\
-5 & 3 & a \\
4 & -2 & -1
\end{bmatrix}
–
\begin{bmatrix}
t & 0 & 0 \\
0 & t & 0 \\
0 & 0 & t
\end{bmatrix}
\right)
\\
&=
\begin{vmatrix}
-2-t & 0 & 1 \\
-5 & 3-t & a \\
4 & -2 & -1-t
\end{vmatrix}
\\
&=
(3-t)
\begin{vmatrix}
-2-t & 1 \\
4 & -1-t
\end{vmatrix}
+2
\begin{vmatrix}
-2-t & 1 \\
-5 & a
\end{vmatrix}
\\
&=
(3-t)
\left[(-2-t)(-1-t)-4\right] +2\left[(-2-t)a+5\right] \\
&=
(3-t)
(2+t+2t+t^{2}-4)
+2(-2a-t a+5)
\\
&=
(3-t)(t^{2}+3t-2)
+(-4a-2t a+10)
\\
&=
3t^{2}+9t-6-t^{3}-3t^{2}+2t
-4a-2t a+10
\\
&=
-t^{3}+11t-2t a+4-4a
\\
&=
-t^{3}+(11-2a)t+4-4a
.
\end{align*}
For the eigenvalues of $A$ to be $0$, $3$, and $-3$, the characteristic polynomial $p(t)$ must have roots at $t=0,3,-3$. This implies
\begin{align*}
p(t)
=
– t(t-3)(t+3)
=
– t(t^{2}-9)
=
– t^{3}+ 9t.
\end{align*}

Therefore,
\[
-t^{3}+(11-2a)t+4-4a
=-t^{3}+9t
.
\]

For this equation to hold, the constant terms on the left and right hand sides of the above equation must be equal. This means that $4-4a=0$, which implies $a=1$. Hence $A$ has eigenvalues $0,3,-3$ precisely when $a=1$.

Let
\[
A=
\begin{bmatrix}
8 & 1 & 6 \\
3 & 5 & 7 \\
4 & 9 & 2
\end{bmatrix}
.
\] Notice that $A$ contains every integer from $1$ to $9$ and that the sums of each row, column, and diagonal of $A$ are equal. Such a grid is sometimes called a magic square.

Compute the determinant of $A$.

Solution.

We compute using the first row cofactor expansion
\begin{align*}
\det(A)
&=
\begin{vmatrix}
8 & 1 & 6 \\
3 & 5 & 7 \\
4 & 9 & 2
\end{vmatrix}\\[6pt] &=
8
\begin{vmatrix}
5 & 7 \\ 9 & 2
\end{vmatrix}
-1
\begin{vmatrix}
3 & 7 \\ 4 & 2
\end{vmatrix}
+6
\begin{vmatrix}
3 & 5 \\ 4 & 9
\end{vmatrix}
\\[6pt] &=
8(10-63)-(6-28)+6(27-20)\\[6pt] &=
8(-53)-(-22)+6(7)
\\[6pt] &=
-424+22+42\\[6pt] &=
-360.
\end{align*}

In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not.

(a) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=0$.

(b) The matrix $A$ is a $3 \times 3$ matrix with eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=-1$.

Hint.

Recall that the product of all the eigenvalues of $A$ is the determinant of $A$.

Solution.

(a) Eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=0$.

The matrix $A$ is not invertible.

The determinant of $A$ is the product of its eigenvalues. In this case, that means that $\det(A) = i\cdot (-i)\cdot 0 = 0$.

Because the determinant is $0$, $A$ is singular and, hence, not invertible.

(b) Eigenvalues $\lambda=i , \lambda=-i$, and $\lambda=-1$

The determinant of $A$ is the product of its eigenvalues.
In this case, that means $\det(A) = i\cdot (-i)\cdot (-1)=-1$. Because the determinant is non-zero, the matrix $A$ is non-singular, and thus is invertible.

To find an expression for $A^{-1}$, we will use the Cayley-Hamilton theorem. First we find the characteristic polynomial of $A$, which is
\[ p(\lambda) = (\lambda-i)(\lambda+i)(\lambda+1) = \lambda^3 + \lambda^2 + \lambda + 1 . \]

The Cayley-Hamilton theorem says that $A$ must satisfy the equality
\[ A^3 + A^2 + A + I = \mathbf{0} , \] where $\mathbf{0}$ is the zero matrix. Rewriting this, we have
\[ I = -A – A^2 – A^3 = A( -I – A – A^2 ) . \]

Multiplying on the left by $A^{-1}$ yields the desired equation,
\[ A^{-1} = -I – A – A^2 . \] Click here if solved 135The post Given the Data of Eigenvalues, Determine if the Matrix is Invertible first appeared on Problems in Mathematics.]]> https://yutsumura.com/given-the-data-of-eigenvalues-determine-if-the-matrix-is-invertible/feed/ 0 6813 https://yutsumura.com/express-the-eigenvalues-of-a-2-by-2-matrix-in-terms-of-the-trace-and-determinant/ https://yutsumura.com/express-the-eigenvalues-of-a-2-by-2-matrix-in-terms-of-the-trace-and-determinant/#respond Mon, 18 Dec 2017 23:51:33 +0000 https://yutsumura.com/?p=6240 Let $A=\begin{bmatrix} a & b\\ c& d \end{bmatrix}$ be an $2\times 2$ matrix. Express the eigenvalues of $A$ in terms of the trace and the determinant of $A$.   Solution. Recall the definitions of...

The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.
There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).
The time limit was 55 minutes.

This post is Part 2 and contains Problem 4, 5, and 6.
Check out Part 1 and Part 3 for the rest of the exam problems.

Problem 4. Let
\[\mathbf{a}_1=\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}, \mathbf{a}_2=\begin{bmatrix}
2 \\
-1 \\
4
\end{bmatrix}, \mathbf{b}=\begin{bmatrix}
0 \\
a \\
2
\end{bmatrix}.\]

Find all the values for $a$ so that the vector $\mathbf{b}$ is a linear combination of vectors $\mathbf{a}_1$ and $\mathbf{a}_2$.

Problem 5.
Find the inverse matrix of
\[A=\begin{bmatrix}
0 & 0 & 2 & 0 \\
0 &1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
1 & 0 & 0 & 1
\end{bmatrix}\] if it exists. If you think there is no inverse matrix of $A$, then give a reason.

Problem 6.
Consider the system of linear equations
\begin{align*}
3x_1+2x_2&=1\\
5x_1+3x_2&=2.
\end{align*}

(a) Find the coefficient matrix $A$ of the system.

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

(c) Using the inverse matrix of $A$, find the solution of the system.

(Linear Algebra Midterm Exam 1, the Ohio State University)
 

Solution of Problem 4.

To be able to express the vector $\mathbf{b}$ as a linear combination of $\mathbf{a}_1$ and $\mathbf{a}_2$, there must be scalars $c_1, c_2$ such that
\[c_1\mathbf{a}+c_2\mathbf{a}_2=\mathbf{b}.\] This is equivalent to the matrix equation
\[A\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}=\mathbf{b},\] where
\[A=[\mathbf{a}_1, \mathbf{a}_2]=\begin{bmatrix}
1 & 2 \\
2 & -1 \\
3 &4
\end{bmatrix}.\] Thus, the vector $\mathbf{b}$ is a linear combination of $\mathbf{a}_1$ and $\mathbf{a}_2$ if and only if the system $A\mathbf{x}=\mathbf{b}$ is consistent.

So let us consider the augmented matrix of the system and reduce it by elementary row operations.
We have
\begin{align*}
[A\mid \mathbf{b}]&=
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
2 &-1 &a \\
3 & 4 & 2
\end{array}\right] \xrightarrow[R_3-3R_1]{R_2-2R_1}
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
0 &-5 &a \\
0 & -2 & 2
\end{array}\right]\\[6pt] &\xrightarrow{-\frac{1}{2}R_3}
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
0 &-5 &a \\
0 & 1 & -1
\end{array}\right] \xrightarrow{R_2 \leftrightarrow R_3}
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
0 & 1 & -1\\
0 &-5 & a
\end{array}\right]\\[6pt] &\xrightarrow[R_3+5R_2]{R_1-2R_2}
\left[\begin{array}{rr|r}
1 & 0 & 2 \\
0 &1 &-1 \\
0 & 0 & a-5
\end{array}\right].
\end{align*}
If $a-5=0$, then we obtain the solution $x_1=2, x_2=-1$. Thus the system is consistent when $a=5$.

On the other hand, if $a-5 \neq 0$, then we divide the third row by $a-5$ and then the third row becomes $ \left[\begin{array}{rr|r}
0 & 0 & 1
\end{array}\right]$, which implies that the system is inconsistent (as we have $0=1$.)

Therefore, the only possible value for $a$ is $a=5$.

Note that when $a=5$, we have
\[2\mathbf{a}_1-\mathbf{a}_2=\mathbf{b}.\]

Solution of Problem 5.

We form the augmented matrix $[A\mid I]$, where $I$ is the $4\times 4$ identity matrix and apply elementary row operations as follows. We have
\begin{align*}
&[A\mid I]= \left[\begin{array}{rrrr|rrrr}
0 & 0 & 2 & 0 &1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
1 &0 & 0 & 1 & 0 & 0 & 0 &1 \\
\end{array} \right]\\[6pt] &\xrightarrow{R_1\leftrightarrow R_3}
\left[\begin{array}{rrrr|rrrr}
1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 &1 & 0 & 0 & 0 \\
1 &0 & 0 & 1 & 0 & 0 & 0 &1 \\
\end{array} \right] \xrightarrow[\frac{1}{2}R_3]{R_4-R_1}
\left[\begin{array}{rrrr|rrrr}
1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 &1/2 & 0 & 0 & 0 \\
0 &0 & 0 & 1 & 0 & 0 & -1 &1 \\
\end{array} \right].
\end{align*}
The left $4\times 4$ part became the identity matrix.
This means that the matrix $A$ is invertible and the inverse matrix of $A$ is given by the right $4\times 4$ part.
Hence
\[A^{-1}=\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1/2 & 0 & 0 & 0 \\
0 & 0 & -1 &1 \\
\end{bmatrix}.\]

Solution of Problem 6.

(a) Find the coefficient matrix $A$ of the system.

The coefficient matrix of the system is
\[A=\begin{bmatrix}
3 & 2\\
5& 3
\end{bmatrix}.\]

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

The determinant of $A$ is given by $\det(A)=3\cdot 3 -2\cdot 5=-1\neq 0$. Thus $A$ is invertible.
Using the formula for the inverse matrix of a $2\times 2$ matrix, we obtain
\[A^{-1}=\frac{1}{-1}\begin{bmatrix}
3 & -2\\
-5& 3
\end{bmatrix}=\begin{bmatrix}
-3 & 2\\
5& -3
\end{bmatrix}.\]

(c) Using $A^{-1}$, find the solution of the system.

The system can be written as
\[A\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}=\begin{bmatrix}
1 \\
2
\end{bmatrix}.\] Multiplying it by $A^{-1}$ on right and using the identity $A^{-1}A=I$, we obtain
\begin{align*}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}&=A^{-1}\begin{bmatrix}
1 \\
2
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
-3 & 2\\
5& -3
\end{bmatrix}\begin{bmatrix}
1 \\
2
\end{bmatrix}
=\begin{bmatrix}
1 \\
-1
\end{bmatrix}.
\end{align*}
Hence the solution of the system is $x_1=1, x_2=-1$.

Go to Part 1 and Part 3

Go to Part 1 for Problem 1, 2, and 3.

Go to Part 3 for Problem 7, 8, and 9.

Let $I$ be the $2\times 2$ identity matrix.
Then prove that $-I$ cannot be a commutator $[A, B]:=ABA^{-1}B^{-1}$ for any $2\times 2$ matrices $A$ and $B$ with determinant $1$.

Proof.

Assume that $[A, B]=-I$. Then $ABA^{-1}B^{-1}=-I$ implies
\[ABA^{-1}=-B. \tag{*}\] Taking the trace, we have
\[-\tr(B)=\tr(-B)=\tr(ABA^{-1})=tr(BAA^{-1})=\tr(B),\] hence the trace $\tr(B)=0$.
Thus, the characteristic polynomial of $B$ is
\[x^2-\tr(B)x+\det(B)=x^2+1.\] Hence the eigenvalues of $B$ are $\pm i$.

Note that the matrix $\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}$ has also eigenvalues $\pm i$.
Thus this matrix is similar to the matrix $B$ as both matrices are similar to the diagonal matrix $\begin{bmatrix}
i & 0\\
0& -i
\end{bmatrix}$.
Let $P$ be a nonsingular matrix such that
\[B’:=P^{-1}BP=\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}.\] Let $A’=P^{-1}AP$.

The relation (*) is equivalent to $AB=-BA$.
Using this we have
\begin{align*}
A’B’&=(P^{-1}AP)(P^{-1}BP)=P^{-1}(AB)P\\
&=P^{-1}(-BA)P=-(P^{-1}BP)(P^{-1}AP)=-B’A’.
\end{align*}

Let $A’=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$.
Then $A’B’=-B’A’$ gives
\begin{align*}
\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}
\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}
=-\begin{bmatrix}
0 & -1\\
1& 0
\end{bmatrix}\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}\\[6pt] \Leftrightarrow
\begin{bmatrix}
b & -a\\
d& -c
\end{bmatrix}=\begin{bmatrix}
c & d\\
-a& -b
\end{bmatrix}.
\end{align*}
Hence we obtain $d=-a$ and $c=b$.

Then
\begin{align*}
1=\det(A)=\det(PA’P^{-1})=\det(A’)=\begin{vmatrix}
a & b\\
b& -a
\end{vmatrix}=-a^2-b^2,
\end{align*}
which is impossible.
Therefore, the matrix $-I$ cannot be written as a commutator $[A, B]$ for any $2\times 2$ matrices $A, B$ with determinant $1$.