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 370

The post Find All Values of $x$ such that the Matrix is Invertible first appeared on Problems in Mathematics.

Determine the values of a real number $a$ such that the matrix
\[A=\begin{bmatrix}
3 & 0 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}\] is nonsingular.
 

Solution.

We apply elementary row operations and obtain:
\begin{align*}
A=\begin{bmatrix}
3 & 0 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & -3 & a \\
2 &3 &0 \\
0 & 18a & a+1
\end{bmatrix}\\[6pt] \xrightarrow{R_2-2R_1}
\begin{bmatrix}
1 & -3 & a \\
0 &9 &-2a \\
0 & 18a & a+1
\end{bmatrix}
\xrightarrow{R_3-(2a)R_2}
\begin{bmatrix}
1 & -3 & a \\
0 &9 &-2a \\
0 & 0 & 4a^2+a+1
\end{bmatrix}.
\end{align*}

From this, we see that the matrix $A$ is nonsingular if and only if the $(3, 3)$-entry $4a^2+a+1$ is not zero.
By the quadratic formula, we see that
\[a=\frac{-1\pm \sqrt{-15}}{8}\] are solutions of $4a^2+a+1=0$.

Note that these are not real numbers. Thus, for any real number $a$, we have $4a^2+a+1\neq 0$.

Hence, we can divide the third row by this number, and eventually we can reduce it to the identity matrix.
So the rank of $A$ is $3$, and $A$ is nonsingular for any real number $a$.

The post For What Values of $a$, Is the Matrix Nonsingular? first appeared on Problems in Mathematics.

Let $A$ be a $2\times 2$ real symmetric matrix.
Prove that all the eigenvalues of $A$ are real numbers by considering the characteristic polynomial of $A$.

Proof.

Let $A=\begin{bmatrix}
a& b \\
c& d
\end{bmatrix}$.
Then as $A$ is a symmetric matrix, we have $A^{\trans}=A$.
This implies that
\[\begin{bmatrix}
a& c \\
b& d
\end{bmatrix}=\begin{bmatrix}
a& b \\
c& d
\end{bmatrix}.\] Hence we have $b=c$ by comparing entries.

Now, we find the characteristic polynomial $p(t)$ of $A$.
We have
\begin{align*}
p(t)&=\det(A-t I)=\begin{vmatrix}
a-t & b\\
b& d-t
\end{vmatrix}\\[6pt] &=(a-t)(d-t)-b^2\\
&=t^2-(a+d)t+ad-b^2.
\end{align*}

Note that the eigenvalues of $A$ are roots of the characteristic polynomial $p(t)$. Hence, it suffices to show that the roots of $p(t)$ are real numbers.
The quadratic polynomial has only real roots if and only if its discriminant is non-negative.
The discriminant of $p(t)$ is given by
\begin{align*}
(a+d)^2-4(ad-b^2)&=a^2+2ad+d^2-4ad+4b^2\\
&=a^2-2ad+d^2+4b^2\\
&=(a-d)^2+4b^2. \end{align*}
Observe that the last expression is the sum of two squares of real numbers. Hence the discriminant of $p(t)$ is nonnegative.

We conclude that every $2\times 2$ symmetric matrix has only real eigenvalues.

Remark

We also could find the eigenvalues directly. By the quadratic formula, the eigenvalues of $A$ are
\[\frac{a+d\pm\sqrt{(a+d)^2-4(ad-b^2)}}{2}=\frac{a+d\pm \sqrt{(a-d)^2+4b^2}}{2}\] and as the number inside the square root (discriminant) is positive, we conclude that the eigenvalues are real.

The post Eigenvalues of \times 2$ Symmetric Matrices are Real by Considering Characteristic Polynomials first appeared on Problems in Mathematics.

Determine all eigenvalues and their algebraic multiplicities of the matrix
\[A=\begin{bmatrix}
1 & a & 1 \\
a &1 &a \\
1 & a & 1
\end{bmatrix},\] where $a$ is a real number.

Proof.

To find eigenvalues we first compute the characteristic polynomial of the matrix $A$ as follows.
\begin{align*}
\det(A-tI)&=\begin{vmatrix}
1-t & a & 1 \\
a &1-t &a \\
1 & a & 1-t
\end{vmatrix}\\
&=(1-t)\begin{vmatrix}
1-t & a\\
a& 1-t
\end{vmatrix}-a\begin{vmatrix}
a & a\\
1& 1-t
\end{vmatrix}+\begin{vmatrix}
a & 1-t\\
1& a
\end{vmatrix}
\end{align*}
We used the first row cofactor expansion in the second equality.

After we compute three $2 \times 2$ determinants and simply, we obtain
\[\det(A-tI)=-t(t^2-3t+2-2a^2).\] The eigenvalues of $A$ are roots of this characteristic polynomial. Thus, eigenvalues are
\[0, \quad \frac{3\pm \sqrt{1+8a^2}}{2}\] by the quadratic formula.

Now the only possible way to obtain a multiplicity $2$ eigenvalue is when
\[\frac{3- \sqrt{1+8a^2}}{2}=0\] and it is straightforward to check that this happens if and only if $a=1$.

Therefore, when $a=1$ eigenvalues of $A$ are $0$ with algebraic multiplicity $2$ and $3$ with algebraic multiplicity $1$.

When $a \neq 1$, eigenvalues are
\[0, \quad \frac{3\pm \sqrt{1+8a^2}}{2}\] and each of them has algebraic multiplicity $1$.

The post Eigenvalues and their Algebraic Multiplicities of a Matrix with a Variable first appeared on Problems in Mathematics.

Prove that the matrix
\[A=\begin{bmatrix}
1 & 1.00001 & 1 \\
1.00001 &1 &1.00001 \\
1 & 1.00001 & 1
\end{bmatrix}\] has one positive eigenvalue and one negative eigenvalue.

(University of California, Berkeley Qualifying Exam Problem)
 

Solution.

Let us put $a=1.00001$. We compute the characteristic polynomial $\det(A-tI)$ of the given matrix $A$ as follows.
We have
\begin{align*}
\det(A-tI)&=\begin{vmatrix}
1-t & a & 1 \\
a &1-t &a \\
1 & a & 1-t
\end{vmatrix}\\
&=(1-t)\begin{vmatrix}
1-t & a\\
a& 1-t
\end{vmatrix}-a\begin{vmatrix}
a & a\\
1& 1-t
\end{vmatrix}+\begin{vmatrix}
a & 1-t\\
1& a
\end{vmatrix}
\end{align*}
by the cofactor expansion corresponding to the first row.

Simplifying this, we obtain
\[\det(A-tI)=-t(t^2-3t+2-2a^2).\]

The eigenvalues of $A$ are roots of this characteristic polynomial.
Hence $0$ is an eigenvalue of $A$. Let $\lambda_1$ and $\lambda_2$ be other two eigenvalues of $A$.

Then we have
\[\det(A-tI)=-t(t-\lambda_1)(t-\lambda_2)=-t(t^2-(\lambda_1+\lambda_2)+\lambda_1 \lambda_2).\] Therefore we have
\[\lambda_1 \lambda_2=2-2a^2.\]

Since $2-2a^2=2(1-a^2)<0$ as $a=1.0001>1$, the product $\lambda_1 \lambda_2$ is negative, and we conclude that one of them is positive and the other is negative.

(Note that if the constant $c$ term of a quadratic polynomial $x^2+bx+c$ is negative, then the roots of the polynomial are real and one is negative and the other is positive.)

In summary, the eigenvalues of the matrix $A$ are $0$ and one positive eigenvalue and one negative eigenvalue.

The post A Matrix Having One Positive Eigenvalue and One Negative Eigenvalue first appeared on Problems in Mathematics.