Determine the values of $x$ so that the matrix
\[A=\begin{bmatrix}
1 & 1 & x \\
1 &x &x \\
x & x & x
\end{bmatrix}\] is invertible.
For those values of $x$, find the inverse matrix $A^{-1}$.

Solution.

We use the fact that a matrix is invertible if and only if its determinant is nonzero.
So we compute the determinant of the matrix $A$.

We have
\begin{align*}
&\det(A)=\begin{vmatrix}
1 & 1 & x \\
1 &x &x \\
x & x & x
\end{vmatrix}\\
&=(1)\begin{vmatrix}
x & x\\
x& x
\end{vmatrix}-(1)\begin{vmatrix}
1 & x\\
x& x
\end{vmatrix}+x\begin{vmatrix}
1 & x\\
x& x
\end{vmatrix} && \text{by the first row cofactor expansion.}\\
&=(x^2-x^2)-(x-x^2)+x(x-x^2)\\
&=(x-1)(x-x^2)\\
&=x(x-1)^2.
\end{align*}

Thus, the determinant $\det(A)$ is zero if and only if $x=0, 1$.
Hence the matrix $A$ is invertible if and only if $x\neq 0, 1$.

Next, we suppose that $x \neq 0, 1$ and find the inverse matrix of $A$.
We reduce the augmented matrix $[A\mid I]$ as follows.
We have
\begin{align*}
&[A\mid I]= \left[\begin{array}{rrr|rrr}
1 & 1 & x & 1 &0 & 0 \\
1 & x & x & 0 & 1 & 0 \\
x & x & x & 0 & 0 & 1 \\
\end{array} \right] \\[6pt] & \xrightarrow{\substack{R_2-R_1 \\ R_3-xR_1}}
\left[\begin{array}{rrr|rrr}
1 & 1 & x & 1 &0 & 0 \\
0 & x-1 & 0 & -1 & 1 & 0 \\
0 & 0 & x-x^2 & -x & 0 & 1 \\
\end{array} \right] \xrightarrow[\frac{1}{x-x^2} R_3]{\frac{1}{x-1}R_2}
\left[\begin{array}{rrr|rrr}
1 & 1 & x & 1 &0 & 0 \\[8pt] 0 & 1 & 0 & \frac{-1}{x-1} & \frac{1}{x-1} & 0 \\[8pt] 0 & 0 & 1 & \frac{-1}{1-x} & 0 & \frac{1}{x-x^2} \\
\end{array} \right]\\[6pt] & \xrightarrow{R_1-R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & x & \frac{x}{x-1} & \frac{-1}{x-1} & 0 \\[8pt] 0 & 1 & 0 & \frac{-1}{x-1} & \frac{1}{x-1} & 0 \\[8pt] 0 & 0 & 1 & \frac{-1}{1-x} & 0 & \frac{1}{x-x^2} \\
\end{array} \right] \xrightarrow{R_1-xR_3}
\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & 0 & \frac{-1}{x-1} & \frac{-x}{x-x^2}\\[8pt] 0 & 1 & 0 & \frac{-1}{x-1} & \frac{1}{x-1} & 0 \\[8pt] 0 & 0 & 1 & \frac{-1}{1-x} & 0 & \frac{1}{x-x^2} \\
\end{array} \right].
\end{align*}

Now that we reduced the left $3\times 3$ matrix into the identity matrix, the right $3\times 3$ matrix is the inverse matrix of $A$.
(Note that when we applied elementary row operations, we divided by $x-1$ and $x-x^2$, and this is where we needed to assume $x \neq 0, 1$.)

We have
\begin{align*}
A^{-1}=\begin{bmatrix}
0 & \frac{-1}{x-1} & \frac{-x}{x-x^2}\\[8pt] \frac{-1}{x-1} & \frac{1}{x-1} & 0 \\[8pt] \frac{-1}{1-x} & 0 & \frac{1}{x-x^2} \\
\end{bmatrix}
=\frac{1}{x(1-x)}\begin{bmatrix}
0 & x & -x \\
x &-x &0 \\
-x & 0 & 1
\end{bmatrix}.
\end{align*}

The post For Which Choices of $x$ is the Given Matrix Invertible? first appeared on Problems in Mathematics.

Let $V$ be the vector space of all $2\times 2$ matrices, and let the subset $S$ of $V$ be defined by $S=\{A_1, A_2, A_3, A_4\}$, where
\begin{align*}
A_1=\begin{bmatrix}
1 & 2 \\
-1 & 3
\end{bmatrix}, \quad
A_2=\begin{bmatrix}
0 & -1 \\
1 & 4
\end{bmatrix}, \quad
A_3=\begin{bmatrix}
-1 & 0 \\
1 & -10
\end{bmatrix}, \quad
A_4=\begin{bmatrix}
3 & 7 \\
-2 & 6
\end{bmatrix}.
\end{align*}
Find a basis of the span $\Span(S)$ consisting of vectors in $S$ and find the dimension of $\Span(S)$.

Proof.

Let $B=\{E_{11}, E_{12}, E_{21}, E_{22}\}$ be the standard basis for the vector space $V$, where
\[E_{11}=\begin{bmatrix}
1 & 0\\
0& 0
\end{bmatrix},
E_{12}=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}, E_{21}=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}, E_{22}=\begin{bmatrix}
0 & 0\\
0& 1
\end{bmatrix}.\] With respect to the basis $B$, we find the coordinate vectors for $A_1, A_2, A_3, A_4$ as follows.
Since we have
\[A_1=E_{11}+2E_{12}-E_{21}+3E_{22},\] the coordinate vector for $A_1$ is
\[[A_1]_B=\begin{bmatrix}
1 \\
2 \\
-1 \\
3
\end{bmatrix}.\] Similarly, we have
\[[A_2]_B=\begin{bmatrix}
0 \\
-1 \\
1 \\
4
\end{bmatrix}, [A_3]_B=\begin{bmatrix}
-1 \\
0 \\
1 \\
-10
\end{bmatrix},
[A_4]_B=\begin{bmatrix}
3 \\
7 \\
-2 \\
6
\end{bmatrix}.\] To find a basis for $\Span(S)$ among vectors in $S$, we first find a basis for $\Span(T)$ among vectors in
\[T=\{[A_1]_B, [A_2]_B, [A_3]_B, [A_4]_B\}.\] Let form a matrix whose columns are vectors in $T$. That is,
\[\begin{bmatrix}
1 & 0 & -1 & 3 \\
2 &-1 & 0 & 7 \\
-1 & 1 & 1 & -2 \\
3 & 4 & -10 & 6
\end{bmatrix}.\]

We apply the elementary row operations as follows and obtain a reduced row echelon form matrix.
\begin{align*}
\begin{bmatrix}
1 & 0 & -1 & 3 \\
2 &-1 & 0 & 7 \\
-1 & 1 & 1 & -2 \\
3 & 4 & -7 & 6
\end{bmatrix}
\xrightarrow{\substack{R_2-2R_1 \\ R_3+R_1\\ R_4-3R_1}}
\begin{bmatrix}
1 & 0 & -1 & 3 \\
0 &-1 & 2 & 1 \\
0 & 1 & 0 & 1 \\
0 & 4 & -7 & -3
\end{bmatrix}
\xrightarrow{\substack{R_3+R_2 \\ R_4+4R_2}}
\begin{bmatrix}
1 & 0 & -1 & 3 \\
0 &-1 & 2 & 1 \\
0 & 0 & 2 & 2 \\
0 & 0& 1 & 1
\end{bmatrix}\\[6pt] \xrightarrow{\substack{R_1+R_4 \\ R_2-2R_4 \\ R_3-2R_4}}
\begin{bmatrix}
1 & 0 & 0 & 4 \\
0 &-1 & 0 & -1 \\
0 & 0 & 0 & 0 \\
0 & 0& 1 & 1
\end{bmatrix}
\xrightarrow{\substack{1R_2\\ R_3 \leftrightarrow R_4}}
\begin{bmatrix}
1 & 0 & 0 & 4 \\
0 &1 & 0 & 1 \\
0 & 0 & 1 & 1 \\
0 & 0& 0 & 0
\end{bmatrix}.
\end{align*}

The the first three columns of the reduced row echelon form contains the leading 1’s.
Thus by, what I call, the leading $1$ method, it follows that
\[\{[A_1]_B, [A_2]_B, [A_3]_B\}\] is a basis for $\Span(T)$.

Therefore, by the correspondence of coordinate vectors, we obtain that
\[\{A_1, A_2, A_3\}\] is a basis of $\Span(S)$.

Since the basis $\{A_1, A_2, A_3\}$ consists of three vectors, the dimension of $\Span(S)$ is $3$.

The post Find a Basis for a Subspace of the Vector Space of \times 2$ Matrices first appeared on Problems in Mathematics.

Find the determinant of the matix
\[A=\begin{bmatrix}
100 & 101 & 102 \\
101 &102 &103 \\
102 & 103 & 104
\end{bmatrix}.\]

Solution.

Note that the determinant does not change if the $i$-th row is added by a scalar multiple of the $j$-th row if $i \neq j$.
We use this fact about the determinant and compute $\det(A)$ as follows.
\begin{align*}
\det(A)&=\begin{vmatrix}
100 & 101 & 102 \\
101 &102 &103 \\
102 & 103 & 104
\end{vmatrix}\\[5 pt] &=\begin{vmatrix}
100 & 101 & 102 \\
101 &102 &103 \\
1 & 1 & 1
\end{vmatrix}
\quad (\text{by } R_3-R_2)\\[5 pt] &=\begin{vmatrix}
100 & 101 & 102 \\
1 &1 &1 \\
1 & 1 & 1
\end{vmatrix}
\quad (\text{by } R_2-R_1)\\[5 pt] &=\begin{vmatrix}
100 & 101 & 102 \\
1 &1 &1 \\
0 & 0 & 0
\end{vmatrix}
\quad (\text{by } R_3-R_1)\\[5 pt] &=0 \quad (\text{by the third row cofactor expansion}.)
\end{align*}
Therefore the determinant $\det(A)$ is zero.

The post How to Find the Determinant of the \times 3$ Matrix first appeared on Problems in Mathematics.

Determine whether the following systems of equations (or matrix equations) described below has no solution, one unique solution or infinitely many solutions and justify your answer.

(a) \[\left\{
\begin{array}{c}
ax+by=c \\
dx+ey=f,
\end{array}
\right.
\] where $a,b,c, d$ are scalars satisfying $a/d=b/e=c/f$.

(b) $A \mathbf{x}=\mathbf{0}$, where $A$ is a singular matrix.

(c) A homogeneous system of $3$ equations in $4$ unknowns.

(d) $A\mathbf{x}=\mathbf{b}$, where the row-reduced echelon form of the augmented matrix $[A|\mathbf{b}]$ looks as follows:
\[\begin{bmatrix}
1 & 0 & -1 & 0 \\
0 &1 & 2 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}.\] (The Ohio State University, Linear Algebra Exam)

Hint.

Recall that possibilities for the solution set of a system of linear equation is either no solution (inconsistent) or one unique solution or infinitely many solutions.

A homogeneous system is a system with zero constant terms.
A homogeneous system always has the zero solution.

Solution.

(a) Note that by the given condition $a/d=b/e=c/f$, the numbers $b, e, c$ are not zero. The augmented matrix of the system is
$\left[\begin{array}{rr|r}
a & b & c \\
d & e & f
\end{array}\right]$.
We apply the elementary row operations as follows.

\begin{align*}
&\left[\begin{array}{rr|r}
a & b & c \\
d & e & f
\end{array}\right] \xrightarrow{R_1-\frac{a}{d}R_2}
\left[\begin{array}{rr|r}
0 & 0 & 0 \\
d & e & f
\end{array}\right] \xrightarrow{R_1\leftrightarrow R_2}
\left[\begin{array}{rr|r}
d & e & f\\
0 & 0 & 0
\end{array}\right]\\
& \xrightarrow{\frac{1}{d}}
\left[\begin{array}{rr|r}
1 & e/d & f/d\\
0 & 0 & 0
\end{array}\right].
\end{align*}
Here in the first step, we used the relation $a/d=b/e=c/f$.
The last matrix is in reduced row echelon form with rank $1$ and it does not have a row of the form $[00|1]$. Therefore the system is consistent and there are $1(=\text{the number of unknowns }-\text{ rank})$ free variable, hence there are infinitely many solutions.

(b) The system is homogeneous, thus it has the zero solution. Since the coefficient matrix $A$ is singular, the system has non-zero solution as well.
Therefore, the only possibility is that the system has infinitely many solutions.

(c) A homogeneous system has the zero solution hence it is consistent. Since there are more unknowns than equations, the system must have infinitely many solutions.

(d) The last row of the reduced row echelon form matrix of the augmented matrix is $[000|1]$.
It corresponds to the equation
\[0x_1+0x_2+0x_3=1.\] Equivalently, this is $0=1$ and this is impossible. Thus the system has no solution (an inconsistent system).

The post Possibilities For the Number of Solutions for a Linear System first appeared on Problems in Mathematics.

Find a basis for the subspace $W$ of all vectors in $\R^4$ which are perpendicular to the columns of the matrix
\[A=\begin{bmatrix}
11 & 12 & 13 & 14 \\
21 &22 & 23 & 24 \\
31 & 32 & 33 & 34 \\
41 & 42 & 43 & 44
\end{bmatrix}.\]

(Harvard University Exam)

Hint.

1. Show that $W=\calN(A^{\trans})$.
2. Find a basis of $\calN(A^{\trans})$ by reducing the matrix $A^{\trans}$.

Solution.

Let us write $A=[A_1 \, A_2 \, A_3 \, A_4]$, where $A_i$ is the $i$-th column vector of $A$ for $i=1,2,3,4$.
First we claim that a vector $\mathbf{x}\in \R^4$ is perpendicular to all column vectors $A_i$ if and only if $\mathbf{x} \in \calN(A^{\trans})$.
To see this, we compute
\begin{align*}
A^{\trans} \mathbf{x} =\begin{bmatrix}
A_1^{\trans} \\
A_2^{\trans} \\
A_3^{\trans} \\
A_4^{\trans}
\end{bmatrix}\mathbf{x}
=\begin{bmatrix}
A_1^{\trans}\mathbf{x} \\
A_2^{\trans} \mathbf{x}\\
A_3^{\trans} \mathbf{x}\\
A_4^{\trans} \mathbf{x}
\end{bmatrix}.
\end{align*}
From this equality the claim follows immediately.

So we proved that $\calN(A^{\trans}) =W$. From this, we see that $W$ is actually a subspace of $\R^4$.

Thus, we need to find a basis for the null space of the transpose $A^{\trans}$.

We apply elementary row operations to $A^{\trans}$ and obtain a reduced row echelon form
\[A^{\trans} \to \begin{bmatrix}
1 & 0 & -1 & -2 \\
0 &1 & 2 & 3 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}.\] The last two columns correspond to two free variables. Let $s$ and $t$ be free variable.
Then $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \in \calN(A^{\trans})$ if and only if $\mathbf{x}$ satisfies
\begin{align*}
x_1 &=s+2t \\
x_2 &=-2s-3t\\
x_3 &=s\\
x_4 &=t,
\end{align*}
equivalently
\begin{align*}
\mathbf{x}=s\begin{bmatrix}
1 \\
-2 \\
1 \\
0
\end{bmatrix}
+t\begin{bmatrix}
2 \\
-3 \\
0 \\
1
\end{bmatrix}.
\end{align*}
Therefore a basis of $W=\calN(A^{\trans})$ is
\[ \begin{bmatrix}
1 \\
-2 \\
1 \\
0
\end{bmatrix} \text{ and } \begin{bmatrix}
2 \\
-3 \\
0 \\
1
\end{bmatrix}.\] Click here if solved 37

The post Find a Basis of the Subspace of All Vectors that are Perpendicular to the Columns of the Matrix first appeared on Problems in Mathematics.