If $A, B$ have the same rank, can we conclude that they are row-equivalent?

If so, then prove it. If not, then provide a counterexample.

Solution.

Having the same rank does not mean they are row-equivalent.

For a simple counterexample, consider $A = \begin{bmatrix} 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \end{bmatrix}$.

Both of these matrices have rank 1, but are not row-equivalent because they are already in reduced row echelon form.

Another solution.

The problem doesn’t specify the sizes of matrices $A$, $B$.

Note that if the sizes of $A$ and $B$ are distinct, then they can never be row-equivalent.
Keeping this in mind, let us consider the following two matrices.

\[A=\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } B=\begin{bmatrix}
1 & 0
\end{bmatrix}.\] Then both matrices are in reduced row echelon form and have rank $1$.
As noted above, they are not row-equivalent because the sizes are distinct.

The post If Two Matrices Have the Same Rank, Are They Row-Equivalent? first appeared on Problems in Mathematics.

For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix.

(a) $A=\begin{bmatrix}
1 & 3 & -2 \\
2 &3 &0 \\
0 & 1 & -1
\end{bmatrix}$
 
(b) $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Elementary Row Operations and Inverse Matrices

Recall the following procedure of testing the invertibility of $A$ as well as finding the inverse matrix if exists.
If the augmented matrix $[A|I]$ is transformed into a matrix of the form $[I|B]$, then the matrix $A$ is invertible and the inverse matrix $A^{-1}$ is given by $B$.
If the reduced row echelon form matrix for $[A|I]$ is not of the form $[I|B]$, then the matrix $A$ is not invertible.

Solution.

(a) $A=\begin{bmatrix}
1 & 3 & -2 \\
2 &3 &0 \\
0 & 1 & -1
\end{bmatrix}$

We apply the elementary row operations as follows.
We have
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
2 & 3 & 0 & 0 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{R_2-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
0 & -3 & 4 & -2 & 1 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
\end{array} \right]\\[6pt] &\xrightarrow{R_2\leftrightarrow R_3}
\left[\begin{array}{rrr|rrr}
1 & 3 & -2 & 1 &0 & 0 \\
0 & 1 & -1 & 0 & 0 & 1 \\
0 & -3 & 4 & -2 & 1 & 0 \\
\end{array} \right] \xrightarrow{\substack{R_1-3R_2\\ R_3+3R_2}}
\left[\begin{array}{rrr|rrr}
1 & 0& 1 & 1 &0 & -3 \\
0 & 1 & -1 & 0 & 0 & 1 \\
0 & 0 & 1 & -2 & 1 & 3 \\
\end{array} \right]\\[6pt] &\xrightarrow{\substack{R_1-R_3\\ R_2+R_3}}
\left[\begin{array}{rrr|rrr}
1 & 0& 0 & 3 & -1 & -6 \\
0 & 1 & 0 & -2 & 1 & 4 \\
0 & 0 & 1 & -2 & 1 & 3 \\
\end{array} \right].
\end{align*}

The left $3\times 3$ part of the last matrix is the identity matrix.
This implies that $A$ is invertible and the inverse matrix is given by the right $3\times 3$ matrix.
Hence
\[A^{-1}=\begin{bmatrix}
3 & -1 & -6 \\
-2 &1 &4 \\
-2 & 1 & 3
\end{bmatrix}.\]

(b) $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Now we consider the matrix $A=\begin{bmatrix}
1 & 0 & 2 \\
-1 &-3 &2 \\
3 & 6 & -2
\end{bmatrix}$.

Applying elementary row operations, we obtain
\begin{align*}
&[A|I]= \left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
-1 & -3 & 2 & 0 & 1 & 0 \\
3 & 6 & -2 & 0 & 0 & 1 \\
\end{array} \right] \xrightarrow{\substack{R_2+R_1\\ R_3-3R_1}}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & -3 & 4 & 1 & 1 & 0 \\
0 & 6 & -8 & -3 & 0 & 1 \\
\end{array} \right]\\[6pt] &\xrightarrow{R_3-2R_1}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & -3 & 4 & 1 & 1 & 0 \\
0 & 0 & 0 & -5 & -2 & 1 \\
\end{array} \right] \xrightarrow{\frac{-1}{3}R_2}
\left[\begin{array}{rrr|rrr}
1 & 0 & 2 & 1 &0 & 0 \\
0 & 1 & -4/3 & -1/3 & -1/3 & 0 \\
0 & 0 & 0 & -5 & -2 & 1 \\
\end{array} \right].
\end{align*}

The last matrix is in reduced row echelon form but the left $3\times 3$ part is not the identity matrix $I$.
It follows that the matrix $A$ is not invertible.

Related Question.

This is a linear algebra exam problem at the Ohio State University.

The solution is given in the post↴
Find the Inverse Matrix of a $3\times 3$ Matrix if Exists

The post Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations first appeared on Problems in Mathematics.

Let $A$ be the following $3 \times 3$ matrix.
\[A=\begin{bmatrix}
1 & 1 & -1 \\
0 &1 &2 \\
1 & 1 & a
\end{bmatrix}.\] Determine the values of $a$ so that the matrix $A$ is nonsingular.

Solution.

We use the fact that a matrix is nonsingular if and only if it is row equivalent to the identity matrix.

We apply the elementary row operations as follows.
\begin{align*}
A \xrightarrow{R_3-R_1} \begin{bmatrix}
1 & 1 & -1 \\
0 &1 &2 \\
0 & 0 & a+1
\end{bmatrix} \xrightarrow{R_1-R_2}
\begin{bmatrix}
1 & 0 & -3 \\
0 &1 &2 \\
0 & 0 & a+1
\end{bmatrix}.
\end{align*}

If $a+1=0$, then the last matrix is in reduced row echelon form.
Thus $A$ is not row equivalent to the identity matrix.

On the other hand, if $a+1 \neq 0$, then we can continue the reduction as follows.
\begin{align*}
\begin{bmatrix}
1 & 0 & -3 \\
0 &1 &2 \\
0 & 0 & a+1
\end{bmatrix}
\xrightarrow{\frac{1}{a+1} R_3}
\begin{bmatrix}
1 & 0 & -3 \\
0 &1 &2 \\
0 & 0 & 1
\end{bmatrix}
\xrightarrow[R_2-2R_3]{R_1+3R_3}
\begin{bmatrix}
1 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}.
\end{align*}
Therefore $A$ is row equivalent to the identity matrix.

We conclude that the matrix $A$ is nonsingular for any values of $a$ except for $a=-1$.

Comment.

If you know how to compute the determinant of a $3 \times 3$ matrix, then you may also solve this using the fact that a matrix is nonsingular if and only if the determinant of it is nonzero.

The post Find Values of $a$ so that the Matrix is Nonsingular first appeared on Problems in Mathematics.

Problem 115

Express the vector $\mathbf{b}=\begin{bmatrix}
2 \\
13 \\
6
\end{bmatrix}$ as a linear combination of the vectors
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
5 \\
-1
\end{bmatrix},
\mathbf{v}_2=
\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix},
\mathbf{v}_3=
\begin{bmatrix}
1 \\
4 \\
3
\end{bmatrix}.\]

 
(The Ohio State University, Linear Algebra Exam)

Solution.

We need to find numbers $x_1, x_2, x_3$ satisfying
\[x_1 \mathbf{v}_1 + x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{b}.\]

This vector equation is equivalent to the following matrix equation.
\[[\mathbf{v}_1 \mathbf{v}_2 \mathbf{v}_3]\mathbf{x}=\mathbf{b}\] or more explicitly we can write it as
\[\begin{bmatrix}
1 & 1 & 1 \\
5 &2 &4 \\
-1 & 1 & 3
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}=
\begin{bmatrix}
2 \\
13 \\
6
\end{bmatrix}.\]

Thus the problem is to find the solution of this matrix equation.
Let us consider the augmented matrix for this system to apply Gauss-Jordan elimination.

The augmented matrix is
\[ \left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
5 &2 & 4 & 13 \\
-1 & 1 & 3 & 6
\end{array} \right].\] We apply elementary row operations and obtain a matrix in reduced row echelon form as follows.
\begin{align*}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
5 &2 & 4 & 13 \\
-1 & 1 & 3 & 6
\end{array} \right] \xrightarrow[R_3+R_1]{R_2-5R_1}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
0 &-3 & -1 & 3 \\
0 & 2 & 4 & 8
\end{array} \right]\\[6pt] \xrightarrow[-R_2]{\frac{1}{2}R_3}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
0 &3 & 1 & -3 \\
0 & 1 & 2 & 4
\end{array} \right] \xrightarrow{R_2 \leftrightarrow R_3}
\left[\begin{array}{rrr|r}
1 & 1 & 1 & 2 \\
0 & 1 & 2 & 4 \\
0 &3 & 1 & -3
\end{array} \right]\\[6pt] \xrightarrow[R_3-3R_2]{R_1-R_2}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & -2 \\
0 & 1 & 2 & 4 \\
0 &0 & -5 & -15
\end{array} \right] \xrightarrow{\frac{-1}{5}R_3}
\left[\begin{array}{rrr|r}
1 & 0 & -1 & -2 \\
0 & 1 & 2 & 4 \\
0 &0 & 1 & 3
\end{array} \right]\\[6pt] \xrightarrow[R_2-2R_3]{R_1+R_3}
\left[\begin{array}{rrr|r}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -2 \\
0 &0 & 1 & 3
\end{array} \right].
\end{align*}

Therefore the solution for the system is
\[x_1=1, x_2=-2, \text{ and } x_3=3\] and we obtain the linear combination
\[\mathbf{b}=\mathbf{v}_1-2\mathbf{v}_2+3\mathbf{v}_3.\]

The Ohio State University Linear Algebra Exam Problems and Solutions

Check out more problems from the Ohio State University Linear Algebra Exams ↴
The Ohio State University Linear Algebra.

The post Express a Vector as a Linear Combination of Other Vectors 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.

Solve the following system of linear equations using Gaussian elimination.
\begin{align*}
x+2y+3z &=4 \\
5x+6y+7z &=8\\
9x+10y+11z &=12
\end{align*}

Elementary row operations

The three elementary row operations on a matrix are defined as follows.

Solution.

First, the augmented matrix of the system is
\[A=\left[ \begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11 & 12
\end{array} \right].\] We apply elementary row operations as follows to reduce the system to row echelon form.

\[A \xrightarrow{R_3 -9R_1}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
0 & -8 & -16 & -24
\end{array}\right] \xrightarrow{-\frac{1}{8}R_3}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
0 & 1 & 2 & 3
\end{array}\right] \] \[\xrightarrow{R_2-5R_1}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 & -4 & -8 & -12 \\
0 & 1 & 2 & 3
\end{array}\right] \xrightarrow{-\frac{1}{4} R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 1 & 2 & 3
\end{array}\right] \] \[\xrightarrow{R_3-R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 0
\end{array}\right] \] The last matrix is in row echelon form.

The corresponding system of linear equations of it is
\begin{align*}
x+2y+3z &=4\\
y+2z&=3 \\
0z&=0
\end{align*}
The last equation $0z=0$ means that $z$ can be any number.
(More systematically, the variables corresponding to leading $1$’s in the echelon form matrix are dependent variables, and the rests are independent (free) variables.)

So let us say that $t$ is a value for $z$, namely $z=t$.
Then from the second equation, we have $y=-2t+3$.
From the first equation, we have
\[x=-2y-3z+4=-2(-2t+3)-3t+4=t-2.\] Thus the solution set is
\[(x,y,z)=(t-2, -2t+3, t)\] for any $t$.

Comment.

You may want to check whether the answer is correct by substituting this solution to the original equations.

Also, if you further reduce the matrix into reduced row echelon form, the last system becomes simpler (and simplest in a sense).
This procedure, to reduce a matrix until reduced row echelon form, is called the Gauss-Jordan elimination.

Related Question.

For a similar problem, you may want to check out Solve a system of linear equations by Gauss-Jordan elimination.

The post Solving a System of Linear Equations Using Gaussian Elimination first appeared on Problems in Mathematics.