1.1Identify a Matrix Entry
For
$$
A =
\begin{bmatrix}
2 & -1 & 4 \\
0 & 5 & 7 \\
3 & 8 & 6
\end{bmatrix},
$$
what is $a_{23}$?
Solution
The entry $a_{23}$ means row $2$, column $3$.
From the matrix, that entry is
$$
a_{23} = 7.
$$
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Difficulty
Direct Practice
1.2Determine a Matrix's Dimensions
What is the size of the matrix below?
$$
\begin{bmatrix}
1 & 0 & 2 & -3 \\
4 & 5 & -1 & 6
\end{bmatrix}
$$
Solution
The matrix has $2$ rows and $4$ columns, so its dimensions are
$$
2 \times 4.
$$
1.3Classify a Matrix
Name every matrix type from the note that applies to
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}.
$$
Solution
This matrix is:
square
diagonal
scalar
identity
upper triangular
lower triangular
symmetric
orthogonal
nonsingular
Each off-diagonal entry is $0$, and each diagonal entry is $1$.
1.4Add Two Matrices
Compute the sum:
$$
\begin{bmatrix}
1 & 3 \\
2 & -4
\end{bmatrix}
+
\begin{bmatrix}
5 & -1 \\
-2 & 7
\end{bmatrix}
$$
Solution
Add corresponding entries:
$$
\begin{bmatrix}
1+5 & 3+(-1) \\
2+(-2) & -4+7
\end{bmatrix}
=
\begin{bmatrix}
6 & 2 \\
0 & 3
\end{bmatrix}
$$
1.5Scale a Matrix
Compute $-3A$ for
$$
A =
\begin{bmatrix}
1 & -2 \\
4 & 0
\end{bmatrix}.
$$
Solution
Multiply every entry by $-3$:
$$
-3A =
\begin{bmatrix}
-3 & 6 \\
-12 & 0
\end{bmatrix}.
$$
1.6Check Matrix Multiplication Compatibility
Suppose $A$ is a $2 \times 3$ matrix and $B$ is a $3 \times 1$ matrix. Is $AB$ defined, and what size is the product?
Solution
Yes. The inner dimensions match because the number of columns of $A$ is $3$ and the number of rows of $B$ is $3$.
So $AB$ is defined, and its size is
$$
2 \times 1.
$$
1.7Transpose a Matrix
Find the transpose of
$$
\begin{bmatrix}
1 & 4 & -2 \\
3 & 0 & 5
\end{bmatrix}.
$$
Solution
Swap rows and columns:
$$
\begin{bmatrix}
1 & 4 & -2 \\
3 & 0 & 5
\end{bmatrix}^T
=
\begin{bmatrix}
1 & 3 \\
4 & 0 \\
-2 & 5
\end{bmatrix}.
$$
1.8Compute a 2 by 2 Determinant
Find the determinant of
$$
\begin{bmatrix}
7 & 2 \\
5 & 3
\end{bmatrix}.
$$
Solution
For a $2 \times 2$ matrix,
$$
\det\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
= ad - bc.
$$
So
$$
\det\begin{bmatrix}
7 & 2 \\
5 & 3
\end{bmatrix}
= 7 \cdot 3 - 2 \cdot 5
= 21 - 10
= 11.
$$
1.9Find the Inverse of a 2 by 2 Matrix
Find the inverse of
$$
A =
\begin{bmatrix}
2 & 1 \\
5 & 3
\end{bmatrix}.
$$
Solution
First compute the determinant:
$$
\det(A) = 2 \cdot 3 - 1 \cdot 5 = 6 - 5 = 1.
$$
Since the determinant is nonzero, the inverse exists. Use the $2 \times 2$ inverse formula:
$$
A^{-1} = \frac{1}{1}
\begin{bmatrix}
3 & -1 \\
-5 & 2
\end{bmatrix}
=
\begin{bmatrix}
3 & -1 \\
-5 & 2
\end{bmatrix}.
$$
1.10Compute the Trace of a Matrix
Find the trace of
$$
\begin{bmatrix}
4 & 1 & 0 \\
2 & -3 & 5 \\
7 & 8 & 6
\end{bmatrix}.
$$
Solution
The trace is the sum of the diagonal entries:
$$
\operatorname{tr}(A) = 4 + (-3) + 6 = 7.
$$
Difficulty
Integrated Practice
2.1Multiply Two Matrices
Compute
$$
\begin{bmatrix}
1 & 2 \\
3 & 0
\end{bmatrix}
\begin{bmatrix}
4 & 1 \\
-2 & 5
\end{bmatrix}.
$$
Solution
Multiply row by column:
$$
\begin{bmatrix}
1 \cdot 4 + 2 \cdot (-2) & 1 \cdot 1 + 2 \cdot 5 \\
3 \cdot 4 + 0 \cdot (-2) & 3 \cdot 1 + 0 \cdot 5
\end{bmatrix}
=
\begin{bmatrix}
0 & 11 \\
12 & 3
\end{bmatrix}.
$$
2.2Use Matrix-Vector Multiplication
Let
$$
A =
\begin{bmatrix}
2 & -1 & 0 \\
1 & 3 & 4
\end{bmatrix}
\quad \text{and} \quad
x =
\begin{bmatrix}
5 \\
2 \\
-1
\end{bmatrix}.
$$
Compute $Ax$.
Solution
Multiply row by column:
$$
Ax =
\begin{bmatrix}
2(5) + (-1)(2) + 0(-1) \\
1(5) + 3(2) + 4(-1)
\end{bmatrix}
=
\begin{bmatrix}
8 \\
7
\end{bmatrix}.
$$
You can also view this as a linear combination of the columns of $A$ weighted by the entries of $x$.
2.3Solve a System by Row Reduction
Solve the system:
$$
\begin{aligned}
x + 2y &= 8 \\
2x - y &= 1
\end{aligned}
$$
Solution
Write the augmented matrix:
$$
\begin{bmatrix}
1 & 2 & \mid & 8 \\
2 & -1 & \mid & 1
\end{bmatrix}.
$$
Eliminate the $2$ below the first pivot:
$$
R_2 \leftarrow R_2 - 2R_1
$$
gives
$$
\begin{bmatrix}
1 & 2 & \mid & 8 \\
0 & -5 & \mid & -15
\end{bmatrix}.
$$
So
$$
y = 3.
$$
Substitute into the first equation:
$$
x + 2(3) = 8
$$
so
$$
x = 2.
$$
2.4Find Rank and Nullity from a Row-Reduced Matrix
A $3 \times 4$ matrix row-reduces to
$$
\begin{bmatrix}
1 & 0 & 2 & -1 \\
0 & 1 & 3 & 4 \\
0 & 0 & 0 & 0
\end{bmatrix}.
$$
What are its rank and nullity?
Solution
There are $2$ pivot columns, so the rank is
$$
\operatorname{rank}(A) = 2.
$$
The matrix has $4$ columns, so by rank-nullity:
$$
\operatorname{rank}(A) + \operatorname{nullity}(A) = 4.
$$
Thus
$$
2 + \operatorname{nullity}(A) = 4
$$
so
$$
\operatorname{nullity}(A) = 2.
$$
2.5Find the Eigenvalues of a Matrix
Find the eigenvalues of
$$
A =
\begin{bmatrix}
3 & 1 \\
0 & 2
\end{bmatrix}.
$$
Solution
Use the characteristic equation:
$$
\det(A - \lambda I) = 0.
$$
Compute
$$
A - \lambda I =
\begin{bmatrix}
3 - \lambda & 1 \\
0 & 2 - \lambda
\end{bmatrix}.
$$
Then
$$
\det(A - \lambda I) = (3 - \lambda)(2 - \lambda).
$$
Set this equal to zero:
$$
(3 - \lambda)(2 - \lambda) = 0.
$$
So the eigenvalues are
$$
\lambda = 3 \quad \text{and} \quad \lambda = 2.
$$
2.6Use Diagonalization to Compute a Power
Suppose
$$
A = P
\begin{bmatrix}
2 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 3
\end{bmatrix}
P^{-1}.
$$
Find $A^4$.
Solution
Use the diagonalization formula:
$$
A^4 = P D^4 P^{-1},
$$
where
$$
D =
\begin{bmatrix}
2 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 3
\end{bmatrix}.
$$
Raise each diagonal entry to the fourth power:
$$
D^4 =
\begin{bmatrix}
16 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 81
\end{bmatrix}.
$$
Therefore
$$
A^4 = P
\begin{bmatrix}
16 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 81
\end{bmatrix}
P^{-1}.
$$
2.7Complete a Symmetric Matrix
Find the values of $a$ and $b$ that make the matrix symmetric:
$$
\begin{bmatrix}
1 & 4 & -2 \\
a & 0 & 7 \\
-2 & b & 5
\end{bmatrix}
$$
Solution
For a symmetric matrix, entries across the main diagonal must match.
So the $(1,2)$ and $(2,1)$ entries must be equal:
$$
a = 4.
$$
Also the $(2,3)$ and $(3,2)$ entries must be equal:
$$
b = 7.
$$
So the matrix is symmetric when
$$
a = 4,\quad b = 7.
$$
2.8Use Determinant Properties
Suppose
$$
\det(A) = -2
\quad \text{and} \quad
\det(B) = 5.
$$
Find $\det(B^T A)$ and determine whether $B^T A$ is singular.
Solution
Use the determinant properties:
$$
\det(B^T A) = \det(B^T)\det(A).
$$
Also,
$$
\det(B^T) = \det(B) = 5.
$$
So
$$
\det(B^T A) = 5(-2) = -10.
$$
Because the determinant is nonzero, $B^T A$ is not singular.
Difficulty
Applied Problems
3.1Model a Purchase with a System
A store sells notebooks and pens.
Two notebooks and three pens cost \$13. One notebook and one pen cost \$5.
What is the price of one notebook?
Solution
Let $n$ be the notebook price and $p$ be the pen price.
The system is
$$
\begin{aligned}
2n + 3p &= 13 \\
n + p &= 5
\end{aligned}
$$
Subtract the second equation from the first after doubling it:
$$
2n + 2p = 10
$$
Then
$$
p = 3.
$$
Substitute into $n + p = 5$:
$$
n + 3 = 5
$$
so
$$
n = 2.
$$
3.2Interpret a Row-Reduced System
A system row-reduces to
$$
\begin{bmatrix}
1 & 0 & 2 & \mid & 4 \\
0 & 1 & -1 & \mid & 3 \\
0 & 0 & 0 & \mid & 0
\end{bmatrix}.
$$
How many solutions does the system have?
Solution
There are pivots in the first two variable columns, but the third variable column has no pivot, so one variable is free.
Because there is no inconsistent row and at least one free variable, the system has infinitely many solutions.
3.3Apply Rank-Nullity
A $4 \times 6$ matrix has rank $4$.
How many free variables does the homogeneous system $Ax = 0$ have?
Solution
Use rank-nullity:
$$
\operatorname{rank}(A) + \operatorname{nullity}(A) = 6.
$$
Since $\operatorname{rank}(A) = 4$,
$$
4 + \operatorname{nullity}(A) = 6.
$$
So
$$
\operatorname{nullity}(A) = 2.
$$
That means there are $2$ free variables.
3.4Recognize an Eigenvector
A nonzero vector $v$ satisfies
$$
Av = -3v.
$$
What does this tell you about $v$ and the scalar $-3$?
Solution
The equation has the form
$$
Av = \lambda v.
$$
So $v$ is an eigenvector of $A$, and the corresponding eigenvalue is
$$
\lambda = -3.
$$
3.5Use an LU Factorization
A matrix $A$ has the factorization $A = LU$, where
$$
L =
\begin{bmatrix}
1 & 0 & 0 \\
-2 & 1 & 0 \\
3 & 4 & 1
\end{bmatrix}
\quad \text{and} \quad
U =
\begin{bmatrix}
2 & -1 & 0 \\
0 & 5 & 3 \\
0 & 0 & -4
\end{bmatrix}.
$$
Without multiplying $L$ and $U$, find $\det(A)$ and say whether $A$ is invertible.
Solution
Use the product property:
$$
\det(A) = \det(L)\det(U).
$$
Because $L$ is lower triangular with diagonal entries all $1$,
$$
\det(L) = 1.
$$
Because $U$ is upper triangular,
$$
\det(U) = 2 \cdot 5 \cdot (-4) = -40.
$$
So
$$
\det(A) = 1 \cdot (-40) = -40.
$$
Since the determinant is nonzero, $A$ is invertible.
Difficulty
Challenge / Synthesis
4.1Use a Determinant Identity
Suppose $A$ is a square matrix and
$$
\det(A) = 3.
$$
Find $\det(A^T A)$ and decide whether $A^T A$ is invertible.
Solution
Use the determinant rules:
$$
\det(A^T A) = \det(A^T)\det(A).
$$
Also,
$$
\det(A^T) = \det(A) = 3.
$$
Therefore
$$
\det(A^T A) = 3 \cdot 3 = 9.
$$
Because the determinant is nonzero, $A^T A$ is invertible.
4.2A Matrix That Is Symmetric and Skew-Symmetric
If a matrix satisfies both
$$
A^T = A
$$
and
$$
A^T = -A,
$$
what must $A$ be?
Solution
Since both equalities hold, we have
$$
A = -A.
$$
Add $A$ to both sides:
$$
2A = 0.
$$
Over the real numbers, this implies
$$
A = 0.
$$
So the matrix must be the zero matrix.
4.3An Invertible Projection Matrix
Suppose a matrix $P$ satisfies
$$
P^2 = P
$$
and also has an inverse. What must $P$ be?
Solution
Because $P$ is invertible, multiply the equation $P^2 = P$ on the left by $P^{-1}$:
$$
P^{-1}P^2 = P^{-1}P.
$$
This simplifies to
$$
P = I.
$$
So an invertible projection matrix must be the identity matrix.
4.4Reason About a QR Factorization
A matrix $A$ is written as
$$
A = QR,
$$
where $Q$ is orthogonal and
$$
R =
\begin{bmatrix}
2 & 1 & 0 \\
0 & -3 & 4 \\
0 & 0 & 5
\end{bmatrix}.
$$
Explain why $A$ is invertible.
Solution
An orthogonal matrix is invertible because
$$
Q^{-1} = Q^T.
$$
The matrix $R$ is upper triangular, so its determinant is the product of its diagonal entries:
$$
\det(R) = 2 \cdot (-3) \cdot 5 = -30.
$$
Since $\det(R) \ne 0$, $R$ is invertible.
Because both $Q$ and $R$ are invertible, their product
$$
A = QR
$$
is also invertible.