101 Best Linear Algebra Problems and Solutions: From Eigenvalues to Vector Spaces

Scale, then subtract component by component:

Check: the first components give $3 \cdot 2 - 2 \cdot 1 = 4$.

The length is $\sqrt{1 + 4 + 4} = 3$, so the unit vector is $\left(\tfrac{1}{3}, \tfrac{2}{3}, \tfrac{2}{3}\right)$.

Check: $\tfrac{1}{9} + \tfrac{4}{9} + \tfrac{4}{9} = 1$.

$2 - 5 + 3 = 0$. The dot product is zero, so yes, they are orthogonal.

Check: $3(0.8) + 2(0.8) - 4 = 2.4 + 1.6 - 4 = 0$.

Check: the result must be orthogonal to both inputs: $-3 + 12 - 9 = 0$ and $-12 + 30 - 18 = 0$.

$\overrightarrow{PQ} = (-1, 2, 0)$ and $\overrightarrow{PR} = (-1, 0, 3)$. Their cross product is $(6, 3, 2)$, whose length is $\sqrt{36 + 9 + 4} = 7$. The triangle is half the parallelogram, so the area is $\tfrac{7}{2} = 3.5$.

Check: $\mathbf{e} \cdot \mathbf{a} = -2.4 + 2.4 = 0$, so the error is perpendicular to the line, as it must be.

$\mathbf{u} \cdot \mathbf{v} = 11$, and $\lVert \mathbf{u} \rVert \, \lVert \mathbf{v} \rVert = \sqrt{5} \cdot 5 \approx 11.18$. Since $11 \le 11.18$, the inequality holds. Equality would require the vectors to be parallel.

Solve $c_1 + 3c_2 = 7$ and $2c_1 + c_2 = 8$. From the second, $c_2 = 8 - 2c_1$; substituting gives $c_1 + 24 - 6c_1 = 7$, so $c_1 = \tfrac{17}{5}$ and $c_2 = \tfrac{6}{5}$.

Check: $\tfrac{17}{5} + \tfrac{18}{5} = 7$ and $\tfrac{34}{5} + \tfrac{6}{5} = 8$.

From the first two components, $c_1 = 1$ and $c_2 = 2$. The third component must then be $c_1 + c_2 = 3$, which matches. Yes: $(1, 2, 3) = 1(1, 0, 1) + 2(0, 1, 1)$.

Put them in the columns of a matrix and take the determinant:

The determinant is nonzero, so the vectors are independent and form a basis of $\mathbb{R}^3$.

The vectors step by the same amount, $(3, 3, 3)$, so the middle one is the average of the other two: $\mathbf{v}_1 - 2\mathbf{v}_2 + \mathbf{v}_3 = \mathbf{0}$.

Check: $(1 - 8 + 7,\; 2 - 10 + 8,\; 3 - 12 + 9) = (0, 0, 0)$.

Any set of more than $n$ vectors in $\mathbb{R}^n$ is dependent, because a matrix with more columns than rows has a free column. Here $(1, 2, 3) = 1\mathbf{e}_1 + 2\mathbf{e}_2 + 3\mathbf{e}_3$.

$\det\begin{bmatrix} 1 & k \\ k & 4 \end{bmatrix} = 4 - k^2 = 0$, so $k = 2$ or $k = -2$.

Check: with $k = 2$, $(2, 4) = 2(1, 2)$.

All multiples $t(1, 2) = (t, 2t)$: the line $y = 2x$ through the origin.

$\det\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} = -2 \ne 0$, so they span the plane. Solving $c_1 + c_2 = 5$ and $c_1 - c_2 = 3$ gives $c_1 = 4$ and $c_2 = 1$.

A normal vector is the cross product: $(1, 0, 2) \times (0, 1, -1) = (-2, 1, 1)$. The plane is $-2x + y + z = 0$.

Check: $(1, 0, 2)$ gives $-2 + 0 + 2 = 0$; $(0, 1, -1)$ gives $0 + 1 - 1 = 0$.

Write their coefficients (constant, $x$, $x^2$) as columns:

The matrix is triangular with nonzero diagonal, so its determinant is $1$ and the polynomials are independent. They form a basis of the polynomials of degree at most 2.

Multiplying by $B$ on the right swaps the columns of $A$; on the left it swaps the rows. $AB \ne BA$.

A $2 \times 3$ matrix times a $3 \times 1$ vector gives a $2 \times 1$ vector:

$(AB)^{T} = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$. Since $B^{T} = B$, $B^{T}A^{T} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix}$. They match. The order reverses.

$A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$, and each further multiplication adds 1 to the top-right entry, so $A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$ and $A^5 = \begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix}$. (Prove the pattern by induction: $A^{n}A$ adds the top-left 1 to the top-right entry.)

$\operatorname{tr}(AB) = 2 + 3 = 5$ and $\operatorname{tr}(BA) = 3 + 2 = 5$. The products differ but their traces agree, which is true for any compatible square products.

Check: $S + K = A$, $S^{T} = S$ and $K^{T} = -K$.

Each multiplication shifts the ones one place further up and to the right:

All eigenvalues of a nilpotent matrix are 0.

This "column picture" is why $A\mathbf{x} = \mathbf{b}$ is solvable exactly when $\mathbf{b}$ is in the column space of $A$.

The second row is twice the first, so the rank is 1. Every nonzero outer product has rank 1.

Check: the length is still 5, and $(3, 4) \cdot (-4, 3) = 0$.

$P^2 = \tfrac{1}{4}\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = P$. Projecting twice is the same as projecting once.

The reflection swaps coordinates: $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, giving $(5, 2)$. Applying it twice returns the original vector, so the matrix is its own inverse.

Adding the equations gives $2x = 6$, so $x = 3$ and $y = 1$.

Subtract twice row 1 from row 2, and row 1 from row 3:

From the second, $y = 2z - 4$. Substituting: $-3(2z - 4) - z = -9$, so $-7z = -21$ and $z = 3$. Then $y = 2$ and $x = 1$.

Check: $1 + 2 + 3 = 6$, $2 - 2 + 3 = 3$, $1 + 4 - 3 = 2$.

Subtracting twice the first equation from the second gives $0 = 1$, which is impossible. There is no solution: the lines are parallel.

Subtracting gives $2y = 2$, so $y = 1$ and $x + z = 2$. Let $z = t$ be free:

The solution set is a line: two planes in $\mathbb{R}^3$ that are not parallel meet in a line.

Pivots are in columns 1 and 3, so $y$ is free. With $y = t$: $(x, y, z) = (4 - 2t,\; t,\; 1)$.

Row 2 is twice row 1, so only $x + 2y + 3z = 0$ remains. With $y = s$ and $z = t$ free:

The null space is a plane with basis $(-2, 1, 0)$ and $(-3, 0, 1)$.

The coefficient determinant is $k - 2$. For $k \ne 2$ there is exactly one solution. For $k = 2$ the second equation says $2x + 2y = 3$, while the first implies $2x + 2y = 2$, so there is no solution.

Replace column 1 with the right-hand side $(6, 3, 2)$:

This matches problem 36. Cramer's rule is useful for theory and tiny systems; elimination is far cheaper for anything larger.

The multiplier for row 2 is $6 / 2 = 3$; subtracting $3 \times$ row 1 leaves $(0, 5)$:

Multipliers 2 and 4 clear column 1, leaving rows $(0, 1, 1)$ and $(0, 3, 5)$. Multiplier 3 then clears column 2, leaving $(0, 0, 2)$:

Check: $\det A = 2 \cdot 1 \cdot 2 = 4$, the product of the pivots.

Exactly: $y = 2 - x$, so $0.0001x + 2 - x = 1$, giving $x = \tfrac{1}{0.9999} \approx 1.0001$ and $y \approx 0.9999$.

Without a swap, the multiplier is $10\,000$, and row 2 becomes $-9999y = -9998$. With three significant digits both numbers round to about $-10\,000$, so $y = 1$, and back-substitution gives $0.0001x = 0$, so $x = 0$, which is badly wrong. Swapping rows first (partial pivoting) uses the multiplier $0.0001$ and gives $x \approx 1$, $y \approx 1$ even at low precision.

Atom counts give a homogeneous system: carbon $3a = c$, hydrogen $8a = 2d$, oxygen $2b = 2c + d$. Setting $a = 1$: $c = 3$, $d = 4$, $b = \tfrac{6 + 4}{2} = 5$.

$3 \cdot 2 - 1 \cdot 0 = 6$. The matrix sends the unit square to the parallelogram spanned by its columns $(3, 0)$ and $(1, 2)$, which has area 6.

Expand along row 1 (the zero saves work):

For a triangular matrix the determinant is the product of the diagonal: $3 \cdot 2 \cdot 4 = 24$.

Multiplying the matrix by 2 multiplies each of the 3 rows by 2, and each row scaling multiplies the determinant by 2: $\det(2A) = 2^3 \cdot 5 = 40$.

$\det(AB) = 3(-2) = -6$. $\det(A^{-1}B^2) = \tfrac{1}{3} \cdot (-2)^2 = \tfrac{4}{3}$. $\det(A^{T}) = \det A = 3$.

$k - 6 = 0$, so $k = 6$. Then row 2 is three times row 1.

This is a Vandermonde matrix with nodes $1, 2, 3$, whose determinant is the product of all differences $x_j - x_i$ with $i < j$:

Because it is nonzero, exactly one parabola passes through any three points with $x = 1, 2, 3$.

The volume is $|6| = 6$.

Adding multiples of one row to another does not change the determinant. Clear column 1 (row 3 minus $2\times$ row 1, row 4 minus row 1), then column 2 using row 2:

Finally row 4 minus $\tfrac{4}{13}$ row 3 leaves $1 + \tfrac{4}{13} = \tfrac{17}{13}$ in the corner. The determinant is $1 \cdot 1 \cdot 13 \cdot \tfrac{17}{13} = 17$.

$\det A = 24 - 14 = 10$. Swap the diagonal, negate the off-diagonal and divide by the determinant:

Check: the top-left entry of $AA^{-1}$ is $\tfrac{24 - 14}{10} = 1$.

Row-reduce $[A \mid I]$ until the left block is $I$. The determinant is $1(0 - 24) - 2(0 - 20) + 3(0 - 5) = 1$, so the inverse has integer entries:

Check: row 1 of $A$ times column 1 of $A^{-1}$: $-24 + 40 - 15 = 1$.

Check: $4(-0.8) + 7(0.6) = 1$ and $2(-0.8) + 6(0.6) = 2$. In software, call a solver (such as numpy.linalg.solve) rather than forming the inverse; it is faster and more accurate.

$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = I$. The order reverses, like taking off socks and shoes.

Its columns are orthonormal, so $R^{-1} = R^{T} = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$, which is rotation by $-\theta$.

The determinant is $4 - 4 = 0$. Also, $(2, -1)$ is a nonzero vector in the null space: $2 \cdot 2 + 4 \cdot (-1) = 0$ and $2 - 2 = 0$. A matrix that sends a nonzero vector to zero cannot be undone.

$I + \mathbf{u}\mathbf{v}^{T} = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}$ and $\mathbf{v}^{T}\mathbf{u} = 1$, so

Check: the 2 × 2 formula gives $\tfrac{1}{2}\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}$, the same matrix. The formula lets software update an inverse after a rank-one change without starting over.

$y = 2x$ is a subspace: it contains $\mathbf{0}$ and is closed under addition and scaling. $y = 2x + 1$ is not, because it does not contain $(0, 0)$.

Sums and multiples of symmetric matrices are symmetric, and the zero matrix is symmetric. Every such matrix is

These three matrices form a basis, so the dimension is 3.

Write $p = a + b(1 + x) + c(1 + x + x^2)$ and match coefficients from the highest power down: $x^2$: $c = 5$; $x$: $b + c = -2$, so $b = -7$; constant: $a + b + c = 3$, so $a = 5$. Coordinates: $(5, -7, 5)$.

Check: $5 - 7(1 + x) + 5(1 + x + x^2) = 3 - 2x + 5x^2$.

Row 2 is twice row 1. Rows 1 and 3 are not multiples of each other. Elimination leaves two nonzero rows, so the rank is 2.

Elimination gives $x - z = 0$ and $y + 2z = 0$. With $z = t$: $\mathbf{x} = t(1, -2, 1)$, so the columns satisfy $\mathbf{c}_1 - 2\mathbf{c}_2 + \mathbf{c}_3 = \mathbf{0}$. Problem 16 found the same coefficients for the rows.

$\text{rank} + \text{nullity} = \text{number of columns}$, so the nullity is $5 - 2 = 3$.

• Column space: multiples of $(1, 2)$.
• Null space: $x + 2y = 0$, so multiples of $(-2, 1)$.
• Row space: multiples of $(1, 2)$ (the matrix is symmetric).
• Left null space (null space of $A^{T}$): multiples of $(-2, 1)$.

Check: the row space is perpendicular to the null space: $(1, 2) \cdot (-2, 1) = 0$.

Solve $2a + b = 4$ and $a + b = 3$: subtracting gives $a = 1$, then $b = 2$. Coordinates: $(1, 2)$. Equivalently, multiply by $P^{-1}$ where $P = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$ has the basis vectors as columns.

One equation of rank 1 on 4 unknowns leaves $4 - 1 = 3$ free variables. A basis is $(-1, 1, 0, 0)$, $(-1, 0, 1, 0)$, $(-1, 0, 0, 1)$.

From the second, $y = x$; then $z = -2x$. The intersection is the line spanned by $(1, 1, -2)$.