The Timpson Lyceum · Linear Algebra (2026)

3.3 Exercises

Problem sheet · 20 points · 4 extra credit

Cramer's Rule, Volume, and Linear Transformations

1. 1. (3 pts) Use Cramer's rule to solve the system. Show each determinant.

   $\begin{aligned} 3x + 2y &= 7 \\ x - y &= -1 \end{aligned}$

2. 2. (3 pts) Use Cramer's rule to solve the system for $x$, $y$, and $z$.

   $\begin{aligned} x + y + z &= 6 \\ 2x - y + z &= 3 \\ x + 2y - z &= 2 \end{aligned}$

3. 3. (3 pts) Use the adjugate (classical adjoint) formula $A^{-1} = \dfrac{1}{\det A}\operatorname{adj}A$ to find the inverse of

   $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}.$

4. 4. (2 pts) Let $\mathbf{u} = (3,1)$ and $\mathbf{v} = (1,4)$. Find the area of the parallelogram determined by $\mathbf{u}$ and $\mathbf{v}$, and the area of the triangle with $\mathbf{u}$ and $\mathbf{v}$ as two of its sides.

5. 5. (2 pts) Find the volume of the parallelepiped determined by the vectors $\mathbf{u} = (1,0,2)$, $\mathbf{v} = (0,3,1)$, and $\mathbf{w} = (2,1,0)$.

6. 6. (2 pts) Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}$. A region $S$ in the plane has area $4$. Find the area of its image $T(S)$, and the area of the image of the unit square.

7. 7. (2 pts) The columns of $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ are the images $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$ under $T(\mathbf{x}) = A\mathbf{x}$. Explain why the unit square maps to a parallelogram of area $|\det A|$, and state the area of the image of a region of area $k$.

8. 8. (3 pts) State whether each statement is true or false, and justify each answer.

   (a) Cramer's rule can be used to solve any system $A\mathbf{x} = \mathbf{b}$.

   (b) For $T(\mathbf{x}) = A\mathbf{x}$ on $\mathbb{R}^2$, the image of the unit disk (area $\pi$) has area $\pi\,|\det A|$.

   (c) The $(i,j)$ entry of $\operatorname{adj}A$ is the $(i,j)$ cofactor $C_{ij}$ of $A$.

9. 9. (4 pts ••) Prove the adjugate inverse formula: for an $n\times n$ matrix $A$, show that $A\,(\operatorname{adj}A) = (\det A)\,I$, and conclude that $A^{-1} = \dfrac{1}{\det A}\operatorname{adj}A$ when $\det A \neq 0$. Recall that $\operatorname{adj}A$ is the transpose of the matrix of cofactors, so its $(i,j)$ entry is $C_{ji}$.