Matrices and determinants of order 2

Unit 10 Matrices and determinants of order 2

Key unit competence: Use matrices and determinants of order 2 in solving simultaneous linear equations.

Key ideas

• A linear transformation $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ satisfies $T(\vec{u}+\vec{v})=T(\vec{u})+T(\vec{v})$ and $T(k\vec{u})=kT(\vec{u})$. Its matrix has columns $T(\vec{i})$ and $T(\vec{j})$.
• Matrices of the same size are added element by element; $kA$ multiplies every element by $k$.
• Product: $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)=\left(\begin{array}{cc}ae+bg& af+bh\\ ce+dg& cf+dh\end{array}\right)$. In general $AB\ne BA$.
• Determinant: $\det \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=ad-bc$. $\det (AB)=\det A\det B$.
• Transpose $A^T$: swap rows and columns. Inverse (if $\det A\ne 0$): $A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$, with $A{A}^{-1}=I$.
• System $AX=B$: if $\det A\ne 0$, $X=A^{-1}B$ (or Cramer's rule).

Formulas to know

Common misconceptions

• "$AB=BA$ for matrices." Not in general: matrix multiplication is not commutative.
• "To multiply matrices, multiply corresponding elements." No: multiply rows by columns.
• "Every square matrix has an inverse." Only if its determinant is non-zero.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Compute $AB$ and $BA$ for $A=\left(\begin{array}{cc}2& 1\\ 0& 3\end{array}\right)$ and $B=\left(\begin{array}{cc}1& -1\\ 2& 0\end{array}\right)$.

   Show answer

   $AB=\left(\begin{array}{cc}4& -2\\ 6& 0\end{array}\right)$; $BA=\left(\begin{array}{cc}2& -2\\ 4& 2\end{array}\right)$. Not equal.

   Common mistake: Multiplying element by element.

2. Find $X$ if $\left(\begin{array}{cc}1& 2\\ 3& 5\end{array}\right)X=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$.

   Show answer

   $X=A^{-1}=\frac{1}{-1}\left(\begin{array}{cc}5& -2\\ -3& 1\end{array}\right)=\left(\begin{array}{cc}-5& 2\\ 3& -1\end{array}\right)$.

   Common mistake: Ignoring the negative determinant.

3. Find the inverse of $\left(\begin{array}{cc}4& 7\\ 2& 6\end{array}\right)$ and check it.

   Show answer

   $\det =10$; $A^{-1}=\frac{1}{10}\left(\begin{array}{cc}6& -7\\ -2& 4\end{array}\right)$. Check $A{A}^{-1}=I$.

   Common mistake: Forgetting to divide by the determinant.

4. Prove that $(A^T{)}^{-1}=(A^{-1}{)}^T$ for $A=\left(\begin{array}{cc}1& 2\\ 1& 3\end{array}\right)$.

   Show answer

   $A^{-1}=\left(\begin{array}{cc}3& -2\\ -1& 1\end{array}\right)$, $(A^{-1}{)}^T=\left(\begin{array}{cc}3& -1\\ -2& 1\end{array}\right)$. $A^T=\left(\begin{array}{cc}1& 1\\ 2& 3\end{array}\right)$, inverse $\left(\begin{array}{cc}3& -1\\ -2& 1\end{array}\right)$. Equal. ✓

   Common mistake: Transposing before computing the cofactors incorrectly.

5. Solve using the inverse matrix: $2x-y=7$, $x+3y=0$.

   Show answer

   $\det =7$; $A^{-1}=\frac{1}{7}\left(\begin{array}{cc}3& 1\\ -1& 2\end{array}\right)$; $X=\frac{1}{7}\left(\begin{array}{c}21\\ -7\end{array}\right)=\left(\begin{array}{c}3\\ -1\end{array}\right)$.

   Common mistake: Computing $B{A}^{-1}$ instead of $A^{-1}B$.

6. For which $k$ is the system $kx+2y=1$, $2x+ky=3$ without a unique solution?

   Show answer

   $\det =k^2-4=0$, $k=\pm 2$.

   Common mistake: Giving only $k=2$.

7. Find the matrix of $T(x,y)=(3x-y,2y)$ and the image of $(1,4)$.

   Show answer

   $\left(\begin{array}{cc}3& -1\\ 0& 2\end{array}\right)$; image $(-1,8)$.

   Common mistake: Writing the coefficients as columns of the wrong rows.

8. Show that $\det (AB)=\det A\det B$ for $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$, $B=\left(\begin{array}{cc}2& 0\\ 1& 1\end{array}\right)$.

   Show answer

   $AB=\left(\begin{array}{cc}4& 2\\ 10& 4\end{array}\right)$, $\det =16-20=-4$. $\det A=-2$, $\det B=2$, product $-4$. ✓

   Common mistake: Arithmetic slips in the product.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Solve $\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)X=\left(\begin{array}{c}5\\ 1\end{array}\right)$.	a. $x=2$, $y=1$
   2. Use Cramer's rule to solve $2x+3y=7$, $x-y=1$.	b. $X=\left(\begin{array}{c}3\\ 2\end{array}\right)$
   3. $A$ is a $2\times 2$ matrix with $\det A=3$. Find $\det (2A)$.	c. $X=\left(\begin{array}{c}1\\ 1\end{array}\right)$
   4. Solve $\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)X=\left(\begin{array}{c}3\\ 2\end{array}\right)$.	d. $12$

   Show answer

   1 → b ($X=\left(\begin{array}{c}3\\ 2\end{array}\right)$) · 2 → a ($x=2$, $y=1$) · 3 → d ($12$) · 4 → c ($X=\left(\begin{array}{c}1\\ 1\end{array}\right)$)

   • Step 1. $\det =-1-1=-2$; $A^{-1}=-\frac{1}{2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$.
     Step 2. $X=\frac{1}{2}\left(\begin{array}{c}6\\ 4\end{array}\right)=\left(\begin{array}{c}3\\ 2\end{array}\right)$.
   • Step 1. $\Delta =2(-1)-3(1)=-5$.
     Step 2. ${\Delta }_x=7(-1)-3(1)=-10$, ${\Delta }_y=2(1)-7(1)=-5$.
     Step 3. $x=\frac{-10}{-5}=2$, $y=\frac{-5}{-5}=1$.
   • Step 1. $2A$ multiplies both rows by 2.
     Step 2. $\det (2A)=2^2\det A=4\times 3=12$.
   • Step 1. $\det =1$, so $A^{-1}=\left(\begin{array}{cc}1& -1\\ -1& 2\end{array}\right)$.
     Step 2. $X=\left(\begin{array}{c}3-2\\ -3+4\end{array}\right)=\left(\begin{array}{c}1\\ 1\end{array}\right)$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\left(\begin{array}{cc}4& -3\\ -1& 1\end{array}\right)$  ·  $a=3$  ·  100 FRW  ·  $\left(\begin{array}{cc}1& -3\\ -1& 4\end{array}\right)$
   1. 2 pens and 3 books cost 2 300 FRW; 4 pens and 1 book cost 1 100 FRW. The price of a pen is .
   2. Find $a$ if $\det \left(\begin{array}{cc}a& 1\\ 2& 3\end{array}\right)=7$. →
   3. Find the inverse of $\left(\begin{array}{cc}4& 3\\ 1& 1\end{array}\right)$. →
   Show answer

   1: 100 FRW · 2: $a=3$ · 3: $\left(\begin{array}{cc}1& -3\\ -1& 4\end{array}\right)$

   • Step 1. $\left(\begin{array}{cc}2& 3\\ 4& 1\end{array}\right)\left(\begin{array}{c}p\\ b\end{array}\right)=\left(\begin{array}{c}2300\\ 1100\end{array}\right)$; $\det =2-12=-10$.
     Step 2. $p=\frac{2300\times 1-3\times 1100}{-10}=\frac{-1000}{-10}=100$ FRW (and $b=700$).
   • Step 1. $3a-2=7$.
     Step 2. $a=3$.
   • Step 1. $\det =4-3=1$.
     Step 2. Swap the diagonal and change the signs of the other entries: $\left(\begin{array}{cc}1& -3\\ -1& 4\end{array}\right)$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$. Find $A^2-3A+I$.	a. $\left(\begin{array}{cc}2& -1\\ -5& 3\end{array}\right)$
   2. Find the inverse of $\left(\begin{array}{cc}2& 4\\ 1& 3\end{array}\right)$.	b. $\frac{1}{2}\left(\begin{array}{cc}3& -4\\ -1& 2\end{array}\right)$
   3. $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$. Find $A+A^T$.	c. The zero matrix
   4. $A=\left(\begin{array}{cc}3& 1\\ 5& 2\end{array}\right)$ and $AB=I$. Find $B$.	d. $\left(\begin{array}{cc}2& 5\\ 5& 8\end{array}\right)$

   Show answer

   1 → c (The zero matrix) · 2 → b ($\frac{1}{2}\left(\begin{array}{cc}3& -4\\ -1& 2\end{array}\right)$) · 3 → d ($\left(\begin{array}{cc}2& 5\\ 5& 8\end{array}\right)$) · 4 → a ($\left(\begin{array}{cc}2& -1\\ -5& 3\end{array}\right)$)

   • Step 1. $A^2=\left(\begin{array}{cc}5& 3\\ 3& 2\end{array}\right)$.
     Step 2. $A^2-3A+I=\left(\begin{array}{cc}5-6+1& 3-3\\ 3-3& 2-3+1\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$.
   • Step 1. $\det =6-4=2$.
     Step 2. $A^{-1}=\frac{1}{2}\left(\begin{array}{cc}3& -4\\ -1& 2\end{array}\right)$.
   • Step 1. $A^T=\left(\begin{array}{cc}1& 3\\ 2& 4\end{array}\right)$ (swap rows and columns).
     Step 2. Add: $\left(\begin{array}{cc}2& 5\\ 5& 8\end{array}\right)$ (a symmetric matrix).
   • Step 1. $B=A^{-1}$; $\det A=6-5=1$.
     Step 2. $A^{-1}=\left(\begin{array}{cc}2& -1\\ -5& 3\end{array}\right)$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $x=1$, $y=2$  ·  $\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$  ·  $\left(\begin{array}{cc}2& 4\\ 0& 2\end{array}\right)$  ·  $\left(\begin{array}{cc}1& 3\\ 0& 1\end{array}\right)$
   1. $A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. Find $A^3$. →
   2. Solve $3x+y=5$, $5x+2y=9$ using matrices. →
   3. $A=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$. Find $A^2$. →
   Show answer

   1: $\left(\begin{array}{cc}1& 3\\ 0& 1\end{array}\right)$ · 2: $x=1$, $y=2$ · 3: $\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$

   • Step 1. $A^2=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$.
     Step 2. $A^3=A^{2}A=\left(\begin{array}{cc}1& 3\\ 0& 1\end{array}\right)$.
   • Step 1. $\det =6-5=1$; $A^{-1}=\left(\begin{array}{cc}2& -1\\ -5& 3\end{array}\right)$.
     Step 2. $X=\left(\begin{array}{c}10-9\\ -25+27\end{array}\right)=\left(\begin{array}{c}1\\ 2\end{array}\right)$.
   • Step 1. Row 1: $(1\cdot 1+2\cdot 0,1\cdot 2+2\cdot 1)=(1,4)$.
     Step 2. Row 2: $(0,1)$.

Take the Unit 10 quiz → · Back to contents