Matrices and determinants of order 3

Unit 6 Matrices and determinants of order 3

Key unit competence: Apply properties of matrices and determinants of order 3 in solving related problems, including systems of three linear equations.

Key ideas

• A square matrix of order 3 has 3 rows and 3 columns; $a_{ij}$ is in row $i$, column $j$. Types: zero, identity $I_3$, diagonal, triangular, symmetric ($A^T=A$), skew-symmetric ($A^T=-A$).
• Addition and scalar multiplication are element by element. Product $AB$: entry $(i,j)$ is row $i$ of $A$ times column $j$ of $B$. In general $AB\ne BA$; $(AB{)}^T=B^T{A}^T$.
• Determinant (expansion along row 1): $\det A=a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|$ (or Sarrus' rule).
• Properties: swapping two rows changes the sign; a row of zeros or two equal rows give 0; multiplying a row by $k$ multiplies $\det$ by $k$; $\det (AB)=\det A\det B$; $\det A^T=\det A$; $\det (kA)=k^3\det A$.
• Cofactor $C_{ij}=(-1{)}^{i+j}{M}_{ij}$; adjoint $\mathrm{adj}A=(C_{ij}{)}^T$; inverse $A^{-1}=\frac{1}{\det A}\mathrm{adj}A$ when $\det A\ne 0$.
• Systems $AX=B$: unique solution if $\det A\ne 0$: $X=A^{-1}B$ or Cramer's rule $x=\frac{{\Delta }_x}{\Delta }$, etc. A linear transformation of ${\mathbb{R}}^3$ has a $3\times 3$ matrix whose columns are the images of $\vec{i},\vec{j},\vec{k}$.

Formulas to know

Common misconceptions

• "In the cofactor expansion all signs are $+$." No: signs alternate $+,-,+$ along row 1 (pattern $(-1{)}^{i+j}$).
• "The adjoint is the matrix of cofactors." It is the transpose of the cofactor matrix.

Worked examples

Practice questions

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

1. Compute $\det \left(\begin{array}{ccc}3& -1& 2\\ 1& 0& 4\\ 2& 5& 1\end{array}\right)$.

   Show answer

   $3(0-20)-(-1)(1-8)+2(5-0)=-60-7+10=-57$.

   Common mistake: Using $+$ for the middle term.

2. Use row properties to show $\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$ when $a=b$.

   Show answer

   If $a=b$, rows 1 and 2 are equal, so the determinant is 0.

   Common mistake: Expanding fully without using the property.

3. Find the inverse of $\left(\begin{array}{ccc}1& 0& 2\\ 0& 1& 0\\ 1& 0& 3\end{array}\right)$.

   Show answer

   $\det =1(3)-0+2(0-1)=1$. Cofactors give $A^{-1}=\left(\begin{array}{ccc}3& 0& -2\\ 0& 1& 0\\ -1& 0& 1\end{array}\right)$. Check $A{A}^{-1}=I$.

   Common mistake: Forgetting to transpose the cofactor matrix.

4. Determine whether the system $x+y+z=1$, $2x+2y+2z=3$, $x-y=0$ has a unique solution.

   Show answer

   Rows 1 and 2 of $A$ are proportional: $\det A=0$, no unique solution. In fact the first two equations contradict ($2\ne 3$): no solution.

   Common mistake: Solving without checking the determinant.

5. Solve by Cramer's rule: $x+2y-z=2$, $2x-y+z=3$, $x+y+z=6$.

   Show answer

   $\Delta =1(-2)-2(1)+(-1)(3)=-7$. ${\Delta }_x=2(-2)-2(3-6)+(-1)(3+6)=-7$, so $x=1$. Similarly $y=2$, $z=3$. Check: $1+4-3=2$ ✓.

   Common mistake: Replacing the wrong column for ${\Delta }_x$.

6. Verify $(AB{)}^T=B^T{A}^T$ with $A=\left(\begin{array}{ccc}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$ and $B=\left(\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 0& 0& 2\end{array}\right)$.

   Show answer

   $AB=\left(\begin{array}{ccc}7& 2& 0\\ 3& 1& 0\\ 0& 0& 2\end{array}\right)$, so $(AB{)}^T=\left(\begin{array}{ccc}7& 3& 0\\ 2& 1& 0\\ 0& 0& 2\end{array}\right)$, which equals $B^T{A}^T$.

   Common mistake: Computing $A^T{B}^T$.

7. Show that $\det (kA)=k^3\det A$ for a $3\times 3$ matrix.

   Show answer

   $kA$ multiplies each of the three rows by $k$; each row multiplies the determinant by $k$, so the factor is $k^3$.

   Common mistake: Answering $k\det A$.

8. Find the matrix of the linear transformation $T(x,y,z)=(2x-z,x+y,3z)$ and the image of $(1,-1,2)$.

   Show answer

   $\left(\begin{array}{ccc}2& 0& -1\\ 1& 1& 0\\ 0& 0& 3\end{array}\right)$; image $(0,0,6)$.

   Common mistake: Writing coefficients as columns instead of rows.

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. Using Cramer's rule for $x+y+z=6$, $2x-y+z=3$, $x+2y-z=2$, find $\Delta$ and $y$.	a. $x=2,y=-1,z=3$
   2. Find $k\ne 1$ such that $\left|\begin{array}{ccc}k& 1& 1\\ 1& k& 1\\ 1& 1& k\end{array}\right|=0$.	b. $k=-2$
   3. Solve $x+2y+z=3$, $2x+y+z=6$, $x+y+2z=7$.	c. $\Delta =7,y=2$
   4. $A=\left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 4\\ 5& 6& 0\end{array}\right)$ has $\det A=1$. Find the entry in row 1, column 1 of $A^{-1}$.	d. $-24$

   Show answer

   1 → c ($\Delta =7,y=2$) · 2 → b ($k=-2$) · 3 → a ($x=2,y=-1,z=3$) · 4 → d ($-24$)

   • Step 1. $\Delta =1(1-2)-1(-2-1)+1(4+1)=-1+3+5=7$.
     Step 2. ${\Delta }_y=\left|\begin{array}{ccc}1& 6& 1\\ 2& 3& 1\\ 1& 2& -1\end{array}\right|=1(-5)-6(-3)+1(1)=14$.
     Step 3. $y=\frac{{\Delta }_y}{\Delta }=\frac{14}{7}=2$.
   • Step 1. Expanding gives $k^3-3k+2$.
     Step 2. Factorise: $(k-1{)}^2(k+2)$.
     Step 3. With $k\ne 1$: $k=-2$.
   • Step 1. Add all three: $4x+4y+4z=16$, so $x+y+z=4$.
     Step 2. Equation 1 minus this: $y=-1$. Equation 2 minus this: $x=2$.
     Step 3. Equation 3 minus this: $z=3$.
   • Step 1. $A^{-1}=\frac{1}{\det A}\mathrm{adj}A$, and entry $(1,1)$ of $\mathrm{adj}A$ is $C_{11}$.
     Step 2. $C_{11}=\left|\begin{array}{cc}1& 4\\ 6& 0\end{array}\right|=0-24=-24$.
     Step 3. Divide by $\det A=1$: $-24$.

2. Drag and drop

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

   $1$  ·  $2$  ·  $k\ne 5$  ·  $\frac{2}{3}$
   1. For the same system $2x+y=3$, $y+z=5$, $x+z=4$, find $x$. →
   2. For which values of $k$ does $x+y+z=1$, $x+2y+3z=2$, $x+3y+kz=3$ have a unique solution? →
   3. $A$ is $3\times 3$, $A^2=A$ and $A$ is invertible. Find $\det A$. →
   Show answer

   1: $\frac{2}{3}$ · 2: $k\ne 5$ · 3: $1$

   • Step 1. ${\Delta }_x=\left|\begin{array}{ccc}3& 1& 0\\ 5& 1& 1\\ 4& 0& 1\end{array}\right|=3(1)-1(5-4)+0=2$.
     Step 2. $x=\frac{{\Delta }_x}{\Delta }=\frac{2}{3}$.
   • Step 1. $\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 3& k\end{array}\right|=1(2k-9)-1(k-3)+1(3-2)$.
     Step 2. $\Delta =k-5$.
     Step 3. A unique solution needs $\Delta \ne 0$: $k\ne 5$.
   • Step 1. $(\det A{)}^2=\det A$, so $\det A=0$ or $1$.
     Step 2. Invertible means $\det A\ne 0$, so $\det A=1$.

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. $\det A=7$. $B$ is obtained by swapping rows 1 and 3 of $A$, then multiplying row 2 by 3. Find $\det B$.	a. $10$
   2. Find the cofactor $C_{32}$ of $\left(\begin{array}{ccc}2& 1& 3\\ 4& 0& 1\\ 5& 2& 6\end{array}\right)$.	b. $\left(\begin{array}{ccc}2& 0& 1\\ 1& 3& 2\\ 1& 1& 1\end{array}\right)$
   3. Which matrix is singular?	c. $40$
   4. $A$ is a $3\times 3$ matrix with $\det A=5$. Find $\det (2A)$.	d. $-21$

   Show answer

   1 → d ($-21$) · 2 → a ($10$) · 3 → b ($\left(\begin{array}{ccc}2& 0& 1\\ 1& 3& 2\\ 1& 1& 1\end{array}\right)$) · 4 → c ($40$)

   • Step 1. Swapping two rows changes the sign: $-7$.
     Step 2. Multiplying one row by 3 multiplies the determinant by 3: $-21$.
   • Step 1. Delete row 3 and column 2: $M_{32}=\left|\begin{array}{cc}2& 3\\ 4& 1\end{array}\right|=2-12=-10$.
     Step 2. $C_{32}=(-1{)}^5(-10)=10$.
   • Step 1. First matrix: $2(3\cdot 1-2\cdot 1)-0+1(1\cdot 1-3\cdot 1)=2-2=0$.
     Step 2. A determinant of 0 means the matrix is singular (no inverse). The others give 6, 6 and 2.
   • Step 1. Each of the 3 rows is multiplied by 2, so $\det (2A)=2^3\det A$.
     Step 2. $8\times 5=40$.

4. Drag and drop

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

   $x=\pm 3$  ·  $6$  ·  $\frac{27}{4}$  ·  $-24$
   1. $\det A=6$. $B$ is obtained by replacing row 2 of $A$ with (row 2) $-4\times$(row 1). Find $\det B$. →
   2. $A$ is $3\times 3$ with $\det A=4$. Find $\det (3A^{-1})$. →
   3. Solve $\left|\begin{array}{ccc}x& 2& 1\\ 0& x& 3\\ 0& 0& 2\end{array}\right|=18$. →
   Show answer

   1: $6$ · 2: $\frac{27}{4}$ · 3: $x=\pm 3$

   • Step 1. Adding a multiple of one row to another row does not change the determinant.
     Step 2. So $\det B=\det A=6$.
   • Step 1. $\det A^{-1}=\frac{1}{\det A}=\frac{1}{4}$.
     Step 2. $\det (3A^{-1})=3^3\times \frac{1}{4}=\frac{27}{4}$.
   • Step 1. The matrix is upper triangular: $\det =x\cdot x\cdot 2=2x^2$.
     Step 2. $2x^2=18\Rightarrow x^2=9\Rightarrow x=\pm 3$.

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