Matrix worksheet

1. Determinant of A: -17

1. Break the 3×3 problem into three small 2×2 ones. Multiply each top-row number by its minor (the 2×2 left when you cover its row and column), using the signs +, −, +

   $\mathrm{det\; A}$$=\text{a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃}$

2. Minor M₁₁: cover row 1 and column 1, which leaves [-1, 4; 2, 0], and find its 2×2 determinant

   $\mathrm{M₁₁}$$=\left(\mathrm{-1}\right)\times 0-4\times 2=\mathrm{-8}$

3. Minor M₁₂: cover row 1 and column 2, which leaves [0, 4; 1, 0], and find its 2×2 determinant

   $\mathrm{M₁₂}$$=0\times 0-4\times 1=\mathrm{-4}$

4. Minor M₁₃: cover row 1 and column 3, which leaves [0, -1; 1, 2], and find its 2×2 determinant

   $\mathrm{M₁₃}$$=0\times 2-\left(\mathrm{-1}\right)\times 1=1$

5. Combine the three parts with their signs to get the determinant

   $\mathrm{det\; A}$$=3\times \left(\mathrm{-8}\right)-1\times \left(\mathrm{-4}\right)+3\times 1$$=\mathrm{-17}$

2. Result A⁻¹: [2/5, -1/5, -1/5; -2/15, 1/15, 2/5; -1/30, 4/15, 1/10]

1. Break the 3×3 problem into three small 2×2 ones. Multiply each top-row number by its minor (the 2×2 left when you cover its row and column), using the signs +, −, +

   $\mathrm{det\; A}$$=\text{a₁₁M₁₁ − a₁₂M₁₂ + a₁₃M₁₃}$

2. Minor M₁₁: cover row 1 and column 1, which leaves [-1, 4; 3, 0], and find its 2×2 determinant

   $\mathrm{M₁₁}$$=\left(\mathrm{-1}\right)\times 0-4\times 3=\mathrm{-12}$

3. Minor M₁₂: cover row 1 and column 2, which leaves [0, 4; 1, 0], and find its 2×2 determinant

   $\mathrm{M₁₂}$$=0\times 0-4\times 1=\mathrm{-4}$

4. Minor M₁₃: cover row 1 and column 3, which leaves [0, -1; 1, 3], and find its 2×2 determinant

   $\mathrm{M₁₃}$$=0\times 3-\left(\mathrm{-1}\right)\times 1=1$

5. Combine the three parts with their signs to get the determinant

   $\mathrm{det\; A}$$=3\times \left(\mathrm{-12}\right)-1\times \left(\mathrm{-4}\right)+2\times 1$$=\mathrm{-30}$

6. Find every cofactor: the minor for each spot, with its sign flipped on a checkerboard pattern (+, −, + across the top row)

   $C$$=\text{[-12, 4, 1; 6, -2, -8; 6, -12, -3]}$

7. Flip the cofactor grid so its rows become columns. This flipped grid is called the adjugate

   $\mathrm{adj\; A}$$=\text{[-12, 6, 6; 4, -2, -12; 1, -8, -3]}$

8. Divide the adjugate by the determinant to get the inverse

   $\mathrm{A⁻¹}$$=\frac{1}{\mathrm{-30}}\mathrm{adj\; A}$

9. Here is the answer, one row at a time (semicolons separate the rows)

   $\text{A⁻¹}$$=\text{[2/5, -1/5, -1/5; -2/15, 1/15, 2/5; -1/30, 4/15, 1/10]}$

3. Result kA: [6, 2, 6; 0, -2, 8; 2, 4, 0]

1. Row 1: multiply every number by k = 2

   $2\times 3=6,2\times 1=2,2\times 3=6$

2. Row 2: multiply every number by k = 2

   $2\times 0=0,2\times \left(\mathrm{-1}\right)=\mathrm{-2},2\times 4=8$

3. Row 3: multiply every number by k = 2

   $2\times 1=2,2\times 2=4,2\times 0=0$

4. Here is the answer, one row at a time (semicolons separate the rows)

   $\text{kA}$$=\text{[6, 2, 6; 0, -2, 8; 2, 4, 0]}$

4. Determinant of A: -1

1. For a 2×2 grid, multiply the two numbers on the main diagonal (top-left to bottom-right), then subtract the product of the other two

   $\mathrm{det\; A}$$=\text{a₁₁a₂₂ − a₁₂a₂₁}$

2. Put in the numbers from A and work it out

   $\mathrm{det\; A}$$=1\times \left(\mathrm{-1}\right)-1\times 0=\mathrm{-1}$

5. Result kA: [3, 2, 2; 0, -1, 5; 1, 3, 0]

1. Row 1: multiply every number by k = 1

   $1\times 3=3,1\times 2=2,1\times 2=2$

2. Row 2: multiply every number by k = 1

   $1\times 0=0,1\times \left(\mathrm{-1}\right)=\mathrm{-1},1\times 5=5$

3. Row 3: multiply every number by k = 1

   $1\times 1=1,1\times 3=3,1\times 0=0$

4. Here is the answer, one row at a time (semicolons separate the rows)

   $\text{kA}$$=\text{[3, 2, 2; 0, -1, 5; 1, 3, 0]}$

6. Determinant of A: -2

1. For a 2×2 grid, multiply the two numbers on the main diagonal (top-left to bottom-right), then subtract the product of the other two

   $\mathrm{det\; A}$$=\text{a₁₁a₂₂ − a₁₂a₂₁}$

2. Put in the numbers from A and work it out

   $\mathrm{det\; A}$$=2\times \left(\mathrm{-1}\right)-1\times 0=\mathrm{-2}$

7. Result A × B: [5, 1; -6, -2]

1. Each answer number cᵢⱼ pairs row i of A with column j of B: multiply the matching numbers and add them up

   $\text{cᵢⱼ = aᵢ₁b₁ⱼ + aᵢ₂b₂ⱼ}$

2. Row 1 of the product

   $\mathrm{c₁₁}=2\times 1+1\times 3=5,\mathrm{c₁₂}=2\times 0+1\times 1=1$

3. Row 2 of the product

   $\mathrm{c₂₁}=0\times 1+\left(\mathrm{-2}\right)\times 3=\mathrm{-6},\mathrm{c₂₂}=0\times 0+\left(\mathrm{-2}\right)\times 1=\mathrm{-2}$

4. Here is the answer, one row at a time (semicolons separate the rows)

   $\text{A × B}$$=\text{[5, 1; -6, -2]}$

8. Result A + B: [4, 1; 3, 0]

1. Row 1: add the numbers in the same spots in A and B

   $2+2=4,1+0=1$

2. Row 2: add the numbers in the same spots in A and B

   $0+3=3,\mathrm{-1}+1=0$

3. Here is the answer, one row at a time (semicolons separate the rows)

   $\text{A + B}$$=\text{[4, 1; 3, 0]}$

9. Result kA: [4, 2; 0, -2]

1. Row 1: multiply every number by k = 2

   $2\times 2=4,2\times 1=2$

2. Row 2: multiply every number by k = 2

   $2\times 0=0,2\times \left(\mathrm{-1}\right)=\mathrm{-2}$

3. Here is the answer, one row at a time (semicolons separate the rows)

   $\text{kA}$$=\text{[4, 2; 0, -2]}$

10. Result A + B: [3, 1, 5; 3, 0, 5; 1, 4, 1]

1. Row 1: add the numbers in the same spots in A and B

   $2+1=3,1+0=1,2+3=5$

2. Row 2: add the numbers in the same spots in A and B

   $0+3=3,\mathrm{-1}+1=0,5+0=5$

3. Row 3: add the numbers in the same spots in A and B

   $1+0=1,2+2=4,0+1=1$

4. Here is the answer, one row at a time (semicolons separate the rows)

   $\text{A + B}$$=\text{[3, 1, 5; 3, 0, 5; 1, 4, 1]}$