Matrix Multiplication

Multiply every entry of \mathbf{A} by -2, taking care with the signs.

1. Each entry is multiplied by -2
2. -2\times4=-8 and -2\times(-1)=2
3. -2\times(-3)=6 and -2\times0=0
4. Answer: \begin{pmatrix}-8&2\\6&0\end{pmatrix}

Each entry of \mathbf{A}\mathbf{B} is a row of \mathbf{A} times a column of \mathbf{B}, with the products added.

1. Top left: row 1 of \mathbf{A} \times column 1 of \mathbf{B}: 3(1)+(-2)(-3)=3+6=9
2. Top right: row 1 \times column 2: 3(6)+(-2)(2)=18-4=14
3. Bottom left: row 2 \times column 1: 5(1)+4(-3)=5-12=-7
4. Bottom right: row 2 \times column 2: 5(6)+4(2)=30+8=38
5. Order matters: \mathbf{B}\mathbf{A} is a different matrix.
6. Answer: \begin{pmatrix}9&14\\-7&38\end{pmatrix}

Remember what the identity matrix looks like, and that the order of a matrix product matters.

1. \mathbf{I}=\begin{pmatrix}1&0\\0&1\end{pmatrix}: 1s on the leading diagonal, 0s elsewhere
2. Multiplying any matrix by \mathbf{I} leaves it unchanged, so \mathbf{A}\mathbf{I}=\mathbf{A}
3. \mathbf{A}\mathbf{B} and \mathbf{B}\mathbf{A} are usually different
4. \mathbf{A}^2 means \mathbf{A}\mathbf{A}, worked out row by column
5. Answer: \mathbf{A}\mathbf{I}=\mathbf{A}

Each entry of the answer is a row of the 2\times2 matrix times the column, with the two products added.

x=

1. Top row of the matrix times the column: 6(4)+(-1)(5)
2. =24-5
3. Answer: x=19

y=

1. Bottom row of the matrix times the column: -2(4)+3(5)
2. =-8+15
3. Answer: y=7

\mathbf{A}^2 means \mathbf{A}\times\mathbf{A}: multiply rows of the first \mathbf{A} by columns of the second.

1. \mathbf{A}^2=\begin{pmatrix}2&-3\\1&4\end{pmatrix}\begin{pmatrix}2&-3\\1&4\end{pmatrix}
2. Top left: 2(2)+(-3)(1)=1
3. Top right: 2(-3)+(-3)(4)=-18
4. Bottom left: 1(2)+4(1)=6
5. Bottom right: 1(-3)+4(4)=13
6. (Squaring each entry, or doubling \mathbf{A}, is not the same.)
7. Answer: \begin{pmatrix}1&-18\\6&13\end{pmatrix}

Multiply the two matrices first, then compare one entry of your answer with the matrix on the right.

1. Multiply the matrices, row by column:
2. \begin{pmatrix}2&1\\-1&3\end{pmatrix}\begin{pmatrix}1&-2\\4&0\end{pmatrix}=\begin{pmatrix}6&-4\\11&2\end{pmatrix}
3. Top left: 6k=-18, so k=-3
4. Check another entry: -3\times(-4)=12 ✓
5. Answer: k=-3

Multiply out the left-hand side to get one equation from each row; the top row has only one unknown. Answer part (a) before attempting part (b).

Part (a)

1. Top row times the column: 4(5)+(-3)q=26
2. 20-3q=26, so -3q=6
3. Answer: q=-2

Part (b)

1. Bottom row times the column: 5p+2q=11
2. With q=-2: 5p-4=11, so 5p=15
3. Answer: p=3

Write down \mathbf{I}, then use a row of \mathbf{A} times the column of \mathbf{B} that contains k.

1. \mathbf{I}=\begin{pmatrix}1&0\\0&1\end{pmatrix} (1s on the leading diagonal, 0s elsewhere).
2. Top right of \mathbf{A}\mathbf{B}: row 1 \times column 2: 2(-1)+1(k)=k-2
3. This must equal the top right of \mathbf{I}, which is 0: k-2=0
4. Check with the bottom right: 5(-1)+3k=-5+6=1 ✓
5. Answer: k=2

k is in the bottom row, so multiply the bottom row of the first matrix by a column of the second.

1. Bottom row \times column 1: 3(4)+k(5)=12+5k
2. This is the bottom-left entry: 12+5k=-8
3. 5k=-20
4. Check bottom right: 3(1)+(-4)(-2)=3+8=11 ✓
5. Answer: k=-4

Write 5\mathbf{I} as a matrix, then multiply \mathbf{A} by each option until you find the one that gives it.

1. 5\mathbf{I}=\begin{pmatrix}5&0\\0&5\end{pmatrix}
2. Try \begin{pmatrix}3&-1\\-4&3\end{pmatrix}
3. Top row: 3(3)+1(-4)=5 and 3(-1)+1(3)=0
4. Bottom row: 4(3)+3(-4)=0 and 4(-1)+3(3)=5
5. So \mathbf{A}\mathbf{B}=\begin{pmatrix}5&0\\0&5\end{pmatrix}=5\mathbf{I}
6. (The second option gives -5\mathbf{I}.)
7. Answer: \begin{pmatrix}3&-1\\-4&3\end{pmatrix}

Let P be (x,\,y): the matrix times the column \begin{pmatrix}x\\y\end{pmatrix} equals the image as a column.

1. \begin{pmatrix}4&3\\3&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}5\\3\end{pmatrix}
2. So 4x+3y=5 and 3x+2y=3
3. Multiply the first by 2: 8x+6y=10
4. Multiply the second by 3: 9x+6y=9
5. Subtract: x=-1
6. 4(-1)+3y=5, so 3y=9 and y=3
7. Answer: P=(-1,\,3)

Multiply out the left-hand side in terms of a and b, then choose entries that contain only one unknown.

Part (a)

1. Top left: row 1 \times column 1: 2a+3(1)=2a+3
2. 2a+3=-5, so 2a=-8
3. Answer: a=-4

Part (b)

1. Bottom right: row 2 \times column 2: -1(b)+2(4)=8-b
2. 8-b=3, so b=5
3. Check top right: ab+3(4)=-20+12=-8 ✓
4. Answer: b=5

Work out both \mathbf{A}\mathbf{B} and \mathbf{B}\mathbf{A} in terms of k, then make one pair of matching entries equal.

1. \mathbf{A}\mathbf{B}=\begin{pmatrix}2+6k&6-4k\\-2&16\end{pmatrix}
2. \mathbf{B}\mathbf{A}=\begin{pmatrix}14&2k-6\\-2&6k+4\end{pmatrix}
3. Top left: 2+6k=14, so 6k=12 and k=2
4. Check top right: 6-4(2)=-2 and 2(2)-6=-2 ✓
5. Answer: k=2

The top row gives two possible values of a; use the bottom row to decide which one is right.

1. Top row: a(a)+4(-2)=a^2-8
2. a^2-8=17, so a^2=25 and a=5 or a=-5
3. Bottom row: 1(a)+a(-2)=-a
4. -a=5, so a=-5
5. Check the top row: (-5)^2-8=17 ✓
6. Answer: a=-5

The right-hand side has two equal entries, so the two entries on the left-hand side must be equal too.

1. Top row: 4x+3x=7x
2. Bottom row: -2(4)+x(x)=x^2-8
3. Both equal k, so x^2-8=7x
4. x^2-7x-8=0
5. (x-8)(x+1)=0, so x=8 or x=-1
6. x is positive
7. Answer: x=8

The matrix for the transformation done first goes on the right of the product. Answer part (a) before attempting part (b).

Part (a)

1. \mathbf{A} acts first, so it goes on the right: the matrix is \mathbf{B}\mathbf{A}
2. \mathbf{B}\mathbf{A}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1&2\\1&3\end{pmatrix}
3. Top row: 0(1)+(-1)(1)=-1 and 0(2)+(-1)(3)=-3
4. Bottom row: 1(1)+0(1)=1 and 1(2)+0(3)=2
5. (\mathbf{A}\mathbf{B} would be \begin{pmatrix}2&-1\\3&-1\end{pmatrix}, which means \mathbf{B} first, then \mathbf{A}.)
6. Answer: \begin{pmatrix}-1&-3\\1&2\end{pmatrix}

Part (b)

1. Let P=(x,\,y): \begin{pmatrix}-1&-3\\1&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-1\\2\end{pmatrix}
2. So -x-3y=-1 and x+2y=2
3. Add the equations: -y=1, so y=-1
4. x+2(-1)=2, so x=4
5. Check: \mathbf{A} maps (4,\,-1) to (2,\,1), and \mathbf{B} maps (2,\,1) to (-1,\,2) ✓
6. Answer: P=(4,\,-1)

Multiply out \mathbf{A}\mathbf{B} and compare it with \begin{pmatrix}k&0\\0&k\end{pmatrix}: start with the entry that contains only one unknown.

Part (a)

1. \mathbf{A}\mathbf{B}=\begin{pmatrix}4p+2&4q-r\\6p-6&6q+3r\end{pmatrix}
2. k\mathbf{I}=\begin{pmatrix}k&0\\0&k\end{pmatrix}
3. Bottom left: 6p-6=0, so p=1
4. Top left: k=4p+2=4+2
5. Answer: k=6

Part (b)

1. Top right: 4q-r=0, so r=4q
2. Bottom right: 6q+3r=k=6
3. Substitute r=4q: 6q+12q=6, so 18q=6
4. Answer: q=\dfrac{1}{3}

Part (c)

1. r=4q=4\times\dfrac{1}{3}
2. Check: \mathbf{A}\begin{pmatrix}1&\frac{1}{3}\\-2&\frac{4}{3}\end{pmatrix}=\begin{pmatrix}6&0\\0&6\end{pmatrix} ✓
3. Answer: r=\dfrac{4}{3}

\mathbf{A}^3 is \mathbf{A}^2 multiplied by \mathbf{A}. Answer part (a) before attempting part (b).

Part (a)

1. \mathbf{A}^2=\begin{pmatrix}a&1\\0&a\end{pmatrix}\begin{pmatrix}a&1\\0&a\end{pmatrix}
2. Top left: a(a)+1(0)=a^2
3. Top right: a(1)+1(a)=2a
4. Bottom row: 0(a)+a(0)=0 and 0(1)+a(a)=a^2
5. Answer: \begin{pmatrix}a^2&2a\\0&a^2\end{pmatrix}

Part (b)

1. \mathbf{A}^3=\mathbf{A}^2\mathbf{A}=\begin{pmatrix}a^2&2a\\0&a^2\end{pmatrix}\begin{pmatrix}a&1\\0&a\end{pmatrix}
2. =\begin{pmatrix}a^3&3a^2\\0&a^3\end{pmatrix}
3. Top left: a^3=-8, so a=\sqrt[3]{-8}=-2 (a cube root, not a square root)
4. Check top right: 3(-2)^2=12 ✓
5. Answer: a=-2

Multiply out to get one equation from each row, then substitute the linear equation into the other one. Answer part (a) before attempting part (b).

Part (a)

1. Top row: a(a)+b(b)=a^2+b^2, so a^2+b^2=25
2. Bottom row: 2a+b=10, so b=10-2a
3. Substitute: a^2+(10-2a)^2=25
4. a^2+100-40a+4a^2=25
5. 5a^2-40a+75=0, so a^2-8a+15=0
6. (a-3)(a-5)=0, so a=3 or a=5
7. a=5 gives b=0, which is not positive
8. Answer: a=3

Part (b)

1. b=10-2(3)=4
2. Check: 3^2+4^2=25 ✓
3. Answer: b=4

Multiply \mathbf{A} by itself row-by-column, then compare entries: the diagonal entries give two possibilities each, and an off-diagonal entry tells you which pair works.

Part (a)

1. \mathbf{A}^2=\mathbf{A}\mathbf{A}=\begin{pmatrix}a&2\\1&b\end{pmatrix}\begin{pmatrix}a&2\\1&b\end{pmatrix}=\begin{pmatrix}a^2+2&2a+2b\\a+b&2+b^2\end{pmatrix}
2. (Not \begin{pmatrix}a^2&4\\1&b^2\end{pmatrix}: you can't square each entry.)
3. Top left: a^2+2=11, so a^2=9 and a=3 or a=-3
4. Bottom right: 2+b^2=6, so b^2=4 and b=2 or b=-2
5. Bottom left: a+b=1 (top right 2a+2b=2 says the same).
6. The only pair that adds to 1 is a=3, b=-2
7. Answer: a=3

Part (b)

1. From part (a), a+b=1 with a=3, so b=-2 (and b^2=4 ✓).
2. Answer: b=-2

Part (c)

1. \mathbf{A}=\begin{pmatrix}3&2\\1&-2\end{pmatrix}
2. \mathbf{A}^2-\mathbf{A}=\begin{pmatrix}11-3&2-2\\1-1&6-(-2)\end{pmatrix}=\begin{pmatrix}8&0\\0&8\end{pmatrix}
3. \begin{pmatrix}8&0\\0&8\end{pmatrix}=8\begin{pmatrix}1&0\\0&1\end{pmatrix}=8\mathbf{I}
4. Answer: k=8