Finding unknowns using matrix multiplication

Solve for a: \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} a \\ 5 \end{bmatrix} = \begin{bmatrix} 11 \\ 23 \end{bmatrix}

• Expand the left-hand side:
  • \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} a \\ 5 \end{bmatrix}
  • = \begin{bmatrix} 1\times a + 2\times 5 \\ 3\times a + 4\times 5 \end{bmatrix}
  • = \begin{bmatrix} a + 10 \\ 3a + 20 \end{bmatrix}
• Set the resulting matrix equal to the right-hand side:
  • \begin{bmatrix} a + 10 \\ 3a + 20 \end{bmatrix} = \begin{bmatrix} 11 \\ 23 \end{bmatrix}
• Now, we can literally just equate the top elements of the left and right-hand matrices, and the bottom elements of the matrices, separately:
  • a + 10 = 11
  • 3a + 20 = 23
• Solving the first equation for a:
  • a = 11 - 10
  • a = 1
• Solving the second equation for a:
  • 3a = 23 - 20
  • 3a = 3
  • a = 1
• Both equations give us the same value for a, so we can be pretty sure our answer is correct.
• Answer: a = 1

Solve for x and y: \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 16 \\ 36 \end{bmatrix}

• Expand the left-hand side:
  • \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}
  • = \begin{bmatrix} 2\times x + 3\times y \\ 4\times x + 5\times y \end{bmatrix}
  • = \begin{bmatrix} 2x + 3y \\ 4x + 5y \end{bmatrix}
• Set the resulting matrix equal to the right-hand side:
  • \begin{bmatrix} 2x + 3y \\ 4x + 5y \end{bmatrix} = \begin{bmatrix} 16 \\ 36 \end{bmatrix}
• Now, we can equate the top elements of the left and right-hand matrices, and the bottom elements of the matrices, separately:
  • 2x + 3y = 16
  • 4x + 5y = 36
• Ah, they’re simultaneous equations! We can solve them using whatever method you like. Here’s substitution, as we know it always works no matter the numbers:
  • Rearrange 2x + 3y = 16:
    • 2x = 16 - 3y
    • x = \frac{16 - 3y}{2}
  • Substitute that value for x into the second equation:
    • 4\left(\frac{16 - 3y}{2}\right) + 5y = 36
    • 2(16 - 3y) + 5y = 36
    • 32 - 6y + 5y = 36
    • 32 - y = 36
    • -y = 36 - 32
    • -y = 4
    • y = -4
  • Substitute that value for y back into the rearranged first equation:
    • x = \frac{16 - 3(-4)}{2}
    • x = \frac{16 + 12}{2}
    • x = \frac{28}{2}
    • x = 14
• Answer: x = 14 and y = -4

Solve for k and c: \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2k & 3k \\ 4c & 5c \end{bmatrix} = \begin{bmatrix} 40 & 54 \\ 96 & 132 \end{bmatrix}

• Expand the left-hand side:
  • \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2k & 3k \\ 4c & 5c \end{bmatrix}
  • = \begin{bmatrix} 1\times 2k + 2\times 4c & 1\times 3k + 2\times 5c \\ 3\times 2k + 4\times 4c & 3\times 3k + 4\times 5c \end{bmatrix}
  • = \begin{bmatrix} 2k + 8c & 3k + 10c \\ 6k + 16c & 9k + 20c \end{bmatrix}
• Set the resulting matrix equal to the right-hand side:
  • \begin{bmatrix} 2k + 8c & 3k + 10c \\ 6k + 16c & 9k + 20c \end{bmatrix} = \begin{bmatrix} 40 & 54 \\ 96 & 132 \end{bmatrix}
• Now, we can equate the top-left elements of the left and right-hand matrices, the top-right elements of the matrices, the bottom-left elements of the matrices, and the bottom-right elements of the matrices, separately:
  • 2k + 8c = 40
  • 3k + 10c = 54
  • 6k + 16c = 96
  • 9k + 20c = 132
• We can rearrange the first two equations to express k in terms of c:
  • 2k = 40 - 8c
  • k = \frac{40 - 8c}{2}
  • 3k = 54 - 10c
  • k = \frac{54 - 10c}{3}
• Set those two expressions for k equal to each other:
  • \frac{40 - 8c}{2} = \frac{54 - 10c}{3}
  • 3(40 - 8c) = 2(54 - 10c)
  • 120 - 24c = 108 - 20c
  • 120 - 108 = -20c + 24c
  • 12 = 4c
  • c = \frac{12}{4}
  • c = 3
• Substitute that value for c back into one of the rearranged equations to find k:
  • k = \frac{40 - 8(3)}{2}
  • k = \frac{40 - 24}{2}
  • k = \frac{16}{2}
  • k = 8
• Answer: k = 8 and c = 3

Solve for a, b and c: \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 14 \\ 21 \\ 76 \end{bmatrix}

• Expand the left-hand side:
  • \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} a \\ b \\ c \end{bmatrix}
  • = \begin{bmatrix} 1\times a + 0\times b + 2\times c \\ 0\times a + 1\times b + 3\times c \\ 4\times a + 5\times b + 6\times c \end{bmatrix}
  • = \begin{bmatrix} a + 2c \\ b + 3c \\ 4a + 5b + 6c \end{bmatrix}
• Set the resulting matrix equal to the right-hand side:
  • \begin{bmatrix} a + 2c \\ b + 3c \\ 4a + 5b + 6c \end{bmatrix} = \begin{bmatrix} 14 \\ 21 \\ 76 \end{bmatrix}
• Now, we can equate the top elements of the left and right-hand matrices, the middle elements of the matrices, and the bottom elements of the matrices, separately:
  • a + 2c = 14
  • b + 3c = 21
  • 4a + 5b + 6c = 76
• We can rearrange the first two equations to express a and b in terms of c:
  • a = 14 - 2c
  • b = 21 - 3c
• Substitute those values for a and b into the third equation:
  • 4(14 - 2c) + 5(21 - 3c) + 6c = 76
  • 56 - 8c + 105 - 15c + 6c = 76
  • 161 - 17c = 76
  • -17c = 76 - 161
  • -17c = -85
  • c = \frac{-85}{-17}
  • c = 5
• Substitute that value for c back into the rearranged first two equations:
  • a = 14 - 2(5) = 14 - 10 = 4
  • b = 21 - 3(5) = 21 - 15 = 6
• Answer: a = 4, b = 6 and c = 5

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