Precalculus Honors | Standards PC.NR.1 & PC.PAFR.7
We will explore the representation and manipulation of data using matrices and using them in solving complex systems.
Identity Matrix (\(I\)): A square matrix with 1s on the main diagonal and 0s elsewhere. Multiplying any matrix \(A\) by \(I\) results in \(A\).
Zero Matrix (\(0\)): A matrix of any dimension where every element is 0. Adding it to \(A\) results in \(A\).
Commutativity: Unlike real numbers, matrix multiplication is not commutative. In general, \(AB \neq BA\).
viewof a11 = Inputs.number({value: 1, label: "A[1,1]"})
viewof a12 = Inputs.number({value: 2, label: "A[1,2]"})
viewof a21 = Inputs.number({value: 3, label: "A[2,1]"})
viewof a22 = Inputs.number({value: 4, label: "A[2,2]"})
// Matrix B Inputs
viewof b11 = Inputs.number({value: 5, label: "B[1,1]"})
viewof b12 = Inputs.number({value: 6, label: "B[1,2]"})
viewof b21 = Inputs.number({value: 7, label: "B[2,1]"})
viewof b22 = Inputs.number({value: 8, label: "B[2,2]"})
// Show/Hide Toggle
viewof showProducts = Inputs.toggle({label: "Show Product Matrices", value: false})ab11 = a11*b11 + a12*b21; ab12 = a11*b12 + a12*b22;
ab21 = a21*b11 + a22*b21; ab22 = a21*b12 + a22*b22;
// Calculations for BA
ba11 = b11*a11 + b12*a21; ba12 = b11*a12 + b12*a22;
ba21 = b21*a11 + b22*a21; ba22 = b21*a12 + b22*a22;
// Comparison Logic
isCommutative = ab11===ba11 && ab12===ba12 && ab21===ba21 && ab22===ba22;
html`
<div style="font-size: 22px; display: grid; grid-template-columns: 1fr 1fr; gap: 20px; margin-bottom: 20px;">
<div style="padding: 10px; border: 1px solid #ccc; border-radius: 5px;">
<b style="color:#332288">Matrix A:</b><br>
[${a11}, ${a12}]<br>[${a21}, ${a22}]
</div>
<div style="padding: 10px; border: 1px solid #ccc; border-radius: 5px;">
<b style="color:#AA4499">Matrix B:</b><br>
[${b11}, ${b12}]<br>[${b21}, ${b22}]
</div>
</div>
${showProducts ? html`
<div style="font-size: 22px; display: grid; grid-template-columns: 1fr 1fr; gap: 20px; animation: fadeIn 0.5s;">
<div style="border: 2px solid #332288; padding: 10px; border-radius: 5px;">
<b style="color:#332288">Product AB:</b><br>
[${ab11}, ${ab12}]<br>[${ab21}, ${ab22}]
</div>
<div style="border: 2px solid #AA4499; padding: 10px; border-radius: 5px;">
<b style="color:#AA4499">Product BA:</b><br>
[${ba11}, ${ba12}]<br>[${ba21}, ${ba22}]
</div>
</div>
<p style="margin-top:20px; font-weight:bold; color: ${isCommutative ? '#117733' : '#CC3311'}">
RESULT: ${isCommutative ? 'COMMUTATIVE (AB = BA)' : 'NOT COMMUTATIVE (AB ≠ BA)'}
</p>` : html`<p style="color: #666; font-style: italic;">Toggle "Show Products" to compare AB and BA.</p>`}
<style>
@keyframes fadeIn { from { opacity: 0; } to { opacity: 1; } }
</style>
`1. Find \(A + B\) if \(A = \begin{bmatrix} 2 & -1 \\ 4 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} 5 & 3 \\ -2 & 1 \end{bmatrix}\).
\(\begin{bmatrix} 2+5 & -1+3 \\ 4-2 & 0+1 \end{bmatrix}\) \(\rightarrow\) \(\begin{bmatrix} 7 & 2 \\ 2 & 1 \end{bmatrix}\)
2. Multiply \(C = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) by scalar \(k = -3\).
\(\begin{bmatrix} -3(1) & -3(2) \\ -3(3) & -3(4) \end{bmatrix}\) \(\rightarrow\) \(\begin{bmatrix} -3 & -6 \\ -9 & -12 \end{bmatrix}\)
3. Find the product \(AB\) if \(A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\) and \(B = \begin{bmatrix} 3 & 1 \\ 4 & 2 \end{bmatrix}\).
\(\begin{bmatrix} (1)(3)+(0)(4) & (1)(1)+(0)(2) \\ (2)(3)+(1)(4) & (2)(1)+(1)(2) \end{bmatrix}\) \(\rightarrow\) \(\begin{bmatrix} 3 & 1 \\ 10 & 4 \end{bmatrix}\)
4. True or False: If \(A\) is \(2 \times 3\) and \(B\) is \(3 \times 2\), then \(AB\) and \(BA\) are both defined.
True. \(AB\) is \(2 \times 2\) and \(BA\) is \(3 \times 3\). However, they cannot be equal because their dimensions differ!
5. Identify the \(3 \times 3\) Identity Matrix \(I_3\).
\(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
For a square matrix \(A\), the determinant (\(\det A\) or \(|A|\)) is a scalar value that determines if an inverse exists.
Invertibility: A square matrix \(A\) has a multiplicative inverse \(A^{-1}\) if and only if \(\det A \neq 0\).
2x2 Formula: If \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), then \(\det A = ad - bc\).
Multiplicative Inverse: \(A^{-1} = \frac{1}{\det A} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\).
1. Find the determinant of \(A = \begin{bmatrix} 3 & 2 \\ 5 & 4 \end{bmatrix}\).
\((3)(4) - (2)(5) = 12 - 10\) \(\rightarrow\) 2
2. Does the matrix \(B = \begin{bmatrix} 6 & 3 \\ 4 & 2 \end{bmatrix}\) have an inverse?
\(\det B = (6)(2) - (3)(4) = 12 - 12 = 0\) \(\rightarrow\) No inverse exists (singular matrix).
3. Find the additive inverse of \(C = \begin{bmatrix} -1 & 5 \\ 0 & -2 \end{bmatrix}\).
\(-C = \begin{bmatrix} 1 & -5 \\ 0 & 2 \end{bmatrix}\)
4. Calculate \(A^{-1}\) for \(A = \begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\).
\(\det A = 8 - 7 = 1\). \(A^{-1} = \frac{1}{1}\begin{bmatrix} 4 & -1 \\ -7 & 2 \end{bmatrix}\)
5. If \(\det(M) = 5\), what is \(\det(M^{-1})\)?
\(\det(M^{-1}) = \frac{1}{\det M}\) \(\rightarrow\) \(\frac{1}{5}\)
A system consisting of a line and a parabola can have 0, 1, or 2 real solutions.
Algebraic Path: Set the equations equal (\(f(x) = g(x)\)) and solve the resulting quadratic equation.
Graphical Path: Identify the points of intersection on the coordinate plane.
Any system of linear equations can be written as a single matrix equation:
\(\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases} \implies \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}\)
To solve: \(X = A^{-1}B\)
1. Solve the system: \(y = x^2\) and \(y = x + 2\).
\(x^2 = x + 2 \rightarrow x^2 - x - 2 = 0 \rightarrow (x-2)(x+1) = 0\) Solutions: \((2, 4)\) and \((-1, 1)\)
2. How many solutions does \(y = x^2 + 5\) and \(y = 2\) have?
\(x^2 + 5 = 2 \rightarrow x^2 = -3\) \(\rightarrow\) 0 real solutions (no intersection).
3. Represent as \(AX=B\): \(\begin{cases} 3x - 2y = 10 \\ 5x + y = 7 \end{cases}\)
\(\begin{bmatrix} 3 & -2 \\ 5 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 7 \end{bmatrix}\)
4. Use a graph to solve \(\sqrt{x+3} = x + 1\).
Points of intersection for \(f(x)=\sqrt{x+3}\) and \(g(x)=x+1\). Intersection at \(x = 1\).
5. Solve \(\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 6 \\ 2 \end{bmatrix}\) using \(X = A^{-1}B\).
\(\det A = -2, A^{-1} = -\frac{1}{2}\begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 0.5 & 0.5 \\ 0.5 & -0.5 \end{bmatrix}\) \(X = \begin{bmatrix} 0.5(6)+0.5(2) \\ 0.5(6)-0.5(2) \end{bmatrix} = \begin{bmatrix} 4 \\ 2 \end{bmatrix}\)
| Feature | Key Concept | Constraint |
|---|---|---|
| Multiplication | \(AB \neq BA\) | Inner dimensions must match |
| Inverse | \(A \cdot A^{-1} = I\) | \(\det A \neq 0\) |
| Systems | \(AX = B\) | \(X = A^{-1}B\) |
| Lin-Quad | Intersection points | Max 2 solutions |
\(ad - bc = \det A\) \(\quad\) \(A^{-1} = \frac{1}{\det A} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\) \(\quad\) \(f(x) = g(x)\)