10.5 — Inverse Matrices & Systems of Equations

The Matrix
Equation Quiz

Ian Kim · 20 Questions · HSA.REI.C.8 / C.9 / CED.A.3
Core Formula
AX = B → X = A⁻¹B
2×2 Inverse
A⁻¹ = 1/det · [d −b; −c a]
Determinant
det = ad − bc
Identity
A · A⁻¹ = I
No Solution
det = 0 → no inverse
01
of 20
Concept
The Matrix Equation · Vocabulary
In the matrix equation AX = B, which matrix is called the coefficient matrix?

💡 MEMORY KEY: "A = Arranges coefficients"
02
of 20
Hard
Determinant · 2×2 Matrix
Calculate the determinant of the coefficient matrix from the example:

\[ A = \begin{bmatrix} 2 & 1 \\ -1 & -3 \end{bmatrix} \]
💡 MEMORY KEY: "det = ad − bc" (diagonal down MINUS diagonal up)
03
of 20
Hard
Inverse Matrix · 2×2 Formula
For a 2×2 matrix \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), what is the correct inverse formula?

💡 MEMORY KEY: "SWAP diagonals, NEGATE off-diagonals, DIVIDE by det"
04
of 20
Hard
Solving System · 2-Variable
Solve the system using the matrix equation: \[\begin{cases} 2x + y = 6 \\ -x - 3y = 2 \end{cases}\] The answer \((x, y)\) is:

💡 MEMORY KEY: "Isolate X: multiply BOTH sides by A⁻¹ on the LEFT"
05
of 20
Concept
Identity Matrix · Properties
Which of the following correctly describes what happens when you multiply a matrix by its inverse?

💡 MEMORY KEY: "A · A⁻¹ = I (Identity) — like multiplying by 1"
06
of 20
Hard
Matrix Setup · Writing System as Matrix
Write the system \(\begin{cases} 3x - 2y = 7 \\ x + 4y = -1 \end{cases}\) as a matrix equation \(AX = B\).
What is the coefficient matrix \(A\)?

💡 MEMORY KEY: "Coefficients go LEFT in A, constants go RIGHT in B"
07
of 20
Hard
Determinant · When Inverse Doesn't Exist
For which matrix does the inverse NOT exist?

💡 MEMORY KEY: "det = 0 → NO inverse → singular matrix"
08
of 20
Hard
Inverse Calculation · 2×2
Find \(A^{-1}\) for \(A = \begin{bmatrix} 2 & 1 \\ -1 & -3 \end{bmatrix}\)

(det = −5)

💡 MEMORY KEY: "SWAP a↔d, FLIP signs of b and c, DIVIDE by det"
09
of 20
Hard
Order Matters · Matrix Multiplication
When solving \(AX = B\) by multiplying both sides by \(A^{-1}\), which is the correct step?

💡 MEMORY KEY: "Matrix multiplication is NOT commutative: AB ≠ BA. Always LEFT-multiply."
10
of 20
Hard
3-Variable System · Matrix Setup
Write the system as a matrix equation and identify the constant matrix \(B\): \[\begin{cases} -4x - 5y + 3z = -9 \\ x + y - z = 2 \\ 4x + 3y - 6z = 14 \end{cases}\]
11
of 20
Hard
3-Variable System · Solution
Using the matrix equation, the 3-variable system from the notes gives the solution: \[\begin{cases} -4x - 5y + 3z = -9 \\ x + y - z = 2 \\ 4x + 3y - 6z = 14 \end{cases}\] What is \((x, y, z)\)?
12
of 20
Hard
Determinant · New Problem
Calculate \(\det(A)\) for \(A = \begin{bmatrix} 5 & -2 \\ 3 & 4 \end{bmatrix}\)

💡 MEMORY KEY: "det = (top-left)(bottom-right) − (top-right)(bottom-left)"
13
of 20
Hard
Solve System · New 2-Variable
Solve using matrix equations: \[\begin{cases} 2x + 3y = 12 \\ x - y = 1 \end{cases}\] 💡 det(A) = 2(−1) − 3(1) = −5

💡 MEMORY KEY: "Find A⁻¹ first, then X = A⁻¹B"
14
of 20
Concept
Identity Matrix · Recognition
When \(A^{-1} \cdot A \cdot X = A^{-1} \cdot B\) is simplified, the left side becomes \(IX\). What does \(IX\) equal?

💡 MEMORY KEY: "I · X = X (identity times anything = itself)"
15
of 20
Hard
Tricky · Sign Errors
Which is the correct inverse of \(A = \begin{bmatrix} 4 & 3 \\ 1 & 1 \end{bmatrix}\)?
(det = 4·1 − 3·1 = 1)

💡 Watch out: students often forget to NEGATE b and c!
16
of 20
Hard
Solve System · Negative Coefficients
Solve using matrix equations: \[\begin{cases} -2x + y = 5 \\ 4x - 3y = -9 \end{cases}\] 💡 det = (−2)(−3) − (1)(4) = 6 − 4 = 2
17
of 20
Hard
Tricky · Reading Matrix Form
The solution to a system is given as: \[\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -3 \\ 7 \end{bmatrix}\] What is the coordinate pair?

💡 MEMORY KEY: "Top row = x, Bottom row = y"
18
of 20
Hard
Verify Solution · Plug Back In
After solving \(\begin{cases} 2x + y = 6 \\ -x - 3y = 2 \end{cases}\), you get \((4, -2)\).

Does this check out in the second equation \(-x - 3y = 2\)?

💡 MEMORY KEY: "Always CHECK by substituting back into BOTH equations"
19
of 20
Hard
Solve System · Challenge
Solve using matrix equations: \[\begin{cases} 3x + y = 11 \\ x + 2y = 7 \end{cases}\] 💡 det = 3(2) − (1)(1) = 6 − 1 = 5
20
of 20
Hard
Final Boss · Full Process
For the system \(\begin{cases} x + 2y = 4 \\ 3x + 7y = 13 \end{cases}\), what is the complete solution process result?
det = 1·7 − 2·3 = 7 − 6 = 1

💡 MEMORY KEY: "When det = 1, A⁻¹ = adjugate matrix (no division needed!)"
Final Score
out of 20