If \(A = \begin{bmatrix}a&b\\c&d\end{bmatrix}\)→\(A^{-1} = \dfrac{1}{|A|}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\)
\(|A| = ad - bc\)•\(AA^{-1} = A^{-1}A = I\)•If \(|A|=0\) → No Inverse!
🔑 SWAP main diagonal➖ NEGATE off diagonal➗ DIVIDE by det🚫 det=0 → NO inverse🔲 Square ONLY✅ AA⁻¹ = I always
🌱 Level 1 — Foundations (Q1–5)
Q 01Easy ⭐
A matrix can have an inverse only if it is a _____ matrix.
💡 Memory: Square → Same rows & columns
📖 Explanation
Only a square matrix (n×n) can have an inverse. A rectangular matrix has no inverse because \(AA^{-1} = I\) requires the identity to be the same size as \(A\).
Q 02Easy ⭐
What is \(AA^{-1}\) always equal to?
💡 IDENTITY rule: like multiplying a number by its reciprocal → 1
📖 Explanation
By definition, \(AA^{-1} = A^{-1}A = I\), the identity matrix. This is exactly like \(5 \times \frac{1}{5} = 1\) in regular numbers.
Q 03Easy ⭐
Find the determinant of \(A = \begin{bmatrix}3 & 1\\ 2 & 4\end{bmatrix}\)
💡 Formula: \(\det = ad - bc\). "Butterfly" ✖ the 2×2!
📖 Explanation
\(\det(A) = (3)(4) - (1)(2) = 12 - 2 = \mathbf{10}\)
Since \(\det \neq 0\), the inverse exists ✓
Q 04Easy ⭐
Which matrix is the 2×2 identity matrix \(I\)?
💡 Identity = 1s on diagonal, 0s everywhere else
📖 Explanation
\(I = \begin{bmatrix}1&0\\0&1\end{bmatrix}\). The main diagonal is all 1s, all other entries are 0. Multiplying any matrix by \(I\) gives back the same matrix: \(AI = A\).
Q 05Easy ⭐
If \(\det(A) = 0\), which statement is TRUE? ⚠ Trap!
💡 det = 0 → you'd be dividing by zero in the formula!
📖 Explanation
Since \(A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A)\), if \(\det(A)=0\) we'd divide by zero — impossible! The matrix is called singular and has no inverse.
💡 Diagonal matrix trick: just flip each diagonal entry to its reciprocal!
📖 Explanation
\(\det(A) = 10\). Apply the formula:
\(A^{-1} = \frac{1}{10}\begin{bmatrix}5&0\\0&2\end{bmatrix} = \begin{bmatrix}\frac{1}{2}&0\\0&\frac{1}{5}\end{bmatrix}\)
For diagonal matrices: just take the reciprocal of each diagonal entry! 🔑
Does \(A = \begin{bmatrix}6 & 4\\ 3 & 2\end{bmatrix}\) have an inverse? ⚠ Classic Trap!
💡 Always compute det FIRST before anything else!
📖 Explanation
\(\det(A) = (6)(2)-(4)(3) = 12-12 = \mathbf{0}\)
The matrix is singular — no inverse exists! Notice row 1 is exactly 2× row 2. Parallel rows/columns always give det = 0.
🌳 Level 3 — Properties & Tricks (Q11–15)
Q 11Hard ⭐⭐⭐
In the transpose \(A^T\), what happens to the main diagonal elements?
💡 T = Turn the matrix sideways. Rows become columns!
📖 Explanation
In \(A^T\), the main diagonal elements don't move! They are on the line of reflection (the diagonal itself). Only the off-diagonal elements swap positions. This is a classic trap question!
Q 12Hard ⭐⭐⭐
In the transpose \(A^T\), the right diagonal (off-diagonal) elements ___.
Wait — the answer is C! Off-diagonal entries only swap positions (\(b\) and \(c\) trade places) but do NOT change sign. Signs only change in the inverse formula, not in transpose. Many students confuse transpose with inverse here!
Transpose: swap positions, no sign change. Inverse: swap diagonal + negate off-diagonal + divide by det.
Q 13Hard ⭐⭐⭐
If \(A = \begin{bmatrix}k & 2\\ 3 & k\end{bmatrix}\) has no inverse, what is \(k\)?
💡 Set det = 0 and solve for k!
📖 Explanation
For no inverse: \(\det(A) = 0\)
\(k \cdot k - 2 \cdot 3 = 0\)
\(k^2 - 6 = 0\)
\(k^2 = 6 \Rightarrow k = \pm\sqrt{6}\)
There are TWO values! Don't forget the negative root 🔑
Q 14Hard ⭐⭐⭐
Which product correctly verifies \(A = \begin{bmatrix}2&3\\5&5\end{bmatrix}\) has inverse \(A^{-1} = \begin{bmatrix}-1 & \frac{3}{5}\\ 1 & -\frac{2}{5}\end{bmatrix}\)?
Let \(A = \begin{bmatrix}4&6\\3&4\end{bmatrix}\) and \(B = \begin{bmatrix}3&-2\\-2&8\end{bmatrix}\). Is \(AB = I\)? ⚠ Are they inverses?
💡 Always MULTIPLY to check — don't just look at the numbers!
📖 Explanation
\(AB = \begin{bmatrix}4(3)+6(-2) & 4(-2)+6(8)\\ 3(3)+4(-2) & 3(-2)+4(8)\end{bmatrix} = \begin{bmatrix}0&40\\1&26\end{bmatrix} \neq I\)
They are NOT inverses. Being 2×2 doesn't automatically make matrices inverses of each other!
Q 17Trap 🎯
If \((AB)^{-1} = ?\) where both A and B are invertible. ⚠ Order matters!
💡 Think of putting on socks and shoes — to undo, reverse the order! 🧦👟
📖 Explanation
\((AB)^{-1} = B^{-1}A^{-1}\) — the order reverses!
Proof: \((AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A \cdot I \cdot A^{-1} = AA^{-1} = I\) ✓
This is called the "socks and shoes" property. 🔑
Q 18Trap 🎯
A student says: "If \(A = \begin{bmatrix}1&2\\2&4\end{bmatrix}\), then \(A^{-1} = \begin{bmatrix}1&-2\\-2&4\end{bmatrix}\)." Is this correct? ⚠ Sign Trap!
💡 Check the det FIRST — is this even possible?
📖 Explanation
\(\det(A) = (1)(4)-(2)(2) = 4-4 = 0\)
The matrix is singular! No inverse exists at all. Row 2 = 2 × Row 1.
The student made the classic mistake of applying the inverse formula without checking det first. Always check det = 0 first! 🚨
Q 19Trap 🎯
Find \(x\) if \(A^{-1} = \begin{bmatrix}2&-3\\-1&x\end{bmatrix}\) and \(A = \begin{bmatrix}x&3\\1&2\end{bmatrix}\)
💡 Use \(\det(A) \cdot \det(A^{-1}) = 1\) or verify \(AA^{-1} = I\)!
📖 Explanation
For \(A = \begin{bmatrix}x&3\\1&2\end{bmatrix}\), using the inverse formula:
\(A^{-1} = \frac{1}{2x-3}\begin{bmatrix}2&-3\\-1&x\end{bmatrix}\)
For this to equal the given \(A^{-1}\), we need \(2x-3 = 1 \Rightarrow x = 2\)
Check: \(\det(A) = 2(2)-3(1) = 1\) ✓ and the formula gives exactly the matrix shown ✓
Q 20Trap 🎯⭐⭐⭐
The system \(2x + 3y = 7\) and \(5x + 5y = 10\) can be written as \(AX = B\). Solving gives \(X = A^{-1}B\). What is the solution \((x, y)\)?
💡 \(A = \begin{bmatrix}2&3\\5&5\end{bmatrix}\), find \(A^{-1}\) first, then multiply \(A^{-1}\begin{bmatrix}7\\10\end{bmatrix}\)
📖 Explanation
\(\det(A) = (2)(5)-(3)(5) = 10-15 = -5\)
\(A^{-1} = \frac{1}{-5}\begin{bmatrix}5&-3\\-5&2\end{bmatrix} = \begin{bmatrix}-1&\frac{3}{5}\\1&-\frac{2}{5}\end{bmatrix}\)
\(X = A^{-1}B = \begin{bmatrix}-1&\frac{3}{5}\\1&-\frac{2}{5}\end{bmatrix}\begin{bmatrix}7\\10\end{bmatrix} = \begin{bmatrix}-7+6\\7-4\end{bmatrix} = \begin{bmatrix}-1\\3\end{bmatrix}\)
So \(x = -1,\; y = 3\) ✓. This is the most powerful use of inverse matrices! 🔑