Inverse of Matrix
Digital Handwritten Lesson
🌟 New Millennium Academy
Pokhara - 17, Biraunta | Excellence in Learning
📐 Additional Maths · Class 10
🔍 Activity - 1: Is There Division of Matrices?
- Question: Is there anything like division of matrices?
- In the number system, we know: \(a \times \frac{1}{a} = 1\) or \(a \cdot a^{-1} = 1\)
Here, \(a^{-1}\) is called the multiplicative inverse of \(a\).
Key Insight: The product of a number and its inverse is always 1.
- Can we apply this to matrices? Do matrices have their own "1"?
🧪 Activity - 2: Discovering Inverse
Let us consider two matrices:
$$A = \begin{bmatrix} 3 & 5 \\ 2 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} -3 & 5 \\ 2 & -3 \end{bmatrix}$$Calculate \(AB\):
$$AB = \begin{bmatrix} 3 & 5 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} -3 & 5 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} -9+10 & 15-15 \\ -6+6 & 10-9 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
Calculate \(BA\):
$$BA = \begin{bmatrix} -3 & 5 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & 5 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} -9+10 & -15+15 \\ 6-6 & 10-9 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$
Observation: \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) is the identity matrix (acts like 1).
\(AB = I\) and \(BA = I\) → \(A\) and \(B\) are inverses of each other.
\(AB = I\) and \(BA = I\) → \(A\) and \(B\) are inverses of each other.
📖 Definition of Inverse of a Matrix
Let \(A\) and \(B\) be square matrices of same order. \(B\) is the inverse of \(A\) if:
$$AB = BA = I$$Notation: \(B = A^{-1}\).
Conclusion: If two matrices satisfy \(AB = BA = I\), they are inverses.
📌 Try this:
\(P = \begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix}\), \(Q = \begin{bmatrix} -3 & 4 \\ 5 & 1 \end{bmatrix}\)
Find \(PQ\), \(QP\). Are they inverses?
\(QP = \begin{bmatrix} -3 & 4 \\ 5 & 1 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} -3+16 & -9+20 \\ 5+4 & 15+5 \end{bmatrix} = \begin{bmatrix} 13 & 11 \\ 9 & 20 \end{bmatrix} \neq I\)
∴ \(P\) and \(Q\) are not inverses.
Find \(PQ\), \(QP\). Are they inverses?
✨ Show Solution
\(PQ = \begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} -3 & 4 \\ 5 & 1 \end{bmatrix} = \begin{bmatrix} -3+15 & 4+3 \\ -12+25 & 16+5 \end{bmatrix} = \begin{bmatrix} 12 & 7 \\ 13 & 21 \end{bmatrix} \neq I\)\(QP = \begin{bmatrix} -3 & 4 \\ 5 & 1 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 4 & 5 \end{bmatrix} = \begin{bmatrix} -3+16 & -9+20 \\ 5+4 & 15+5 \end{bmatrix} = \begin{bmatrix} 13 & 11 \\ 9 & 20 \end{bmatrix} \neq I\)
∴ \(P\) and \(Q\) are not inverses.
📐 Formula for Inverse (2×2)
$$A^{-1} = \frac{1}{|A|} \text{Adj}(A) = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
Condition: \(|A| \neq 0\) (non-singular) → inverse exists.
\(|A| = 0\) (singular) → no inverse.
\(|A| = 0\) (singular) → no inverse.
🧠 Adjoint of a 2×2 Matrix
For \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), \(\text{Adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)
Example: \(A = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}\) → \(\text{Adj}(A) = \begin{bmatrix} 8 & -6 \\ -7 & 5 \end{bmatrix}\)
Example: \(A = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}\) → \(\text{Adj}(A) = \begin{bmatrix} 8 & -6 \\ -7 & 5 \end{bmatrix}\)
📝 Step-by-Step: Finding \(A^{-1}\)
Given: \(A = \begin{bmatrix} 3 & 5 \\ 2 & 3 \end{bmatrix}\)
Step 1: \(|A| = 3\times3 - 5\times2 = 9 - 10 = -1 \neq 0\) → inverse exists.
Step 2: \(\text{Adj}(A) = \begin{bmatrix} 3 & -5 \\ -2 & 3 \end{bmatrix}\)
Step 3: \(A^{-1} = \frac{1}{-1}\begin{bmatrix} 3 & -5 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -3 & 5 \\ 2 & -3 \end{bmatrix}\)
Verification: \(AA^{-1} = \begin{bmatrix} 3 & 5 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} -3 & 5 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) ✓
Step 1: \(|A| = 3\times3 - 5\times2 = 9 - 10 = -1 \neq 0\) → inverse exists.
Step 2: \(\text{Adj}(A) = \begin{bmatrix} 3 & -5 \\ -2 & 3 \end{bmatrix}\)
Step 3: \(A^{-1} = \frac{1}{-1}\begin{bmatrix} 3 & -5 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} -3 & 5 \\ 2 & -3 \end{bmatrix}\)
Verification: \(AA^{-1} = \begin{bmatrix} 3 & 5 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} -3 & 5 \\ 2 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) ✓
⚡ Properties of Inverse (with Numerical Proofs)
👇 Click/Tap on any property to see the complete step-by-step numerical verification using concrete 2×2 matrices.
❶ Property I: \((AB)^{-1} = B^{-1}A^{-1}\) 🔍 Click for numerical proof
📌 Step 1: Choose numerical matrices
Let \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}\).
Let \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}\).
📌 Step 2: Compute \(AB\)
\(AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1\cdot2+2\cdot1 & 1\cdot0+2\cdot2 \\ 3\cdot2+4\cdot1 & 3\cdot0+4\cdot2 \end{bmatrix} = \begin{bmatrix} 2+2 & 0+4 \\ 6+4 & 0+8 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 10 & 8 \end{bmatrix}\)
\(AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1\cdot2+2\cdot1 & 1\cdot0+2\cdot2 \\ 3\cdot2+4\cdot1 & 3\cdot0+4\cdot2 \end{bmatrix} = \begin{bmatrix} 2+2 & 0+4 \\ 6+4 & 0+8 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 10 & 8 \end{bmatrix}\)
📌 Step 3: Find \((AB)^{-1}\) using formula
\(|AB| = 4\cdot8 - 4\cdot10 = 32 - 40 = -8\)
\(\text{Adj}(AB) = \begin{bmatrix} 8 & -4 \\ -10 & 4 \end{bmatrix}\)
\((AB)^{-1} = \frac{1}{-8}\begin{bmatrix} 8 & -4 \\ -10 & 4 \end{bmatrix} = \begin{bmatrix} -1 & 0.5 \\ 1.25 & -0.5 \end{bmatrix}\)
\(|AB| = 4\cdot8 - 4\cdot10 = 32 - 40 = -8\)
\(\text{Adj}(AB) = \begin{bmatrix} 8 & -4 \\ -10 & 4 \end{bmatrix}\)
\((AB)^{-1} = \frac{1}{-8}\begin{bmatrix} 8 & -4 \\ -10 & 4 \end{bmatrix} = \begin{bmatrix} -1 & 0.5 \\ 1.25 & -0.5 \end{bmatrix}\)
📌 Step 4: Find \(A^{-1}\) and \(B^{-1}\) separately
\(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), \(|A| = 1\cdot4 - 2\cdot3 = 4-6 = -2\)
\(A^{-1} = \frac{1}{-2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}\)
\(B = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}\), \(|B| = 2\cdot2 - 0\cdot1 = 4\)
\(B^{-1} = \frac{1}{4}\begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 0.5 & 0 \\ -0.25 & 0.5 \end{bmatrix}\)
\(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), \(|A| = 1\cdot4 - 2\cdot3 = 4-6 = -2\)
\(A^{-1} = \frac{1}{-2}\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}\)
\(B = \begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}\), \(|B| = 2\cdot2 - 0\cdot1 = 4\)
\(B^{-1} = \frac{1}{4}\begin{bmatrix} 2 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 0.5 & 0 \\ -0.25 & 0.5 \end{bmatrix}\)
📌 Step 5: Compute \(B^{-1}A^{-1}\)
\(B^{-1}A^{-1} = \begin{bmatrix} 0.5 & 0 \\ -0.25 & 0.5 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}\)
\(= \begin{bmatrix} 0.5\cdot(-2)+0\cdot1.5 & 0.5\cdot1+0\cdot(-0.5) \\ (-0.25)\cdot(-2)+0.5\cdot1.5 & (-0.25)\cdot1+0.5\cdot(-0.5) \end{bmatrix}\)
\(= \begin{bmatrix} -1+0 & 0.5+0 \\ 0.5+0.75 & -0.25-0.25 \end{bmatrix} = \begin{bmatrix} -1 & 0.5 \\ 1.25 & -0.5 \end{bmatrix}\)
\(B^{-1}A^{-1} = \begin{bmatrix} 0.5 & 0 \\ -0.25 & 0.5 \end{bmatrix} \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}\)
\(= \begin{bmatrix} 0.5\cdot(-2)+0\cdot1.5 & 0.5\cdot1+0\cdot(-0.5) \\ (-0.25)\cdot(-2)+0.5\cdot1.5 & (-0.25)\cdot1+0.5\cdot(-0.5) \end{bmatrix}\)
\(= \begin{bmatrix} -1+0 & 0.5+0 \\ 0.5+0.75 & -0.25-0.25 \end{bmatrix} = \begin{bmatrix} -1 & 0.5 \\ 1.25 & -0.5 \end{bmatrix}\)
✅ Conclusion: \((AB)^{-1} = \begin{bmatrix} -1 & 0.5 \\ 1.25 & -0.5 \end{bmatrix}\) and \(B^{-1}A^{-1} = \begin{bmatrix} -1 & 0.5 \\ 1.25 & -0.5 \end{bmatrix}\)
Therefore, \((AB)^{-1} = B^{-1}A^{-1}\). ✓
Therefore, \((AB)^{-1} = B^{-1}A^{-1}\). ✓
❷ Property II: \(AA^{-1} = I\) 🔍 Click for numerical proof
📌 Step 1: Take a concrete matrix
Let \(A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\)
Let \(A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\)
📌 Step 2: Compute \(A^{-1}\) using formula
\(|A| = 2\cdot3 - 5\cdot1 = 6 - 5 = 1\)
\(A^{-1} = \frac{1}{1}\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\)
\(|A| = 2\cdot3 - 5\cdot1 = 6 - 5 = 1\)
\(A^{-1} = \frac{1}{1}\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\)
📌 Step 3: Multiply \(A \times A^{-1}\)
\(AA^{-1} = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\)
First row, first column: \(2\cdot3 + 5\cdot(-1) = 6 - 5 = 1\)
First row, second column: \(2\cdot(-5) + 5\cdot2 = -10 + 10 = 0\)
Second row, first column: \(1\cdot3 + 3\cdot(-1) = 3 - 3 = 0\)
Second row, second column: \(1\cdot(-5) + 3\cdot2 = -5 + 6 = 1\)
\(AA^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I\)
\(AA^{-1} = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}\)
First row, first column: \(2\cdot3 + 5\cdot(-1) = 6 - 5 = 1\)
First row, second column: \(2\cdot(-5) + 5\cdot2 = -10 + 10 = 0\)
Second row, first column: \(1\cdot3 + 3\cdot(-1) = 3 - 3 = 0\)
Second row, second column: \(1\cdot(-5) + 3\cdot2 = -5 + 6 = 1\)
\(AA^{-1} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I\)
✅ Conclusion: \(AA^{-1} = I\) for any non-singular matrix. ✓
❸ Property III: \(|A| \cdot |A^{-1}| = 1\) 🔍 Click for numerical proof
📌 Step 1: Choose a matrix
Let \(A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}\)
Let \(A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}\)
📌 Step 2: Compute \(|A|\)
\(|A| = 4\cdot6 - 7\cdot2 = 24 - 14 = 10\)
\(|A| = 4\cdot6 - 7\cdot2 = 24 - 14 = 10\)
📌 Step 3: Compute \(A^{-1}\) and its determinant
\(A^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}\)
\(|A^{-1}| = (0.6)(0.4) - (-0.7)(-0.2) = 0.24 - 0.14 = 0.1\)
\(A^{-1} = \frac{1}{10}\begin{bmatrix} 6 & -7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 0.6 & -0.7 \\ -0.2 & 0.4 \end{bmatrix}\)
\(|A^{-1}| = (0.6)(0.4) - (-0.7)(-0.2) = 0.24 - 0.14 = 0.1\)
📌 Step 4: Multiply determinants
\(|A| \times |A^{-1}| = 10 \times 0.1 = 1\)
\(|A| \times |A^{-1}| = 10 \times 0.1 = 1\)
✅ Conclusion: \(|A| \cdot |A^{-1}| = 1\) → \(|A^{-1}| = \frac{1}{|A|}\). ✓
❹ Property IV: \(I^{-1} = I\) 🔍 Click for numerical proof
📌 Step 1: Identity matrix of order 2
\(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
📌 Step 2: Check if \(I \times I = I\)
\(I \times I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1\cdot1+0\cdot0 & 1\cdot0+0\cdot1 \\ 0\cdot1+1\cdot0 & 0\cdot0+1\cdot1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I\)
\(I \times I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1\cdot1+0\cdot0 & 1\cdot0+0\cdot1 \\ 0\cdot1+1\cdot0 & 0\cdot0+1\cdot1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I\)
📌 Step 3: By definition of inverse
Since \(I \cdot I = I\), the matrix \(I\) satisfies the inverse condition. Therefore, \(I\) is its own inverse.
Since \(I \cdot I = I\), the matrix \(I\) satisfies the inverse condition. Therefore, \(I\) is its own inverse.
✅ Conclusion: \(I^{-1} = I\). ✓
📋 Summary Table
| Concept | Formula / Condition |
|---|---|
| Inverse definition | \(AA^{-1} = A^{-1}A = I\) |
| 2×2 Inverse formula | \(A^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\) |
| Singular matrix | \(|A| = 0\) → no inverse |
| Non-singular matrix | \(|A| \neq 0\) → inverse exists |
| Property \((AB)^{-1}\) | \(B^{-1}A^{-1}\) |
| Determinant property | \(|A^{-1}| = \frac{1}{|A|}\) |
🌟 Key takeaway: The inverse "undoes" the original matrix. Only square matrices with non-zero determinant have inverses.
© New Millennium Academy, Pokhara-17, Birauta | Additional Mathematics, Class 10 — Matrix Inverse & Properties
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Course material curated by Mr. Nripendraswar Acharya