Applications of Matrices and Determinants
ICSE · Class 12 · Mathematics
Flashcards for Applications of Matrices and Determinants — ICSE Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.
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Explore the full setSolve the system by testing consistency: 5x + 2y = 4 and 7x + 3y = 5.
Answer
Step 1: Write in matrix form AX = B, where A = [[5, 2], [7, 3]], X = [x, y]^T, B = [4, 5]^T. Step 2: Find |A| = (5)(3) - (2)(7) = 15 - 14 = 1. Step 3: Since |A| ≠ 0, the system has a unique solution.
Solve the system by testing consistency: x + 3y = 5 and 2x + 6y = 8.
Answer
Step 1: Write A = [[1, 3], [2, 6]], X = [x, y]^T, B = [5, 8]^T. Step 2: Find |A| = (1)(6) - (3)(2) = 6 - 6 = 0. Step 3: Compute adj A = [[6, -3], [-2, 1]]. Step 4: Compute (adj A)B = [[6, -3], [-2, 1]…
Solve the system by testing consistency: 4x + 3y = 5 and 8x + 6y = 10.
Answer
Step 1: Write A = [[4, 3], [8, 6]], X = [x, y]^T, B = [5, 10]^T. Step 2: Find |A| = (4)(6) - (3)(8) = 24 - 24 = 0. Step 3: Compute adj A = [[6, -3], [-8, 4]]. Step 4: Compute (adj A)B = [[6, -3], [-8,…
For the system x + y + z = 2, 2x + y - z = 3, 3x + 2y + kz = 4, find the value of k for a unique solution.
Answer
Step 1: A = [[1, 1, 1], [2, 1, -1], [3, 2, k]]. Step 2: For a unique solution, |A| must not be zero. Step 3: |A| = 1(k + 2) - 1(2k + 3) + 1(4 - 3) = k + 2 - 2k - 3 + 1 = -k. Step 4: Set -k ≠ 0. Answer…
Why does a non-singular coefficient matrix give a unique solution?
Answer
If |A| ≠ 0, then A has an inverse. The system AX = B can be solved as X = A^{-1}B. That gives exactly one value of X, so the solution is unique. Quick check: For 5x + 2y = 4 and 7x + 3y = 5, |A| = 1 ≠…
Solve using the matrix method: 5x + 2y = 4 and 7x + 3y = 5.
Answer
Step 1: Write AX = B with A = [[5, 2], [7, 3]], X = [x, y]^T, B = [4, 5]^T. Step 2: |A| = 1, so A is invertible. Step 3: adj A = [[3, -2], [-7, 5]]. Step 4: A^{-1} = (1/|A|)(adj A) = [[3, -2], [-7, 5]…
Solve using the matrix method: 3x - 2y + 3z = 8, 2x + y - z = 1, 4x - 3y + 2z = 4.
Answer
Step 1: Write A = [[3, -2, 3], [2, 1, -1], [4, -3, 2]], X = [x, y, z]^T, B = [8, 1, 4]^T. Step 2: |A| = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = -3 + 16 - 30 = -17. Step 3: Since |A| ≠ 0, use X = A^{-1}B. St…
Solve using the matrix method: x + y + z = 1, 2x + 2y + 2z = 2, 3x + 3y + 3z = 3.
Answer
Step 1: A = [[1, 1, 1], [2, 2, 2], [3, 3, 3]], B = [1, 2, 3]^T. Step 2: |A| = 0. Step 3: adj A = O because every cofactor is zero. Step 4: Since (adj A)B = O, the system is consistent and dependent. S…
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