Q5. Examine the consistency of x + y + z = 1, 2x + 3y + 2z = 2, ax + ay + 2az = 4 (NCERT Exercise 4.5, Q4).
|A| = 1(6a − 2a) − 1(4a − 2a) + 1(2a − 3a) = 4a − 2a − a = a. So for a ≠ 0 the system is consistent, with exactly one solution. NCERT slip (answers, page 214): the key says only “Consistent”; that holds only for a ≠ 0, since for a = 0 the third equation reads 0 = 4.
Q6. Examine the consistency of 5x − y + 4z = 5, 2x + 3y + 5z = 2, 5x − 2y + 6z = −1 (NCERT Exercise 4.5, Q6).
|A| = 5(18 + 10) + 1(12 − 25) + 4(−4 − 15) = 140 − 13 − 76 = 51 ≠ 0, so the system is consistent, with the one solution x = 3, y = 2, z = −2.
Q7. For A with rows 2, −3, 5; 3, 2, −4; 1, 1, −2, find A and use it to solve 2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y − 2z = −3 (NCERT Exercise 4.5, Q15).
|A| = 2(−4 + 4) + 3(−6 + 4) + 5(3 − 2) = 0 − 6 + 5 = −1. adj A has rows 0, −1, 2; 2, −9, 23; 1, −5, 13, so A = −adj A, with rows 0, 1, −2; −2, 9, −23; −1, 5, −13. Then X = AB with B = 11, −5, −3 gives x = 0 − 5 + 6 = 1, y = −22 − 45 + 69 = 2, z = −11 − 25 + 39 = 3.
Q8. Use the product of the matrices with rows 1, −1, 2; 0, 2, −3; 3, −2, 4 and rows −2, 0, 1; 9, 2, −3; 6, 1, −2 to solve x − y + 2z = 1, 2y − 3z = 1, 3x − 2y + 4z = 2 (NCERT Example 19, pages 98–99). (In NCERT’s PDF the minus signs in these two matrices use a font glyph that the file does not define, so they can show as dotted bars and are lost when the text is copied. This is a defect of the PDF file, not a misprint: use the signs in NCERT’s solution on page 98.)
The product is I (for example, row 1 times column 1 is −2 − 9 + 12 = 1), so the second matrix is the inverse of the first, which is the coefficient matrix. X = AB with B = 1, 1, 2 gives −2 + 0 + 2 = 0, 9 + 2 − 6 = 5 and 6 + 1 − 4 = 3: x = 0, y = 5, z = 3.
Q9. Solve 2x + y + z = 1, x − 2y − z = 3/2, 3y − 5z = 9 by the matrix method (NCERT Exercise 4.5, Q11).
|A| = 2(10 + 3) − 1(−5 − 0) + 1(3 − 0) = 26 + 5 + 3 = 34. The matrix method gives x = 1, y = 1/2, z = −3/2. Check: 2 + 1/2 − 3/2 = 1, 1 − 1 + 3/2 = 3/2 and 3/2 + 15/2 = 9.
Q10. For which values of k does the system x + 2y = 3, 2x + ky = 5 have exactly one solution? What happens for the other value?
|A| = k − 4, so there is exactly one solution when k ≠ 4. For k = 4, adj A has rows 4, −2 and −2, 1, and (adj A)B has entries 12 − 10 = 2 and −6 + 5 = −1, which is not O: the system is inconsistent (the lines x + 2y = 3 and x + 2y = 5/2 are parallel).