Questions: Determinants
Determinants26 min readLesson 5 of 6
Q1. Find the determinant of the 2 by 2 matrix with rows [3, 4] and [2, 5].
Answer (English): The determinant is the product of the main diagonal minus the product of the other diagonal, so 3 times 5 minus 4 times 2. That is 15 minus 8, which equals 7. A non-zero determinant also means this matrix is invertible and the corresponding homogeneous system has only the trivial solution.
Hinglish: Determinant = 3 x 5 - 4 x 2 = 15 - 8 = 7. Non-zero hai, isliye matrix invertible hai.
Q2. Which row or column operations leave a determinant unchanged, and which change it?
Answer (English): Adding a multiple of one row to another row leaves the determinant unchanged, and the same is true for columns. Multiplying one row by a non-zero scalar k multiplies the whole determinant by k, so the value changes unless k equals 1. Interchanging two rows or two columns multiplies the determinant by minus one. These three facts are the standard toolkit for property based questions.
Hinglish: Ek row me doosri row ka multiple add karne se value same rehti hai, columns ke liye bhi yahi rule hai. Row ko scalar k se multiply karne par determinant k times ho jaata hai. Rows ya columns swap karne par sign badal jaata hai. Yeh teen facts property questions ka base hain.
Q3. Why is a matrix with a zero determinant not invertible?
Answer (English): The inverse formula divides the adjoint by the determinant, so a zero determinant makes the division impossible. Geometrically, a zero determinant means the matrix collapses area or volume to nothing, and the transformation cannot be reversed. The matrix is then called singular.
Hinglish: Inverse ka formula adjoint divided by determinant hota hai, isliye determinant zero hone par division possible nahi. Aise matrix ko singular kehte hain aur wapas reverse nahi kar sakte.
Q4. Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).
Answer (English): Write the determinant with rows [0, 0, 1], [4, 0, 1] and [0, 3, 1]. Expanding along the first row gives 0 times a minor minus 0 times a minor plus 1 times (12 minus 0), which is 12. The area is half of the absolute value, so 6 square units. The answer also matches the right triangle formula with legs 4 and 3.
Hinglish: Determinant se value 12 aati hai, aur area = 12/2 = 6 square units. Right triangle ke formula se bhi same answer aata hai.
Q5. Find the minor and cofactor of the entry 2 in the matrix [[1, 2], [3, 4]].
Answer (English): The entry 2 sits in row 1 and column 2. Deleting row 1 and column 2 removes both 1 and 2 in the first row and the entry 4 in the second row, leaving only the entry 3 in row 2, column 1. So the minor is 3. The cofactor carries the sign for position row 1, column 2, and that sign is minus, so the cofactor is minus 3. The same checkerboard rule gives plus for the other two corners.
Hinglish: Entry 2 row 1 column 2 me hai. Row 1 aur column 2 delete karne par sirf entry 3 (row 2 column 1) bachta hai, isliye minor 3 hai. Position ka sign minus hota hai, isliye cofactor -3 hai. Baaki corners ka sign plus hota hai.
Q6. Write the adjoint of the 2 by 2 matrix [[1, 2], [3, 4]].
Answer (English): For a 2 by 2 matrix, the cofactor matrix is formed by swapping the diagonal entries and negating the off-diagonal ones, and the transpose of that matrix is the same matrix. So the adjoint is [[4, minus 2], [minus 3, 1]]. Multiplying the original matrix by this adjoint gives the determinant times the identity, which here is minus 2 times the identity.
Hinglish: 2 by 2 me diagonal swap karo, off-diagonal ke sign badlo, transpose lo. Adjoint = [[4, -2], [-3, 1]].
Q7. Use the adjoint to find the inverse of [[2, 1], [5, 3]].
Answer (English): The determinant is 6 minus 5, which is 1. The adjoint is [[3, minus 1], [minus 5, 2]]. Dividing by the determinant leaves the same matrix, so the inverse is [[3, minus 1], [minus 5, 2]]. Multiply the original matrix by this inverse to confirm the identity matrix.
Hinglish: Determinant = 6 - 5 = 1, adjoint = [[3, -1], [-5, 2]]. Inverse = adjoint/1 = [[3, -1], [-5, 2]].
Q8. For what value of k are the points (1, 2), (3, 4) and (k, 6) collinear?
Answer (English): Collinearity means the area determinant is zero. The determinant with rows [1, 2, 1], [3, 4, 1] and [k, 6, 1] expands to 1(4 minus 6) minus 2(3 minus k) plus 1(18 minus 4k). Setting it to zero gives minus 2 minus 6 plus 2k plus 18 minus 4k = 0, so 10 minus 2k = 0 and k = 5.
Hinglish: Collinear matlab area zero, isliye determinant zero rakho. Solve karne par 10 - 2k = 0, yaani k = 5.
Q9. Solve the system x + y = 3 and 2x minus y = 0 using matrices.
Answer (English): Write the system as AX = B with A = [[1, 1], [2, minus 1]], X = [x, y] and B = [3, 0]. The determinant of A is minus 1 minus 2, which is minus 3, so the inverse exists. The adjoint is [[minus 1, minus 1], [minus 2, 1]], and dividing by minus 3 gives the inverse. Multiplying by B gives x = 1 and y = 2, which satisfies both equations.
Hinglish: System ko AX = B likho. Determinant -3, inverse nikaalo, phir X = A⁻¹B se x = 1, y = 2 milta hai. Dono equations me substitute karke check karo.
Q10. If A is a 3 by 3 matrix with determinant 2, find the determinant of its adjoint.
Answer (English): For an n by n matrix, the determinant of the adjoint equals the determinant of the matrix raised to the power n minus 1. Here n is 3 and the determinant is 2, so the determinant of the adjoint is 2 squared, which is 4. This identity also shows that the adjoint is invertible exactly when the original matrix is.
Hinglish: |adj A| = |A|^(n-1). Yahan n = 3 aur |A| = 2, isliye |adj A| = 2² = 4.
Marks pattern on this chapter
Objective questions test determinant values, properties and the adjoint identities in one line. Two and three mark questions ask for an area, a cofactor or a short inverse. Five mark questions combine the matrix method with a system of equations. Write the determinant first, then the adjoint or inverse, then the final answer, so each stage earns its mark.
Marks pattern on this chapter
Objective questions test determinant values, properties and adjoint identities in one line. Two and three mark questions ask for an area, a cofactor or a short inverse. Five mark questions combine the matrix method with a system of equations. Write the determinant first, then the adjoint or inverse, then the final answer, so each stage earns its mark.
Quiz below: five multiple choice questions on determinant values, properties, area, adjoint and inverse. Attempt them after revising the formula sheet.