CBSE Class 12 Mathematics Determinants Worksheet Set 02

CBSE Class 12 Mathematics Determinants Worksheet (2). The Determinants questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practise them to clear their Determinants concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Determinants worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Determinants chapter and other subjects too. Use them for better understanding of the subjects.

Question. Solve \( \begin{vmatrix} a + x & a - x & a - x \\ a - x & a + x & a - x \\ a - x & a - x & a + x \end{vmatrix} = 0 \)
Answer: We begin by applying the column transformation \( C_1 \to C_1 + C_2 + C_3 \):
\( \implies \begin{vmatrix} 3a - x & a - x & a - x \\ 3a - x & a + x & a - x \\ 3a - x & a - x & a + x \end{vmatrix} = 0 \)
Factoring out the term \( (3a - x) \) from the first column \( C_1 \):
\( \implies (3a - x) \begin{vmatrix} 1 & a - x & a - x \\ 1 & a + x & a - x \\ 1 & a - x & a + x \end{vmatrix} = 0 \)
Next, we perform the row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \) to simplify the determinant:
\( \implies (3a - x) \begin{vmatrix} 1 & a - x & a - x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \end{vmatrix} = 0 \)
Now, factoring out \( 2x \) from both the second row \( R_2 \) and third row \( R_3 \):
\( \implies (3a - x)(2x)(2x) \begin{vmatrix} 1 & a - x & a - x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 0 \)
Expanding the simplified determinant along the first row \( R_1 \):
\( \implies (3a - x)(4x^2)[1(1 - 0) - 0 + 0] = 0 \)
\( \implies 4x^2(3a - x) = 0 \)
Solving for the roots, we get:
\( \implies x = 3a \quad \text{or} \quad x = 0 \)

Question. Solve \( \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ x - 4 & 2x - 9 & 3x - 16 \\ x - 8 & 2x - 27 & 3x - 64 \end{vmatrix} = 0 \)
Answer: We are given the determinant equation:
\( \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ x - 4 & 2x - 9 & 3x - 16 \\ x - 8 & 2x - 27 & 3x - 64 \end{vmatrix} = 0 \)
We apply row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \implies \begin{vmatrix} x - 2 & 2x - 3 & 3x - 4 \\ -2 & -6 & -12 \\ -6 & -24 & -60 \end{vmatrix} = 0 \)
Now, performing column operations \( C_2 \to C_2 - 2C_1 \) and \( C_3 \to C_3 - 3C_1 \):
\( \implies \begin{vmatrix} x - 2 & 1 & 2 \\ -2 & -2 & -6 \\ -6 & -12 & -42 \end{vmatrix} = 0 \)
By extracting the common factors \( -2 \) from \( R_2 \) and \( -6 \) from \( R_3 \), we obtain:
\( \implies (-2)(-6) \begin{vmatrix} x - 2 & 1 & 2 \\ 1 & 1 & 3 \\ 1 & 2 & 7 \end{vmatrix} = 0 \)
\( \implies 12 \begin{vmatrix} x - 2 & 1 & 2 \\ 1 & 1 & 3 \\ 1 & 2 & 7 \end{vmatrix} = 0 \)
Expanding the determinant along the first row \( R_1 \):
\( \implies 12[(x - 2)(7 - 6) - 1(7 - 3) + 2(2 - 1)] = 0 \)
\( \implies 12[x - 2 - 4 + 2] = 0 \)
\( \implies 12(x - 4) = 0 \)
\( \implies x = 4 \)

Proving Questions

Question. Show that \( \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = (a - b)(b - c)(c - a) \)
Answer: Let us define the determinant as:
\( \Delta = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \)
Performing the row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = \begin{vmatrix} 1 & a & a^2 \\ 0 & b - a & b^2 - a^2 \\ 0 & c - a & c^2 - a^2 \end{vmatrix} \)
We take out \( (b - a) \) from the second row \( R_2 \) and \( (c - a) \) from the third row \( R_3 \) as common factors:
\( \Delta = (b - a)(c - a) \begin{vmatrix} 1 & a & a^2 \\ 0 & 1 & b + a \\ 0 & 1 & c + a \end{vmatrix} \)
Now, we expand the determinant along the first row \( R_1 \):
\( \Delta = (b - a)(c - a) [1 \cdot (c + a - (b + a)) - a(0) + a^2(0)] \)
\( \implies \Delta = (b - a)(c - a)(c - b) \)
By rearranging the signs to fit the standard cyclic order, we get:
\( \implies \Delta = (a - b)(b - c)(c - a) = \text{R.H.S.} \)

Question. Show that \( \begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ bc & ca & ab \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} = (a - b)(b - c)(c - a)(ab + bc + ca) \)
Answer: Let the first determinant be:
\( \Delta = \begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ bc & ca & ab \end{vmatrix} \)
Multiplying the columns \( C_1 \), \( C_2 \), and \( C_3 \) by \( a \), \( b \), and \( c \) respectively, and dividing by \( abc \) on the outside:
\( \Delta = \frac{1}{abc} \begin{vmatrix} a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \\ abc & abc & abc \end{vmatrix} \)
Factoring out the term \( abc \) from the third row \( R_3 \):
\( \Delta = \frac{abc}{abc} \begin{vmatrix} a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \\ 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \\ 1 & 1 & 1 \end{vmatrix} \)
Interchanging row operations to rearrange the order: first swap \( R_2 \leftrightarrow R_3 \) (introducing a negative sign):
\( \Delta = - \begin{vmatrix} a^2 & b^2 & c^2 \\ 1 & 1 & 1 \\ a^3 & b^3 & c^3 \end{vmatrix} \)
Next, swap \( R_1 \leftrightarrow R_2 \) (bringing another negative sign, making it positive):
\( \Delta = \begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} \)
This proves the first equality.
Now, we perform column transformations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = \begin{vmatrix} 1 & 0 & 0 \\ a^2 & b^2 - a^2 & c^2 - a^2 \\ a^3 & b^3 - a^3 & c^3 - a^3 \end{vmatrix} \)
Writing the expressions in factored form:
\( \Delta = \begin{vmatrix} 1 & 0 & 0 \\ a^2 & (b - a)(b + a) & (c - a)(c + a) \\ a^3 & (b - a)(b^2 + ab + a^2) & (c - a)(c^2 + ac + a^2) \end{vmatrix} \)
Factoring out \( (b - a) \) from column \( C_2 \) and \( (c - a) \) from column \( C_3 \):
\( \Delta = (b - a)(c - a) \begin{vmatrix} 1 & 0 & 0 \\ a^2 & b + a & c + a \\ a^3 & b^2 + ab + a^2 & c^2 + ac + a^2 \end{vmatrix} \)
Expanding along the first row \( R_1 \):
\( \Delta = (b - a)(c - a) [(b + a)(c^2 + ac + a^2) - (c + a)(b^2 + ab + a^2)] \)
Multiplying out the terms within the bracket:
\( = (b - a)(c - a) [bc^2 + abc + a^2b + ac^2 + a^2c + a^3 - (b^2c + abc + a^2c + ab^2 + a^2b + a^3)] \)
Canceling out identical terms simplifies the expression to:
\( = (b - a)(c - a) [bc^2 + ac^2 - b^2c - ab^2] \)
Grouping the remaining terms:
\( = (b - a)(c - a) [bc(c - b) + a(c^2 - b^2)] \)
\( = (b - a)(c - a) [bc(c - b) + a(c - b)(c + b)] \)
Factoring out \( (c - b) \):
\( = (b - a)(c - a)(c - b) [bc + a(c + b)] \)
\( = (b - a)(c - a)(c - b)(ab + bc + ca) \)
Finally, rearranging the signs to obtain the standard cyclic order:
\( \Delta = (a - b)(b - c)(c - a)(ab + bc + ca) = \text{R.H.S.} \)

Question. Show that \( \begin{vmatrix} a + b + 2c & a & b \\ c & b + c + 2a & b \\ c & a & c + a + 2b \end{vmatrix} = 2(a + b + c)^3 \)
Answer: We apply the column transformation \( C_1 \to C_1 + C_2 + C_3 \):
\( \Delta = \begin{vmatrix} 2(a + b + c) & a & b \\ 2(a + b + c) & b + c + 2a & b \\ 2(a + b + c) & a & c + a + 2b \end{vmatrix} \)
Factoring out \( 2(a + b + c) \) from the first column \( C_1 \):
\( \Delta = 2(a + b + c) \begin{vmatrix} 1 & a & b \\ 1 & b + c + 2a & b \\ 1 & a & c + a + 2b \end{vmatrix} \)
Now, we perform row subtractions \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = 2(a + b + c) \begin{vmatrix} 1 & a & b \\ 0 & a + b + c & 0 \\ 0 & 0 & a + b + c \end{vmatrix} \)
Expanding the determinant along the first column \( C_1 \):
\( \Delta = 2(a + b + c) [1 \cdot ((a + b + c)(a + b + c) - 0)] \)
\( \implies \Delta = 2(a + b + c)^3 = \text{R.H.S.} \)

Question. Show \( \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta + \gamma & \gamma + \alpha & \alpha + \beta \end{vmatrix} = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)(\alpha + \beta + \gamma) \)
Answer: We perform the row operation \( R_3 \to R_3 + R_1 \) to create a common algebraic term:
\( \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \alpha + \beta + \gamma & \alpha + \beta + \gamma & \alpha + \beta + \gamma \end{vmatrix} \)
Factoring out \( (\alpha + \beta + \gamma) \) from the third row \( R_3 \):
\( \Delta = (\alpha + \beta + \gamma) \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ 1 & 1 & 1 \end{vmatrix} \)
Performing column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = (\alpha + \beta + \gamma) \begin{vmatrix} \alpha & \beta - \alpha & \gamma - \alpha \\ \alpha^2 & \beta^2 - \alpha^2 & \gamma^2 - \alpha^2 \\ 1 & 0 & 0 \end{vmatrix} \)
Taking \( (\beta - \alpha) \) as a common factor from column \( C_2 \), and \( (\gamma - \alpha) \) from column \( C_3 \):
\( \Delta = (\alpha + \beta + \gamma)(\beta - \alpha)(\gamma - \alpha) \begin{vmatrix} \alpha & 1 & 1 \\ \alpha^2 & \beta + \alpha & \gamma + \alpha \\ 1 & 0 & 0 \end{vmatrix} \)
Expanding the determinant along the third row \( R_3 \):
\( \Delta = (\alpha + \beta + \gamma)(\beta - \alpha)(\gamma - \alpha) [1 \cdot ((\gamma + \alpha) - (\beta + \alpha))] \)
\( \implies \Delta = (\alpha + \beta + \gamma)(\beta - \alpha)(\gamma - \alpha)(\gamma - \beta) \)
By rewriting the differences in standard cyclic order:
\( \implies \Delta = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)(\alpha + \beta + \gamma) = \text{R.H.S.} \)

Question. Show \( \begin{vmatrix} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix} = 3abc - a^3 - b^3 - c^3 \)
Answer: We begin by performing the row addition \( R_1 \to R_1 + R_2 + R_3 \):
\( \Delta = \begin{vmatrix} 2(a + b + c) & 0 & a + b + c \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix} \)
Factoring out \( (a + b + c) \) from the first row \( R_1 \):
\( \Delta = (a + b + c) \begin{vmatrix} 2 & 0 & 1 \\ c + a & b - c & b \\ a + b & c - a & c \end{vmatrix} \)
Applying the column operation \( C_1 \to C_1 - 2C_3 \) to create zeros in \( R_1 \):
\( \Delta = (a + b + c) \begin{vmatrix} 0 & 0 & 1 \\ c + a - 2b & b - c & b \\ a + b - 2c & c - a & c \end{vmatrix} \)
Expanding along the first row \( R_1 \):
\( \Delta = (a + b + c) [1 \cdot ((c + a - 2b)(c - a) - (a + b - 2c)(b - c))] \)
Multiplying out the algebraic terms:
\( = (a + b + c) [ (c^2 - ac + ac - a^2 - 2bc + 2ab) - (ab - ac + b^2 - bc - 2bc + 2c^2) ] \)
\( = (a + b + c) [ c^2 - a^2 - 2bc + 2ab - ab + ac - b^2 + bc + 2bc - 2c^2 ] \)
Combining similar terms gives:
\( = (a + b + c) [-a^2 - b^2 - c^2 + ab + bc + ca] \)
\( = -(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \)
Using the standard algebraic expansion of the sum of cubes:
\( \implies \Delta = -(a^3 + b^3 + c^3 - 3abc) \)
\( \implies \Delta = 3abc - a^3 - b^3 - c^3 = \text{R.H.S.} \)

Question. Show \( \begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} = (x^3 - 1)^2 \)
Answer: Performing the column operation \( C_1 \to C_1 + C_2 + C_3 \):
\( \Delta = \begin{vmatrix} 1 + x + x^2 & x & x^2 \\ 1 + x + x^2 & 1 & x \\ 1 + x + x^2 & x^2 & 1 \end{vmatrix} \)
Factoring out \( (1 + x + x^2) \) from the first column \( C_1 \):
\( \Delta = (1 + x + x^2) \begin{vmatrix} 1 & x & x^2 \\ 1 & 1 & x \\ 1 & x^2 & 1 \end{vmatrix} \)
Applying row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = (1 + x + x^2) \begin{vmatrix} 1 & x & x^2 \\ 0 & 1 - x & x - x^2 \\ 0 & x^2 - x & 1 - x^2 \end{vmatrix} \)
Factoring \( (1 - x) \) from row \( R_2 \) and row \( R_3 \):
\( \Delta = (1 + x + x^2)(1 - x)^2 \begin{vmatrix} 1 & x & x^2 \\ 0 & 1 & x \\ 0 & -x & 1 + x \end{vmatrix} \)
Expanding the determinant along the first column \( C_1 \):
\( \Delta = (1 + x + x^2)(1 - x)^2 [1 \cdot (1(1 + x) - x(-x))] \)
\( \implies \Delta = (1 + x + x^2)(1 - x)^2 (1 + x + x^2) \)
\( \implies \Delta = [ (1 - x)(1 + x + x^2) ]^2 \)
Using the difference of cubes identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( \implies \Delta = (1 - x^3)^2 \)
Since \( (1 - x^3)^2 = (x^3 - 1)^2 \):
\( \implies \Delta = (x^3 - 1)^2 = \text{R.H.S.} \)

Question. Show \( \begin{vmatrix} a^2 & 2ab & b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2 \end{vmatrix} = (a^3 + b^3)^2 \)
Answer: Let us apply column transformation \( C_1 \to C_1 + C_2 + C_3 \):
\( \Delta = \begin{vmatrix} a^2 + 2ab + b^2 & 2ab & b^2 \\ a^2 + 2ab + b^2 & a^2 & 2ab \\ a^2 + 2ab + b^2 & b^2 & a^2 \end{vmatrix} \)
Replacing \( a^2 + 2ab + b^2 \) with \( (a + b)^2 \), we factor it out from the first column \( C_1 \):
\( \Delta = (a + b)^2 \begin{vmatrix} 1 & 2ab & b^2 \\ 1 & a^2 & 2ab \\ 1 & b^2 & a^2 \end{vmatrix} \)
Now, performing row subtractions \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = (a + b)^2 \begin{vmatrix} 1 & 2ab & b^2 \\ 0 & a^2 - 2ab & 2ab - b^2 \\ 0 & b^2 - 2ab & a^2 - b^2 \end{vmatrix} \)
Expanding along the first column \( C_1 \):
\( \Delta = (a + b)^2 [ (a^2 - 2ab)(a^2 - b^2) - (b^2 - 2ab)(2ab - b^2) ] \)
Multiplying out the algebraic products inside the bracket:
\( = (a + b)^2 [ (a^4 - a^2b^2 - 2a^3b + 2ab^3) - (2ab^3 - b^4 - 4a^2b^2 + 2ab^3) ] \)
\( = (a + b)^2 [ a^4 - a^2b^2 - 2a^3b + 2ab^3 - 4ab^3 + b^4 + 4a^2b^2 ] \)
\( = (a + b)^2 [ a^4 + b^4 + 3a^2b^2 - 2a^3b - 2ab^3 ] \)
This trinomial expression can be factored as:
\( = (a + b)^2 (a^2 - ab + b^2)^2 \)
Group the terms together under a single exponent:
\( \implies \Delta = [ (a + b)(a^2 - ab + b^2) ]^2 \)
Using the identity for the sum of cubes \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( \implies \Delta = (a^3 + b^3)^2 = \text{R.H.S.} \)

Question. Show \( \begin{vmatrix} a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix} = 1 + a^2 + b^2 + c^2 \)
Answer: Multiplying row \( R_1 \) by \( a \), row \( R_2 \) by \( b \), and row \( R_3 \) by \( c \), and dividing the determinant by \( abc \) on the outside:
\( \Delta = \frac{1}{abc} \begin{vmatrix} a(a^2 + 1) & a^2b & a^2c \\ ab^2 & b(b^2 + 1) & b^2c \\ c^2a & c^2b & c(c^2 + 1) \end{vmatrix} \)
Next, factoring out \( a \), \( b \), and \( c \) from columns \( C_1 \), \( C_2 \), and \( C_3 \) respectively:
\( \Delta = \frac{abc}{abc} \begin{vmatrix} a^2 + 1 & a^2 & a^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} = \begin{vmatrix} a^2 + 1 & a^2 & a^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \)
Applying the row addition operation \( R_1 \to R_1 + R_2 + R_3 \):
\( \Delta = \begin{vmatrix} 1 + a^2 + b^2 + c^2 & 1 + a^2 + b^2 + c^2 & 1 + a^2 + b^2 + c^2 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \)
Factoring the common term \( (1 + a^2 + b^2 + c^2) \) from the first row \( R_1 \):
\( \Delta = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & 1 & 1 \\ b^2 & b^2 + 1 & b^2 \\ c^2 & c^2 & c^2 + 1 \end{vmatrix} \)
Applying column transformations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \) to simplify:
\( \Delta = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & 0 & 0 \\ b^2 & 1 & 0 \\ c^2 & 0 & 1 \end{vmatrix} \)
Expanding the determinant along the first row \( R_1 \):
\( \Delta = (1 + a^2 + b^2 + c^2) \times [1 \cdot (1 - 0)] \)
\( \implies \Delta = 1 + a^2 + b^2 + c^2 = \text{R.H.S.} \)

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