CBSE Class 12 Mathematics Determinants Important Questions Set 01

Fill in the Blanks

Question. The value of the determinant \( \begin{vmatrix} \sin^2 23^\circ & \sin^2 67^\circ & \cos 180^\circ \\ -\sin^2 67^\circ & -\sin^2 23^\circ & \cos^2 180^\circ \\ \cos 180^\circ & \sin^2 23^\circ & \sin^2 67^\circ \end{vmatrix} \) is _____________ .
Answer: 0

Question. The cofactor of element \( a_{12} \) in the matrix \( \begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix} \) is _____________ .
Answer: 46

Question. If \( A \) is a skew-symmetric matrix of order 3, then the value of \( |A| = \) _____________ .
Answer: 0

Question. If \( \cos 2\theta = 0 \), then \( \begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2 = \) _____________ .
Answer: \( \frac{1}{2} \)

Question. If \( x = -9 \) is a root of \( \begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix} = 0 \), then other two roots are _____________ .
Answer: \( x = 2, x = 7 \)

Very Short Answer Questions

Question. If \( A \) and \( B \) are square matrices of the same order 3, such that \( |A| = 2 \) and \( AB = 2I \), write the value of \( |B| \). 
Answer: \( |A| = 2 \) and \( AB = 2I \)
\( \Rightarrow |AB| = |2I| = 8 \)
\( \Rightarrow |A| |B| = 8 \Rightarrow 2 |B| = 8 \Rightarrow |B| = 4 \)

Question. Find \( |AB| \), if \( A = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix} \). 
Answer: We have, \( AB = \begin{bmatrix} 0 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 5 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
\( \therefore |AB| = \begin{vmatrix} 0 & 0 \\ 0 & 0 \end{vmatrix} = 0 \)

Question. Let \( A \) be a square matric of order 3 × 3. Write the value of \( |2A| \), where \( |A| = 4 \). 
Answer: \(\because |kA| = k^n |A|\), where \( n \) is order of matrix \( A \).
Here \( |A| = 4 \) and \( n = 3 \)
\( |2A| = 2^3 \times 4 = 32 \)

Question. If \( \begin{vmatrix} x + 1 & x - 1 \\ x - 3 & x + 2 \end{vmatrix} = \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} \), then write the value of \( x \).
Answer: Given \( \begin{vmatrix} x + 1 & x - 1 \\ x - 3 & x + 2 \end{vmatrix} = \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} \)
\( \Rightarrow (x + 1) (x + 2) - (x - 1) (x - 3) = 12 + 1 \)
\( \Rightarrow x^2 + 2x + x + 2 - (x^2 - 3x - x + 3) = 13 \)
\( \Rightarrow 7x - 1 = 13 \)
\( \Rightarrow 7x = 14 \)
\( \Rightarrow x = 2 \)

Question. If \( A = [a_{ij}] \) is a matrix of order 2 × 2 , such that \( |A| = -15 \) and \( C_{ij} \) represents the cofactor of \( a_{ij} \), then find \( a_{21}C_{21} + a_{22}C_{22} \).  
Answer: Given, \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \therefore |A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} \)
Expanding along \( R_2 \)
\( \Rightarrow -15 = a_{21} \cdot C_{21} + a_{22} \cdot C_{22} \) [\(C_{ij} = \text{Cofactor of } a_{ij}\)]
\( \Rightarrow a_{21}C_{21} + a_{22}C_{22} = -15 \)

Question. Write the value of the following determinant: \( \begin{vmatrix} a - b & b - c & c - a \\ b - c & c - a & a - b \\ c - a & a - b & b - c \end{vmatrix} \)
Answer: Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get
\( \begin{vmatrix} 0 & b - c & c - a \\ 0 & c - a & a - b \\ 0 & a - b & b - c \end{vmatrix} = 0 \) [\( \because \) All elements of \( C_1 \) are zero]

Question. Show that the points (1, 0), (6, 0), (0, 0) are collinear. [CBSE (AI) 2008]
Answer: Since \( \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0 \)
Hence, (1, 0), (6, 0) and (0, 0) are collinear.

Question. What positive value of \( x \) makes the following pair of determinants equal? \( \begin{vmatrix} 2x & 3 \\ 5 & x \end{vmatrix} \), \( \begin{vmatrix} 16 & 3 \\ 5 & 2 \end{vmatrix} \)   
Answer: \(\because \begin{vmatrix} 2x & 3 \\ 5 & x \end{vmatrix} = \begin{vmatrix} 16 & 3 \\ 5 & 2 \end{vmatrix}\)
\( \Rightarrow 2x^2 - 15 = 32 - 15 \Rightarrow 2x^2 = 32 \)
\( \Rightarrow x^2 = 16 \Rightarrow x = \pm 4 \)
\( \Rightarrow x = 4 \) (+ve value).

Question. Evaluate: \( \begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} \) 
Answer: Expanding the determinant, we get
\( \cos 15^\circ \cdot \cos 75^\circ - \sin 15^\circ \cdot \sin 75^\circ = \cos (15^\circ + 75^\circ) = \cos 90^\circ = 0 \)
[Note : \( \cos (A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B \)]

Question. Write the value of the following determinant: \( \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \) 
Answer: Let \( \Delta = \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 - 6R_3 \), we get
\( \Delta = \begin{vmatrix} 0 & 0 & 0 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} = 0 \) [\(\because\) Each element of \( R_1 \) is zero]

Question. If \( A \) is a square matrix and \( |A| = 2 \), then write the value of \( |AA'| \), where \( A' \) is the transpose of matrix \( A \).
Answer: \( |AA'| = |A| \cdot |A'| = |A| \cdot |A| = |A|^2 = 2^2 = 4 \)
[Note: \( |AB| = |A| \cdot |B| \) and \( |A| = |A^T| \), where \( A \) and \( B \) are square matrices.]

Question. If \( A \) is a 3 × 3 invertible matrix, then what will be the value of \( k \) if \( \text{det } (A^{-1}) = (\text{det } A)^k \). 
Answer: Given, \( \text{det } (A^{-1}) = (\text{det } A)^k \)
\( \Rightarrow |A^{-1}| = |A|^k \Rightarrow k = -1 \)

Short Answer Questions-I and II

Question. Find the equations of line joining (1,2) and (3,6) using determinants.
Answer: Let given points are \( A(1, 2) \) and \( B(3, 6) \) and \( P(x, y) \) lies on the line joining points \( A \) and \( B \).
\( \therefore \) Points \( A, P \) and \( B \) are collinear
\( \therefore \) Area of \( \Delta APB = 0 \)
\( \Rightarrow \frac{1}{2} \begin{vmatrix} 1 & 2 & 1 \\ x & y & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} 1 & 2 & 1 \\ x & y & 1 \\ 3 & 6 & 1 \end{vmatrix} = 0 \)
\( \Rightarrow 1(y - 6) - 2(x - 3) + 1(6x - 3y) = 0 \)
\( \Rightarrow y - 6 - 2x + 6 + 6x - 3y = 0 \Rightarrow 4x - 2y = 0 \)
\( \Rightarrow 2(2x - y) = 0 \)
\( \Rightarrow 2x - y = 0 \)
\( \therefore \) Equation of line be \( 2x - y = 0 \)

Question. Evaluate the determinant: \( \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix} \) 
Answer: Let \( \Delta = \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix} \)
\( = (x + 1) \begin{vmatrix} x^2 - x + 1 & x - 1 \\ 1 & 1 \end{vmatrix} \) [Taking common \( (x + 1) \) from \( R_2 \)]
\( = (x + 1) \{x^2 - x + 1 - (x - 1)\} = (x + 1) (x^2 - 2x + 2) \)
\( = x^3 - 2x^2 + 2x + x^2 - 2x + 2 = x^3 - x^2 + 2 \)

Question. What is the value of the following determinant: \( \begin{vmatrix} 4 & a & b + c \\ 4 & b & c + a \\ 4 & c & a + b \end{vmatrix} \) 
Answer: Here, \( \Delta = \begin{vmatrix} 4 & a & b + c \\ 4 & b & c + a \\ 4 & c & a + b \end{vmatrix} = \begin{vmatrix} 4 & a & a + b + c \\ 4 & b & a + b + c \\ 4 & c & a + b + c \end{vmatrix} \) [Applying \( C_3 \rightarrow C_3 + C_2 \)]
\( \Delta = 4(a + b + c) \begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix} \) [Taking out common 4 from \( C_1 \) and \( a + b + c \) from \( C_3 \)]
\( \Delta = 4(a + b + c) \cdot 0 = 0 \) [\(\because C_1 = C_3\)]

Question. Write the value of \( \Delta = \begin{vmatrix} x + y & y + z & z + x \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix} \). 
Answer: Here, \( \Delta = \begin{vmatrix} x + y & y + z & z + x \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 + R_2 \), we get
\( = \begin{vmatrix} x + y + z & x + y + z & x + y + z \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix} \)
Taking \( (x + y + z) \) common from \( R_1 \), we get
\( = (x + y + z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix} \)
Applying \( R_3 \rightarrow R_3 + 3R_1 \), we get
\( = (x + y + z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 0 & 0 & 0 \end{vmatrix} = 0 \) [\(\because\) Each element of \( R_3 \) is zero]

Question. Without expanding evaluate the determinant: \( \begin{vmatrix} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (a^y + a^{-y})^2 & (a^y - a^{-y})^2 & 1 \\ (a^z + a^{-z})^2 & (a^z - a^{-z})^2 & 1 \end{vmatrix} \), where \( a > 0 \) and \( x, y, z \in R \). 
Answer: Here \( \Delta = \begin{vmatrix} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (a^y + a^{-y})^2 & (a^y - a^{-y})^2 & 1 \\ (a^z + a^{-z})^2 & (a^z - a^{-z})^2 & 1 \end{vmatrix} \)
Applying \( C_1 \rightarrow C_1 - C_2 \), we get
\( \Delta = \begin{vmatrix} 4 & (a^x - a^{-x})^2 & 1 \\ 4 & (a^y - a^{-y})^2 & 1 \\ 4 & (a^z - a^{-z})^2 & 1 \end{vmatrix} \) [Using \( (a + b)^2 - (a - b)^2 = 4ab \)]
Taking out 4 from \( C_1 \), we get
\( \Delta = 4 \begin{vmatrix} 1 & (a^x - a^{-x})^2 & 1 \\ 1 & (a^y - a^{-y})^2 & 1 \\ 1 & (a^z - a^{-z})^2 & 1 \end{vmatrix} \Rightarrow \Delta = 4 \times 0 = 0 \). [\(\because C_1 \text{ and } C_3 \text{ are identical}\)]

Long Answer Questions

Question. If \( a, b, c \) are \( p \)th, \( q \)th and \( r \)th terms respectively of a G.P, then prove that \( \begin{vmatrix} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{vmatrix} = 0 \). 
Answer: Let \( A \) be the first term and \( R \) be the common ratio of the G.P respectively.
\( \therefore a = AR^{p-1}, b = AR^{q-1}, c = AR^{r-1} \)
Now, we have,
L.H.S. \( \Delta = \begin{vmatrix} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{vmatrix} = \begin{vmatrix} \log AR^{p-1} & p & 1 \\ \log AR^{q-1} & q & 1 \\ \log AR^{r-1} & r & 1 \end{vmatrix} \)
\( \Rightarrow \Delta = \begin{vmatrix} \log A + (p - 1) \log R & p & 1 \\ \log A + (q - 1) \log R & q & 1 \\ \log A + (r - 1) \log R & r & 1 \end{vmatrix} \)
[\(\because \log(ab) = \log a + \log b \text{ and } \log a^m = m \log a\)]
\( \Rightarrow \Delta = \begin{vmatrix} \log A & p & 1 \\ \log A & q & 1 \\ \log A & r & 1 \end{vmatrix} + \log R \begin{vmatrix} p - 1 & p & 1 \\ q - 1 & q & 1 \\ r - 1 & r & 1 \end{vmatrix} \)
Apply \( C_2 \rightarrow C_2 - C_3 \)
\( \Rightarrow \Delta = \log A \begin{vmatrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{vmatrix} + \log R \begin{vmatrix} p - 1 & p - 1 & 1 \\ q - 1 & q - 1 & 1 \\ r - 1 & r - 1 & 1 \end{vmatrix} \)
\( (C_1 = C_3) \text{ and } (C_2 = C_3) \)
\( \Delta = \log A \times 0 + \log R \times 0 \)
\( \therefore \Delta = 0 + 0 = R.H.S \)
\( \therefore \Delta = 0 \)

Question. Using properties of determinants, find the value of \( x \) for which \[ \begin{vmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} = 0. \] 
Answer: We have, \[ \begin{vmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} = 0 \] Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get \[ \begin{vmatrix} 12+x & 4+x & 4+x \\ 12+x & 4-x & 4+x \\ 12+x & 4+x & 4-x \end{vmatrix} = 0 \] \( \Rightarrow (12+x) \begin{vmatrix} 1 & 4+x & 4+x \\ 1 & 4-x & 4+x \\ 1 & 4+x & 4-x \end{vmatrix} = 0 \) [ Taking \( (12+x) \) common from \( C_1 \) ] Applying \( R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1 \), we get \[ (12+x) \begin{vmatrix} 1 & 4+x & 4+x \\ 0 & -2x & 0 \\ 0 & 0 & -2x \end{vmatrix} = 0 \] \( \Rightarrow (x+12)(4x^2) = 0 \Rightarrow x = 0, -12 \)

Question. Using properties of determinants, prove that \[ \begin{vmatrix} 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{vmatrix} = (x+y+z)^3. \] 
Answer: LHS = \[ \begin{vmatrix} 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{vmatrix} \] Applying \( R_2 \leftrightarrow R_3 \), then \( R_1 \leftrightarrow R_2 \), we get \[ = \begin{vmatrix} x-y-z & 2x & 2x \\ 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \end{vmatrix} \] Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \), we get \[ = \begin{vmatrix} x+y+z & y+z+x & z+x+y \\ 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \end{vmatrix} \] Taking out \( (x+y+z) \) from first row, we get \[ = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \end{vmatrix} \] Applying \( C_1 \rightarrow C_1 - C_3 \) and \( C_2 \rightarrow C_2 - C_3 \), we get \[ = (x+y+z) \begin{vmatrix} 0 & 0 & 1 \\ 0 & -(y+z+x) & 2y \\ (x+y+z) & x+y+z & z-x-y \end{vmatrix} \] Expanding along first row, we get \( = (x+y+z) \{ (x+y+z)^2 \} = (x+y+z)^3 = \text{RHS} \)

Question. If \( x, y, z \) are different and \( \Delta = \begin{vmatrix} x & x^2 & x^3-1 \\ y & y^2 & y^3-1 \\ z & z^2 & z^3-1 \end{vmatrix} = 0 \), then using properties of determinants, show that \( xyz = 1 \). 
Answer: We have \[ \begin{vmatrix} x & x^2 & x^3-1 \\ y & y^2 & y^3-1 \\ z & z^2 & z^3-1 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} + \begin{vmatrix} x & x^2 & -1 \\ y & y^2 & -1 \\ z & z^2 & -1 \end{vmatrix} = 0 \] \( \Rightarrow xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} + (-1) \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} = 0 \) [In det 1 taking \( x, y, z \) common from each row and in det 2 using \( C_1 \leftrightarrow C_3 \) and applying \( C_2 \leftrightarrow C_3 \) in det 2] \( \Rightarrow xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} - \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0 \) \( \Rightarrow \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} (xyz - 1) = 0 \) If \( \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} 1 & x & x^2 \\ 0 & y-x & y^2-x^2 \\ 0 & z-x & z^2-x^2 \end{vmatrix} = 0 \) [Applying \( R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1 \)] \( \Rightarrow \begin{vmatrix} y-x & y^2-x^2 \\ z-x & z^2-x^2 \end{vmatrix} = 0 \Rightarrow (y-x)(z-x) \begin{vmatrix} 1 & y+x \\ 1 & z+x \end{vmatrix} = 0 \) \( \Rightarrow (y-x)(z-x)(z+x-y-x) = 0 \) \( \Rightarrow (y-x)(z-x)(z-y) = 0 \) \( \Rightarrow x = y \) or \( z = x \) or \( y = z \), which is a contradiction. Hence, \( (xyz - 1) = 0 \Rightarrow xyz = 1 \).

Question. Prove 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: LHS = \[ \begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} \] Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get \[ = \begin{vmatrix} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2a & b \\ 2(a+b+c) & a & c+a+2b \end{vmatrix} \] Taking \( 2(a+b+c) \) common from \( C_1 \), we get \[ = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix} \] Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \), we get \[ = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & a+b+c & 0 \\ 0 & 0 & a+b+c \end{vmatrix} \] Taking \( (a+b+c) \) common from \( R_2 \) and \( R_3 \), we get \[ = 2(a+b+c)^3 \begin{vmatrix} 1 & a & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \] Expanding along \( C_1 \), we get \( = 2(a+b+c)^3 [1-0] = 2(a+b+c)^3 = \text{RHS} \)

Question. If \( \Delta = \begin{vmatrix} 1 & a & a^2 \\ a & a^2 & 1 \\ a^2 & 1 & a \end{vmatrix} = -4 \) then find the value of \[ \begin{vmatrix} a^3-1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix} \]. 
Answer: We have \( \Delta = \begin{vmatrix} 1 & a & a^2 \\ a & a^2 & 1 \\ a^2 & 1 & a \end{vmatrix} \). \( C_{11} = a^3 - 1; \quad C_{12} = 0; \quad C_{13} = a - a^4 \) \( C_{21} = 0; \quad C_{22} = a - a^4; \quad C_{23} = a^3 - 1 \) \( C_{31} = a - a^4; \quad C_{32} = a^3 - 1; \quad C_{33} = 0 \) Where \( C_{ij} = \) Co-factor of \( a_{ij} \)(\( i, j \))th element of determinant \( \Delta \). Let \( \Delta_1 \) be the determinant made by corresponding co-factor of each element of determinant \( \Delta \). i.e., \( \Delta_1 = \begin{vmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{vmatrix} \) We know that \( \Delta_1 = \Delta^{n-1} \) where \( n \) is order of the determinant. \( \therefore \Delta_1 = (-4)^2 = 16 \) \( \Rightarrow \begin{vmatrix} a^3-1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix} = 16 \)

Question. Using property of determinant, prove the following: \[ \begin{vmatrix} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} = 9b^2(a+b) \]  
Answer: LHS = \[ \begin{vmatrix} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} \] Applying \( R_1 = R_1 + R_2 + R_3 \), we get \[ = \begin{vmatrix} 3(a+b) & 3(a+b) & 3(a+b) \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} \] Taking \( 3(a+b) \) common from \( R_1 \), we get \[ = 3(a+b) \begin{vmatrix} 1 & 1 & 1 \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} \] Applying \( C_1 \rightarrow C_1 - C_3, C_2 \rightarrow C_2 - C_3 \), we get \[ = 3(a+b) \begin{vmatrix} 0 & 0 & 1 \\ b & -b & a+b \\ b & 2b & a \end{vmatrix} \] Expanding along \( R_1 \), we get \( = 3(a+b) \{ 1(2b^2 + b^2) \} = 9b^2(a+b) = \text{RHS} \)

Question. Using properties of determinants, find the value of \( k \) if \[ \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix} = k(x^3+y^3). \] 
Answer: We have \( k(x^3+y^3) = \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix} \) Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get \[ = \begin{vmatrix} 2x+2y & y & x+y \\ 2x+2y & x+y & x \\ 2x+2y & x & y \end{vmatrix} \] \( \Rightarrow (2x+2y) \begin{vmatrix} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{vmatrix} \) [Taking \( (2x+2y) \) common from \( C_1 \)] Applying \( R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1 \), we get \[ = 2(x+y) \begin{vmatrix} 1 & y & x+y \\ 0 & x & -y \\ 0 & x-y & -x \end{vmatrix} \] \( = 2(x+y) \begin{vmatrix} x & -y \\ x-y & -x \end{vmatrix} = 2(x+y)(-x^2 + xy - y^2) = -2(x+y)(x^2 - xy + y^2) \) \( \Rightarrow k(x^3+y^3) = -2(x^3+y^3) \) Comparing the coefficient of \( (x^3+y^3) \) on both the sides, we get \( k = -2 \)

Question. Using properties of determinants show that \[ \begin{vmatrix} 1 & 1 & 1+x \\ 1 & 1+y & 1 \\ 1+z & 1 & 1 \end{vmatrix} = -(xyz + yz + zx + xy) \] 
Answer: LHS = \[ \begin{vmatrix} 1 & 1 & 1+x \\ 1 & 1+y & 1 \\ 1+z & 1 & 1 \end{vmatrix} \] Apply \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \), we get \[ = \begin{vmatrix} 1 & 1 & 1+x \\ 0 & y & -x \\ z & 0 & -x \end{vmatrix} \] Expanding by \( R_1 \), we get \( = 1[-yx - 0] - 1[0 - (-zx)] + (1+x)(0 - zy) = -yx - zx - zy - xyz \) \( = -(xy + xz + zy + xyz) = \text{RHS} \)

Question. Using properties of determinant, prove that: \[ \begin{vmatrix} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{vmatrix} = a^2(a+x+y+z) \] 
Answer: LHS = \[ \begin{vmatrix} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{vmatrix} \] Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get \[ = \begin{vmatrix} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{vmatrix} \] \( = (a+x+y+z) \begin{vmatrix} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{vmatrix} \) [Taking out \( (a+x+y+z) \) common from \( C_1 \)] Apply \( R_1 \rightarrow R_1 - R_2 \), we get \[ = (a+x+y+z) \begin{vmatrix} a & -a & 0 \\ 1 & a+y & z \\ 1 & y & a+z \end{vmatrix} \] Expanding along \( R_1 \), we get \( = (a+x+y+z) \{ a(a+z-z) + a(a+z-z) \} \) \( = a^2(a+x+y+z) = \text{RHS} \)

Question. Using properties of determinant, prove the following: \[ \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} = x^3 \] 
Answer: LHS = \[ \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} \] \( = x^2 \begin{vmatrix} x+y & 1 & 1 \\ 5x+4y & 4 & 2 \\ 10x+8y & 8 & 3 \end{vmatrix} \) [Taking out \( x \) from \( C_2 \) and \( C_3 \)] Applying \( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 3R_1 \), we get \[ = x^2 \begin{vmatrix} x+y & 1 & 1 \\ 3x+2y & 2 & 0 \\ 7x+5y & 5 & 0 \end{vmatrix} \] Expanding along \( C_3 \), we get \( = x^2 [1 \{ (3x+2y)5 - 2(7x+5y) \} - 0 + 0] = x^2 (15x + 10y - 14x - 10y) \) \( = x^2 (x) = x^3 = \text{RHS} \)

Question. Without expanding, show that: \[ \begin{vmatrix} \text{cosec}^2\theta & \cot^2\theta & 1 \\ \cot^2\theta & \text{cosec}^2\theta & -1 \\ 42 & 40 & 2 \end{vmatrix} = 0 \] 
Answer: Given, \( \Delta = \begin{vmatrix} \text{cosec}^2\theta & \cot^2\theta & 1 \\ \cot^2\theta & \text{cosec}^2\theta & -1 \\ 42 & 40 & 2 \end{vmatrix} \) Applying \( C_1 \rightarrow C_1 - C_2 - C_3 \), we get \[ = \begin{vmatrix} \text{cosec}^2\theta - \cot^2\theta - 1 & \cot^2\theta & 1 \\ \cot^2\theta - \text{cosec}^2\theta + 1 & \text{cosec}^2\theta & -1 \\ 0 & 40 & 2 \end{vmatrix} \] \( = \begin{vmatrix} 1-1 & \cot^2\theta & 1 \\ -1+1 & \text{cosec}^2\theta & -1 \\ 0 & 40 & 2 \end{vmatrix} \) [\(\because \text{cosec}^2\theta - \cot^2\theta = 1\)] \( = \begin{vmatrix} 0 & \cot^2\theta & 1 \\ 0 & \text{cosec}^2\theta & -1 \\ 0 & 40 & 2 \end{vmatrix} = 0 \) [\(\because\) All elements of \( C_1 \) are 0]

Question. Prove the following using properties of determinant: \[ \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 2(3abc - a^3 - b^3 - c^3) \] 
Answer: LHS = \[ \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \] Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \), we get \[ = \begin{vmatrix} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \] Taking \( 2(a+b+c) \) common from \( R_1 \), we get \[ = 2(a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \] Applying \( C_2 \rightarrow C_2 - C_1; C_3 \rightarrow C_3 - C_1 \), we get \[ = 2(a+b+c) \begin{vmatrix} c+a & b-c & b-a \\ a+b & c-a & c-b \end{vmatrix} \] \( = 2(a+b+c) [1(bc - b^2 - c^2 + bc - bc + ac + ab - a^2)] \) [Expanding along \( R_1 \)] \( = 2(a+b+c) (bc + ac + ab - a^2 - b^2 - c^2) \) \( = -2(a+b+c) (a^2 + b^2 + c^2 - ab - bc - ca) = -2(a^3 + b^3 + c^3 - 3abc) \) \( = 2(3abc - a^3 - b^3 - c^3) = \text{RHS} \)

Question. If \( a + b + c \neq 0 \) and \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \), then using properties of determinants, prove that \( a = b = c \).
Answer: We have \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \) Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get \[ \begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{vmatrix} = 0 \] Taking \( (a+b+c) \) common from \( C_1 \), we get \( (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0 \) [\(\because a + b + c \neq 0\)] Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \), we get \[ \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} = 0 \] Expanding along \( C_1 \), we get \( 1\{(c-b)(b-c) - (a-c)(a-b)\} - 0 + 0 = 0 \Rightarrow -(b-c)^2 - (a-c)(a-b) = 0 \) \( \Rightarrow -b^2 - c^2 + 2bc - a^2 + ab + ac - bc = 0 \Rightarrow a^2 + b^2 + c^2 - bc - ab - ac = 0 \) \( \Rightarrow \frac{1}{2} [2a^2 + 2b^2 + 2c^2 - 2bc - 2ab - 2ac] = 0 \Rightarrow (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 \) \( \Rightarrow (a-b)^2 = 0; (b-c)^2 = 0; (c-a)^2 = 0 \Rightarrow a-b = 0; b-c = 0; c-a = 0; \) \( \Rightarrow a = b = c \)

Question. Using properties of determinants, show that \(\triangle ABC\) is an isosceles if: \[ \begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^2 A + \cos A & \cos^2 B + \cos B & \cos^2 C + \cos C \end{vmatrix} = 0 \] 
Answer: We have \[ \begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^2 A + \cos A & \cos^2 B + \cos B & \cos^2 C + \cos C \end{vmatrix} = 0 \] Applying \( C_1 \rightarrow C_1 - C_3 \) and \( C_2 \rightarrow C_2 - C_3 \), we get \[ \begin{vmatrix} 0 & 0 & 1 \\ \cos A - \cos C & \cos B - \cos C & 1 + \cos C \\ \cos^2 A + \cos A - \cos^2 C - \cos C & \cos^2 B + \cos B - \cos^2 C - \cos C & \cos^2 C + \cos C \end{vmatrix} = 0 \] \( \Rightarrow \begin{vmatrix} 0 & 0 & 1 \\ \cos A - \cos C & \cos B - \cos C & 1 + \cos C \\ (\cos A - \cos C)(\cos A + \cos C + 1) & (\cos B - \cos C)(\cos B + \cos C + 1) & \cos^2 C + \cos C \end{vmatrix} = 0 \) Taking common \( (\cos A - \cos C) \) from \( C_1 \) and \( (\cos B - \cos C) \) from \( C_2 \), we get \( (\cos A - \cos C)(\cos B - \cos C) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & 1 + \cos C \\ \cos A + \cos C + 1 & \cos B + \cos C + 1 & \cos^2 C + \cos C \end{vmatrix} = 0 \) Applying \( C_1 \rightarrow C_1 - C_2 \), we get \( \Rightarrow (\cos A - \cos C)(\cos B - \cos C) \begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 + \cos C \\ \cos A - \cos B & \cos B + \cos C + 1 & \cos^2 C + \cos C \end{vmatrix} = 0 \) Expanding along \( R_1 \), we get \( \Rightarrow (\cos A - \cos C) (\cos B - \cos C) (\cos B - \cos A) = 0 \) \( \Rightarrow \cos A - \cos C = 0 \text{ i.e., } \cos A = \cos C \)