CBSE Class 12 Maths Exemplar Chapter 4: Determinants NCERT Solutions

We are asked to evaluate the determinant:

\begin{vmatrix}

x^2 - x + 1 & x - 1 \\

x + 1 & x + 1

\end{vmatrix}

To simplify the calculation, we can apply column operations. Let's perform the operation C_1 \rightarrow C_1 - C_2. This means we subtract the elements of the second column from the corresponding elements of the first column.

The determinant becomes:

\begin{vmatrix}

(x^2 - x + 1) - (x - 1) & x - 1 \\

(x + 1) - (x + 1) & x + 1

\end{vmatrix} = \begin{vmatrix}

x^2 - 2x + 2 & x - 1 \\

0 & x + 1

\end{vmatrix}

Now, we can expand this determinant. For a 2x2 determinant \begin{vmatrix} a & b \\ c & d \end{vmatrix}, the value is ad - bc.

Expanding the determinant, we get:

(x^2 - 2x + 2)(x + 1) - (x - 1)(0)

This simplifies to:

(x^2 - 2x + 2)(x + 1)

Let's expand this product:

x(x^2 - 2x + 2) + 1(x^2 - 2x + 2)

= (x^3 - 2x^2 + 2x) + (x^2 - 2x + 2)

= x^3 - 2x^2 + x^2 + 2x - 2x + 2

= x^3 - x^2 + 2

Thus, the value of the determinant is x^3 - x^2 + 2.

We are given the determinant:

\begin{vmatrix}

a+x & y & z \\

x & a+y & z \\

x & y & a+z

\end{vmatrix}

Let's apply row operations to simplify the determinant. First, perform R_1 \rightarrow R_1 - R_2 and R_2 \rightarrow R_2 - R_3.

Applying R_1 \rightarrow R_1 - R_2:

\begin{vmatrix}

(a+x) - x & y - (a+y) & z - z \\

x & a+y & z \\

x & y & a+z

\end{vmatrix} = \begin{vmatrix}

a & -a & 0 \\

x & a+y & z \\

x & y & a+z

\end{vmatrix}

Applying R_2 \rightarrow R_2 - R_3 to the original determinant:

\begin{vmatrix}

a+x & y & z \\

x - x & (a+y) - y & z - (a+z) \\

x & y & a+z

\end{vmatrix} = \begin{vmatrix}

a+x & y & z \\

0 & a & -a \\

x & y & a+z

\end{vmatrix}

Combining these operations, we can perform them simultaneously or sequentially. Let's use the result after R_1 \rightarrow R_1 - R_2 and R_2 \rightarrow R_2 - R_3 on the original determinant:

Original determinant:

\begin{vmatrix}

a+x & y & z \\

x & a+y & z \\

x & y & a+z

\end{vmatrix}

Apply R_1 \rightarrow R_1 - R_2 and R_2 \rightarrow R_2 - R_3:

New R_1: (a+x)-x = a, y-(a+y) = -a, z-z = 0.

New R_2: x-x = 0, (a+y)-y = a, z-(a+z) = -a.

The determinant becomes:

\begin{vmatrix}

a & -a & 0 \\

0 & a & -a \\

x & y & a+z

\end{vmatrix}

Now, let's expand along the first row:

a \begin{vmatrix} a & -a \\ y & a+z \end{vmatrix} - (-a) \begin{vmatrix} 0 & -a \\ x & a+z \end{vmatrix} + 0 \begin{vmatrix} 0 & a \\ x & y \end{vmatrix}

= a [a(a+z) - (-a)y] + a [0(a+z) - (-a)x] + 0

= a [a^2 + az + ay] + a [ax]

= a^3 + a^2z + a^2y + a^2x

= a^2 (a + z + y + x)

= a^2 (a + x + y + z)

Thus, the value of the determinant is a^2(a + x + y + z).

We need to evaluate the determinant:

\begin{vmatrix}

0 & xy^2 & xz^2 \\

x^2y & 0 & yz^2 \\

x^2z & zy^2 & 0

\end{vmatrix}

We can take common factors from the columns. From the first column (C_1), we can take out x. From the second column (C_2), we can take out y. From the third column (C_3), we can take out z.

Let's re-examine the source. The source suggests taking x^2, y^2, and z^2 from C_1, C_2, and C_3 respectively, which seems incorrect based on the structure. Let's proceed by taking common factors from rows instead, or by direct expansion.

Let's try taking common factors from rows. From R_1, we can take x. From R_2, we can take y. From R_3, we can take z.

Determinant = xyz \begin{vmatrix}

0 & y^2 & z^2 \\

x^2 & 0 & z^2 \\

x^2 & y^2 & 0

\end{vmatrix}

Now, let's expand this determinant along the first row:

xyz \left[ 0 \begin{vmatrix} 0 & z^2 \\ y^2 & 0 \end{vmatrix} - y^2 \begin{vmatrix} x^2 & z^2 \\ x^2 & 0 \end{vmatrix} + z^2 \begin{vmatrix} x^2 & 0 \\ x^2 & y^2 \end{vmatrix} \right]

= xyz \left[ 0 - y^2 (x^2 \cdot 0 - z^2 \cdot x^2) + z^2 (x^2 \cdot y^2 - 0 \cdot x^2) \right]

= xyz \left[ -y^2 (-x^2z^2) + z^2 (x^2y^2) \right]

= xyz \left[ x^2y^2z^2 + x^2y^2z^2 \right]

= xyz [2x^2y^2z^2]

= 2x^3y^3z^3

Alternatively, following the source's approach which seems to have a typo in common factor extraction:

Original determinant:

\begin{vmatrix}

0 & xy^2 & xz^2 \\

x^2y & 0 & yz^2 \\

x^2z & zy^2 & 0

\end{vmatrix}

Let's expand directly:

0 \begin{vmatrix} 0 & yz^2 \\ zy^2 & 0 \end{vmatrix} - xy^2 \begin{vmatrix} x^2y & yz^2 \\ x^2z & 0 \end{vmatrix} + xz^2 \begin{vmatrix} x^2y & 0 \\ x^2z & zy^2 \end{vmatrix}

= 0 - xy^2 ( (x^2y)(0) - (yz^2)(x^2z) ) + xz^2 ( (x^2y)(zy^2) - (0)(x^2z) )

= - xy^2 ( -x^2y^2z^2 ) + xz^2 ( x^2yz^3y^2 )

= x^3y^4z^2 + x^3y^3z^3

This does not match the expected result. Let's re-examine the source's claim of taking common factors from columns.

Source claims: x^2y^2z^2 \begin{vmatrix} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{vmatrix}. This extraction is incorrect. Let's assume the source meant taking x from C_1, y from C_2, and z from C_3 after dividing the elements appropriately.

Let's follow the source's final calculation step, assuming the intermediate determinant was correct:

If x^2 y^2 z^2 \begin{vmatrix} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{vmatrix} was the intermediate step (which is derived incorrectly from the original), then expanding the inner determinant:

0 \begin{vmatrix} 0 & y \\ z & 0 \end{vmatrix} - x \begin{vmatrix} y & y \\ z & 0 \end{vmatrix} + x \begin{vmatrix} y & 0 \\ z & z \end{vmatrix}

= 0 - x(y \cdot 0 - y \cdot z) + x(y \cdot z - 0 \cdot z)

= -x(-yz) + x(yz)

= xyz + xyz = 2xyz

Multiplying by the factor x^2y^2z^2 (from the incorrect extraction):

x^2y^2z^2 \cdot 2xyz = 2x^3y^3z^3

The final result 2x^3y^3z^3 is correct, but the intermediate step of extracting common factors from columns as shown in the source is mathematically flawed.

We are asked to evaluate the determinant:

\begin{vmatrix}

3x & -x+y & -x+z \\

x-y & 3y & z-y \\

x-z & y-z & 3z

\end{vmatrix}

Let's apply the column operation C_1 \rightarrow C_1 + C_2 + C_3. This means we add the elements of the second and third columns to the corresponding elements of the first column.

The elements of the first column will become:

Row 1: 3x + (-x+y) + (-x+z) = 3x - x + y - x + z = x + y + z

Row 2: (x-y) + 3y + (z-y) = x - y + 3y + z - y = x + y + z

Row 3: (x-z) + (y-z) + 3z = x - z + y - z + 3z = x + y + z

So, the determinant becomes:

\begin{vmatrix}

x+y+z & -x+y & -x+z \\

x+y+z & 3y & z-y \\

x+y+z & y-z & 3z

\end{vmatrix}

Now, we can take the common factor (x+y+z) from the first column:

(x+y+z) \begin{vmatrix}

1 & -x+y & -x+z \\

1 & 3y & z-y \\

1 & y-z & 3z

\end{vmatrix}

Next, let's apply row operations to simplify further. Perform R_2 \rightarrow R_2 - R_1 and R_3 \rightarrow R_3 - R_1.

New R_2: 1 - 1 = 0, 3y - (-x+y) = 3y + x - y = x + 2y, (z-y) - (-x+z) = z - y + x - z = x - y.

New R_3: 1 - 1 = 0, (y-z) - (-x+y) = y - z + x - y = x - z, 3z - (-x+z) = 3z + x - z = x + 2z.

The determinant becomes:

(x+y+z) \begin{vmatrix}

1 & -x+y & -x+z \\

0 & x+2y & x-y \\

0 & x-z & x+2z

\end{vmatrix}

Now, expand the determinant along the first column:

(x+y+z) \left[ 1 \begin{vmatrix} x+2y & x-y \\ x-z & x+2z \end{vmatrix} - 0 + 0 \right]

= (x+y+z) [(x+2y)(x+2z) - (x-y)(x-z)]

Let's expand the products inside the brackets:

(x+2y)(x+2z) = x^2 + 2xz + 2xy + 4yz

(x-y)(x-z) = x^2 - xz - xy + yz

Subtracting the second from the first:

(x^2 + 2xz + 2xy + 4yz) - (x^2 - xz - xy + yz)

= x^2 + 2xz + 2xy + 4yz - x^2 + xz + xy - yz

= 3xz + 3xy + 3yz

= 3(xy + yz + xz)

So, the value of the determinant is:

(x+y+z) [3(xy + yz + xz)]

= 3(x+y+z)(xy + yz + xz)

Thus, the value of the determinant is 3(x+y+z)(xy + yz + xz).