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May 3, 201939 Insightful Publications

**Differentiation of determinants:** We know that the equations in two variables can be solved quickly. But if the number of variables is three or more, our task becomes lengthier because we must first eliminate the unknown. It is usually done using two equations and then solving for the same unknown using the other two equations. This continues till we get the solution.

Imagine how tedious it will be for an increased number of unknowns. Instead, we can use determinants to solve for unknowns. Determinants have several applications, such as checking the matrix is invertible, finding the equation of a line and checking the collinearity of three points.

Corresponding to every square matrix \(A,\) there exists a number called the determinant of the matrix. The determinant is defined as a function \(f:M→D,\) where \(M\) is the set of all square matrices and \(D\) is a set of values of the determinant of that matrix.

If \(f(A) = \lambda \) then \(A∈M\) and \(λ∈D\) then we can write \(A=λ.\)

As for two different square matrices, the same value of their determinants is possible, so the function \(f\) is many-one.

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}& \cdots &{{a_{1n}}} \\ \vdots & \ddots & \vdots \\ {{a_{n1}}}& \cdots &{{a_{nn}}} \end{array}} \right]\) is a square matrix of order n, then

det \(A\) or \(|A| = \left| {\begin{array}{*{20}{c}} {{a_{11}}}& \cdots &{{a_{1n}}} \\ \vdots & \ddots & \vdots \\ {{a_{n1}}}& \cdots &{{a_{nn}}} \end{array}} \right|\)

**Note: **The determinant of a square matrix is a scalar quantity, whereas a matrix is an arrangement of numbers and has no value.

- The value of a determinant remains the same if the rows and columns are interchanged.
- If all the elements in a row or a column are zero, then the value of the determinant will be zero.
- If any two rows or columns of the determinant are interchanged, the sign of the determinant is changed, while the numerical value remains unaltered.
- If two rows or columns of the determinant are identical or proportional, the value of the determinant is zero.
- If all the elements of a row or column are multiplied by a non-zero constant \(\lambda ,\) then the value of the resulting determinant is \(\lambda \) times the original determinant.
- If each element in any row or column can be expressed as the sum of two other terms, the determinant can be expressed as the sum of two determinants.
**Example:**

\(\left| {\begin{array}{*{20}{c}} {{a_1} + {k_1}}&{{a_2} + {k_2}}&{{a_3} + {k_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {{k_1}}&{{k_2}}&{{k_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right|\) - If a determinant is diagonal, upper triangular, or lower triangular, the value of the determinant is the product of the diagonal elements.
**Addition or difference of determinants**: The sum of the determinant of each of two (or more) matrices generally does not equal the determinant of the sum.**Proof**:

Let \(A = \left[ {\begin{array}{*{20}{l}} {{a_{11}}}&{{a_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{l}} {{b_{11}}}&{{b_{12}}} \\ {{b_{21}}}&{{b_{22}}} \end{array}} \right]\)

Then, \(|A| + |B| = \left| {\begin{array}{*{20}{l}} {{a_{11}}}&{{a_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right| + \left| {\begin{array}{*{20}{l}} {{b_{11}}}&{{b_{12}}} \\ {{b_{21}}}&{{b_{22}}} \end{array}} \right|\)

\( = \left( {{a_{11}}{a_{22}} – {a_{12}}{a_{21}}} \right) + \left( {{b_{11}}{b_{22}} – {b_{12}}{b_{21}}} \right)\)

and

\(|A + B| = \left| {\left[ {\begin{array}{*{20}{l}} {{a_{11}}}&{{a_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right] + \left[ {\begin{array}{*{20}{l}} {{b_{11}}}&{{b_{12}}} \\ {{b_{21}}}&{{b_{22}}} \end{array}} \right]} \right|\)

\( = \left| {\begin{array}{*{20}{l}} {{a_{11}} + {b_{11}}}&{{a_{12}} + {b_{12}}} \\ {{a_{21}} + {b_{21}}}&{{a_{22}} + {b_{22}}} \end{array}} \right|\)

\( = \left| {\begin{array}{*{20}{l}} {{a_{11}}}&{{a_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right| + \left| {\begin{array}{*{20}{l}} {{a_{11}}}&{{a_{12}}} \\ {{b_{21}}}&{{b_{22}}} \end{array}} \right| + \left| {\begin{array}{*{20}{l}} {{b_{11}}}&{{b_{12}}} \\ {{a_{21}}}&{{a_{22}}} \end{array}} \right| + \left| {\begin{array}{*{20}{l}} {{b_{11}}}&{{b_{12}}} \\ {{b_{21}}}&{{b_{22}}} \end{array}} \right|\)

\(\therefore |A + B| \ne |A| + |B|\)

The same can be applied for difference of two matrices also.

\( \Rightarrow |A| – |B| \ne |A – B|\)

**Practice 10th CBSE Exam Questions**

If \(∆(x)\) is a determinant whose elements are a function of \(x,\) we can find the derivative of the determinant using two methods.

- First, expand the determinant \(∆(x)\) with the help of properties. Then, differentiate both sides of the equation with respect to \(x\) and find \(\frac{{d\Delta (x)}}{{dx}}.\)
- \(\frac{{d\Delta (x)}}{{dx}}\) is the sum of the n determinant obtained by differentiating the elements of one row or column of the determinant \(∆(x),\) while leaving the other \((n−1)\) rows or columns unaltered.

**Proof:**

Let \(\Delta (x) = \left| {\begin{array}{*{20}{l}} {{f_1}(x)}&{{f_2}(x)} \\ {{g_1}(x)}&{{g_2}(x)} \end{array}} \right|\) of order \(2,\) which is a function of \(x.\)

Then,

\(\frac{{d\Delta (x)}}{{dx}} = \frac{d}{{dx}}\left| {\begin{array}{*{20}{l}} {{f_1}(x)}&{{f_2}(x)} \\ {{g_1}(x)}&{{g_2}(x)} \end{array}} \right|\)

\(\frac{{d\Delta (x)}}{{dx}} = \frac{d}{{dx}}\left( {{f_1}(x){g_2}(x) – {g_1}(x){f_2}(x)} \right)\)

\(\frac{{d\Delta (x)}}{{dx}} = {f_1}(x)g_2^\prime (x) + {g_2}(x)f_1^\prime (x) – {g_1}(x)f_2^\prime (x) – g_1^\prime (x){f_2}(x)\)

\(\therefore \frac{{d\Delta (x)}}{{dx}} = \left| {\begin{array}{*{20}{l}} {f_1^\prime (x)}&{f_2^\prime (x)} \\ {{g_1}(x)}&{{g_2}(x)} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {{f_1}(x)}&{{f_2}(x)} \\ {g_1^\prime (x)}&{g_2^\prime (x)} \end{array}} \right|\)

Similarly, we can write for the determinant of order \(3\)

if \(\Delta (x) = \left| {\begin{array}{*{20}{l}} {{f_1}(x)}&{{f_2}(x)}&{{f_3}(x)} \\ {{g_1}(x)}&{{g_2}(x)}&{{g_3}(x)} \\ {{h_1}(x)}&{{h_2}(x)}&{{h_3}(x)} \end{array}} \right|\) of order \(3,\) which is a function of \(x,\)

Then, \(\frac{{d\Delta (x)}}{{dx}} = \frac{d}{{dx}}\left| {\begin{array}{*{20}{l}} {{f_1}(x)}&{{f_2}(x)}&{{f_3}(x)} \\ {{g_1}(x)}&{{g_2}(x)}&{{g_3}(x)} \\ {{h_1}(x)}&{{h_2}(x)}&{{h_3}(x)} \end{array}} \right|\)

\( \Rightarrow \frac{{d\Delta (x)}}{{dx}} = \left| {\begin{array}{*{20}{l}} {f_1^\prime (x)}&{f_2^\prime (x)}&{{f^\prime }_3(x)} \\ {{g_1}(x)}&{{g_2}(x)}&{{g_3}(x)} \\ {{h_1}(x)}&{{h_2}(x)}&{{h_3}(x)} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {{f_1}(x)}&{{f_2}(x)}&{{f_3}(x)} \\ {g_1^\prime (x)}&{g_2^\prime (x)}&{g_3^\prime (x)} \\ {{h_1}(x)}&{{h_2}(x)}&{{h_3}(x)} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {{f_1}(x)}&{{f_2}(x)}&{{f_3}(x)} \\ {{g_1}(x)}&{{g_2}(x)}&{{g_3}(x)} \\ {h_1^\prime (x)}&{h_2^\prime (x)}&{h_3^\prime (x)} \end{array}} \right|\)

We can also apply the same process to the columns in the determinants. To differentiate a determinant, we differentiate one row or column at a time, keeping the other rows and columns intact.

**Note**: To find out the coefficient of \({x^r}\) in any polynomial \(f(x),\) differentiate the given polynomial \(f(x),\) \(r\) times successively and then substitute \(x=0.\) i.e. the coefficient of \({x^r} = \left[ {\frac{{{f^r}(0)}}{{r!}}} \right],\) where \({f^r}(0) = \left( {\frac{{{d^r}f(x)}}{{d{x^r}}}} \right)\) at \(x=0.\)

If \(∆(x)\) is a determinant whose all or some elements are functions of \(x,\) then the integration of that determinant can be done by two methods.

- If there are elements that are functions of \(?\) in more than one row or one column, then to integrate the determinant, first, we expand the determinant with the help of properties. The result is now integrated with respect to \(x\) between applicable limits.
- If there are elements that are functions of \(x,\) only in one row or one column and elements of other rows or columns are constant, then to integrate the determinant, we have to integrate only that row or column whose elements are functions of \(x\) while other rows or columns remains unaffected.

**Example:** Given: \(\Delta (x) = \left| {\begin{array}{*{20}{c}}
{f(x)}&{g(x)}&{h(x)} \\
a&b&c \\
l&m&n
\end{array}} \right|\)

(Where \(a, b, c, l, m,\) and \(n\) are constants) as a function of \(x.\)

So, \(\int_a^b \Delta (x)dx = \left| {\begin{array}{*{20}{c}} {\int_a^b f (x)dx}&{\int_a^b g (x)dx}&{\int_a^b h (x)dx} \\ a&b&c \\ l&m&n \end{array}} \right|\)

**Proof**: \(\because \Delta (x) = \left| {\begin{array}{*{20}{c}}
{f(x)}&{g(x)}&{h(x)} \\
a&b&c \\
l&m&n
\end{array}} \right| = f(x) \cdot {C_{11}} + g(x){C_{12}} + h(x) \cdot {C_{13}}\)

Where \({C_{11}} = bn – mc;{C_{12}} = cl – an;{C_{13}} = am – bl\)

\(\int_a^b \Delta (x)dx = {C_{11}}\int_a^b f (x)dx + {C_{12}}\int_a^b g (x)dx + {C_{13}}\int_a^b h (x)dx\)

\( = \left| {\begin{array}{*{20}{c}} {\int_a^b f (x)dx}&{\int_a^b g (x)dx}&{\int_a^b h (x)dx} \\ a&b&c \\ l&m&n \end{array}} \right|\)

Hence, \(\int_a^b \Delta (x)dx = \left| {\begin{array}{*{20}{c}} {\int_a^b f (x)dx}&{\int_a^b g (x)dx}&{\int_a^b h (x)dx} \\ a&b&c \\ l&m&n \end{array}} \right|\)

**Q1. If **\(f(x), g(x),\)** and **\(h(x)\)** are three polynomials of degree **\(2,\)** then prove that **\(\Delta (x) = \left| {\begin{array}{*{20}{l}} {f(x)}&{g(x)}&{h(x)} \\ {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \end{array}} \right|\) **is a constant polynomial.****Sol:**

Let \(f(x) = {a_1}{x^2} + {a_2}x + {a_3},\)

\(g(x) = {b_1}{x^2} + {b_2}x + {b_3},\) and

\(h(x) = {c_1}{x^2} + {c_2}x + {c_3}\)

Then, \({f^\prime }(x) = 2{a_1}x + {a_2},\)

\({g^\prime }(x) = 2{b_1}x + {b_2}\)

\({h^\prime }(x) = 2{c_1}x + {c_2}\)

\({f^{\prime \prime }}(x) = 2{a_1}\)

\({g^{\prime \prime }}(x) = 2{b_1}\)

\({h^{\prime \prime }}(x) = 2{c_1}\)

Also, we can write

\({f^{\prime \prime \prime (x)}} = 0\)

\({g^{\prime \prime \prime }}(x) = 0\)

\({h^{\prime \prime \prime }}(x) = 0\)

To prove that \(∆x\) is a constant polynomial, it is sufficient to show that \(∆’x=0\) for all values of \(x.\)

Given: \(\Delta (x) = \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \end{array}} \right|\)

\(\because \frac{{d\Delta (x)}}{{dx}}\) is the sum of the \(n\) determinant obtained by differentiating the elements of one row of \(∆(x)\) and leaving the elements of other \((n−1)\) rows the same.

\(\therefore {\Delta ^\prime }(x) = \left| {\begin{array}{*{20}{l}} {{f^{\prime (x)}}}&{{g^{\prime (x)}}}&{{h^{\prime (x)}}} \\ {{f^{\prime (x)}}}&{{g^{\prime (x)}}}&{{h^{\prime (x)}}} \\ {{f^{\prime \prime (x)}}}&{{g^{\prime \prime (x)}}}&{{h^{\prime \prime (x)}}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\ {{f^{\prime \prime \prime }}(x)}&{{g^{\prime \prime \prime }}(x)}&{{h^{\prime \prime \prime }}(x)} \end{array}} \right|\)

We know that if two rows of a determinant are identical, then the value of the determinant is zero.

So, \(\left| {\begin{array}{*{20}{l}} {{f^{\prime (x)}}}&{{g^{\prime (x)}}}&{{h^{\prime (x)}}} \\ {{f^{\prime (x)}}}&{{g^{\prime (x)}}}&{{h^{\prime (x)}}} \\ {{f^{\prime \prime (x)}}}&{{g^{\prime \prime (x)}}}&{{h^{\prime \prime (x)}}} \end{array}} \right| = 0\)

And

\(\left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \end{array}} \right| = 0\)

We know that if all the elements of a row are zero, then the value of the determinant is zero.

\(\therefore \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\ {{f^{\prime \prime \prime }}(x)}&{{g^{\prime \prime \prime }}(x)}&{{h^{\prime \prime \prime }}(x)} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\ 0&0&0 \end{array}} \right| = 0\)

\(\left[ {\because {f^{\prime \prime \prime }}(x) = {g^{\prime \prime \prime }}(x) = {h^{\prime \prime \prime }}(x) = 0} \right]\)

Hence,

\(\therefore {\Delta ^\prime }(x) = \left| {\begin{array}{*{20}{l}} {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\ {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \\ {{f^{\prime \prime }}(x)}&{{g^{\prime \prime }}(x)}&{{h^{\prime \prime }}(x)} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {f(x)}&{g(x)}&{h(x)} \\ {{f^\prime }(x)}&{{g^\prime }(x)}&{{h^\prime }(x)} \\ {{f^{\prime \prime \prime }}(x)}&{{g^{\prime \prime \prime }}(x)}&{{h^{\prime \prime \prime }}(x)} \end{array}} \right|\)

\( \Rightarrow {\Delta ^\prime }(x) = 0 + 0 + 0 = 0\) for all values of \(x\)

Hence, \(\Delta (x)\) is a constant polynomial.

**Q2. If **\(f(x) = \left| {\begin{array}{*{20}{c}} 1&{2a} \\ x&{2{x^2}} \end{array}} \right|,\), **then find the value of **\(f’(a).\)**Sol**:

Given \(f(x) = \left| {\begin{array}{*{20}{c}} 1&{2a} \\ x&{2{x^2}} \end{array}} \right|\)

\(∵ \frac{d∆(x)}{dx}\) is the sum of the \(n\) determinant obtained by differentiating the elements of one row of \(∆(x)\) and leaving the elements of other \((n−1)\) rows unaltered.

\(\therefore {f^\prime }(x) = \left| {\begin{array}{*{20}{c}} 0&0 \\ x&{2{x^2}} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} 1&{2a} \\ 1&{4x} \end{array}} \right|\)

\( \Rightarrow {f^\prime }(x) = 0 + 4x – 2a\)

\( \Rightarrow {f^\prime }(x) = 4x – 2a\)

\( \Rightarrow {f^\prime }(a) = 4a – 2a\)

\(\therefore {f^\prime }(a) = 2a\)

**Q3. If **\(\Delta (x) = \left| {\begin{array}{*{20}{c}} x&{{x^2}}&{{x^3}} \\ 1&2&{3x} \\ 0&2&{5x} \end{array}} \right|,\)** where **\(a\)** is any constant, then find the derivative of **\(\Delta (x)\)**Sol**:

Given:\(\Delta (x) = \left| {\begin{array}{*{20}{c}} x&{{x^2}}&{{x^3}} \\ 1&2&{3x} \\ 0&2&{5x} \end{array}} \right|,\)

\(\because \frac{{d\Delta (x)}}{{dx}}\) is the sum of the n determinant obtained by differentiating the elements of one row of \(∆(x)\) and leaving the elements of other \((n−1)\) rows unaltered.

\(\therefore \frac{{d\Delta (x)}}{{dx}} = \left| {\begin{array}{*{20}{c}} 1&{2x}&{3{x^2}} \\ 1&2&{3x} \\ 0&2&{5x} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} x&{{x^2}}&{{x^3}} \\ 0&0&3 \\ 0&2&{5x} \end{array}} \right| + \left| {\begin{array}{*{20}{c}} x&{{x^2}}&{{x^3}} \\ 1&2&{3x} \\ 0&0&5 \end{array}} \right|\)

\( \Rightarrow \frac{{d\Delta (x)}}{{dx}} = 4x(1 – x) + ( – 6x) + 5x(2 – x)\)

\( \Rightarrow \frac{{d\Delta (x)}}{{dx}} = 4x – 4{x^2} – 6x + 10x – 5{x^2}\)

\(\therefore \frac{{d\Delta (x)}}{{dx}} = – 9{x^2} + 8x\)

**Q4. Find the coefficient of **\(x\)** in the expansion of **\(\left| {\begin{array}{*{20}{l}} {{{(1 – x)}^{22}}}&{{{(1 – x)}^{44}}}&{{{(1 – x)}^{66}}} \\ {{{(1 – x)}^{33}}}&{{{(1 – x)}^{66}}}&{{{(1 – x)}^{99}}} \\ {{{(1 – x)}^{44}}}&{{{(1 – x)}^{88}}}&{{{(1 – x)}^{144}}} \end{array}} \right|.\)**Sol:**

Let \( f(x) = \left| {\begin{array}{*{20}{l}} {{{(1 – x)}^{22}}}&{{{(1 – x)}^{44}}}&{{{(1 – x)}^{66}}} \\ {{{(1 – x)}^{33}}}&{{{(1 – x)}^{66}}}&{{{(1 – x)}^{99}}} \\ {{{(1 – x)}^{44}}}&{{{(1 – x)}^{88}}}&{{{(1 – x)}^{144}}} \end{array}} \right|\)

Since \(f(x)\) is a polynomial, therefore \(f(x) = {A_0} + {A_1}x + {A_2}{x^2} + \cdots + {A_{232}}{x^{232}}\)

Here, the coefficient of \(x = {f^\prime }(0) = {A_1}\)

Now, differentiating both sides with respect to \(x\) and then putting \(x=0\) on both sides, we get

\({f^\prime }(0) = \left| {\begin{array}{*{20}{c}} {22}&{44}&{66} \\ 1&1&1 \\ 1&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} 1&1&1 \\ {33}&{66}&{99} \\ 1&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&1&1 \\ {44}&{88}&{144} \end{array}} \right|\)

We know that if two rows of a determinant are identical, then the value of the determinant is zero.

\({f^\prime }(0) = 0 + 0 + 0 = 0\)

Hence, the coefficient of \(x\) in the given determinant\(=0\)

**Q5. Let **\(f(x) = \left| {\begin{array}{*{20}{c}} 1&a&{{a^2}} \\ {\sin (n – 1)x}&{\sin nx}&{\sin (n + 1)x} \\ {\cos (n – 1)x}&{\cos nx}&{\cos (n + 1)x} \end{array}} \right|,\) **then what **\(\int_0^{\frac{\pi }{2}} f (x)dx\)**?****Sol:** Since there are elements that are functions of \(x\) in the second and third rows, to integrate the determinant,

- Expand the determinant with the help of properties.
- Integrate the result with respect to x between applicable limits.

Given: \(f(x) = \left| {\begin{array}{*{20}{c}} 1&a&{{a^2}} \\ {\sin (n – 1)x}&{\sin nx}&{\sin (n + 1)x} \\ {\cos (n – 1)x}&{\cos nx}&{\cos (n + 1)x} \end{array}} \right|,\)

\({C_1} \to {C_1} + {C_3} – 2\cos x{C_2}\)

\( \Rightarrow f(x) = \left| {\begin{array}{*{20}{c}}
{1 + {a^2} – 2a\cos x}&a&{{a^2}} \\
0&{\sin nx}&{\sin (n + 1)x} \\
0&{\cos nx}&{\cos (n + 1)x}
\end{array}} \right|\) ** **

\( = \left( {1 + {a^2} – 2a\cos x} \right)\{ \sin nx\cos (n + 1)x – \cos nx\sin (n + 1)x\}\)

\( = \left( {1 + {a^2} – 2a\cos x} \right)\sin ( – x)\)

\( = – \left( {1 + {a^2}} \right)\sin x + 2a\cos x\sin x = – \left( {1 + {a^2}} \right)\sin x + a\sin 2x\)

\(\therefore f(x) = – \left( {1 + {a^2}} \right)\sin x + a\sin 2x\)

Hence,

\(\int_0^{\frac{\pi }{2}} f (x)dx = \int_0^{\frac{\pi }{2}} {\left( { – \left( {1 + {a^2}} \right)\sin x + a\sin 2x} \right)} dx\)

\( = – \left( {1 + {a^2}} \right)\int_0^{\frac{\pi }{2}} {\sin }\,x\,dx + a\int_0^{\frac{\pi }{2}} {\sin }\,2x\,dx\)

\( = – \left( {1 + {a^2}} \right)[ – \cos x]_0^{\frac{\pi }{2}} + \frac{a}{2}[ – \cos 2x]_0^{\frac{\pi }{2}}\)

\( = – \left( {1 + {a^2}} \right)\{ – 0 + 1\} + \frac{a}{2}\{ 1 + 1\} \)

\( = a – {a^2} – 1\)

\(\therefore \int_0^{\frac{\pi }{2}} f (x)dx = – {a^2} + a – 1\)

The determinant of a matrix is defined as \(f:M→D,\) where \(M\) is the set of all square matrices, and \(D\) is the set of values of the determinant. The derivative of the determinant can be calculated by expanding the determinant with the help of properties and differentiating both sides of the equation w.r.t \(x\) to find \(\frac{d∆(x)}{dx}\). Alternatively, write it as the sum of the \(n\) determinant obtained by differentiating one row or column while leaving other \((n−1)\) rows or columns unaltered. If \(∆x\) has \(f(x)\) in more than one row or column, expand the determinant using properties, then integrate between applicable limits. If it has \(f(x)\) in only one row or column, integrate that row or column while other rows or columns remain unaffected.

Students must have many questions with respect to the Differentiation of Determinants. Here are a few commonly asked questions and answers.

** Q.1. How do you differentiate a determinant?** If \(∆(x)\) is a determinant whose all or some elements are functions of \(x,\) we can find the derivative of the determinant by two methods.

Ans:

1. Expand the determinant \(∆(x)\) with the help of properties. Then differentiate both sides of the equation with respect to \(x\) and find \(\frac{{d\Delta (x)}}{{dx}}.\)

2. \(\frac{{d\Delta (x)}}{{dx}}\) is the sum of the n determinant obtained by differentiating the elements of one row (or column) of \(∆(x)\) while leaving the elements of other \((n−1)\) rows or columns unaltered.

** Q.2. Is determinant differentiable?** Yes, a determinant is differentiable if each of its entries are differentiable.

Ans:

** Q.3. What is the formula for determinants?** If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}& \cdots &{{a_{1n}}} \\ \vdots & \ddots & \vdots \\ {{a_{n1}}}& \cdots &{{a_{nn}}} \end{array}} \right]\) is a square matrix of order \(n,\)

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then det \(A\) or \(|A| = \left| {\begin{array}{*{20}{c}} {{a_{11}}}& \cdots &{{a_{1n}}} \\ \vdots & \ddots & \vdots \\ {{a_{n1}}}& \cdots &{{a_{nn}}} \end{array}} \right|\)

The expansion of \(|A|\) with respect to \({i^{{\text{th}}}}\) row is given by

\(|A| = \sum\limits_{j = 1}^n {{a_{ij}}} \cdot {C_{ij}}\)

where \({C_{ij}}\) is cofactor of \({{a_{ij}}}.\)

** Q.4. What are the types of determinants?** The determinant is mainly of three types:

Ans:

1. First-order determinant

2. Second-order determinant

3. Third-order determinant.

** Q.5. What is a determinant? Give an example.** Corresponding to every square matrix \(A,\) there exists a number called the determinant of the matrix \(A.\) A determinant is defined as \(f:M→D,\) where \(M\) is the set of all square matrices and \(D\) is the set of values of the determinant of that matrix.

Ans:

Example: \(\left| {\begin{array}{*{20}{c}} 0&{ – 6}&9 \\ 7&3&{ – 4} \\ { – 1}&0&3 \end{array}} \right|\)

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