Determinants

Determinants in Mathematics

Determinants in Mathematics are used to study square matrices and solve systems of linear equations efficiently. They play an important role in higher mathematics and have practical applications in science, engineering, and economics.

What are Determinants?

The determinant of a matrix A is a number that is calculated from the matrix. For a determinant to exist, the matrix $A$ must be a square matrix. A matrix is only a representation, while the determinant is the value of the matrix. The determinant of a matrix is denoted by $\operatorname{det} \mathrm{A}$ or $|\mathrm{A}|$.

To every square matrix $A=\left[a_{i j}\right]$ of order $n$, we can associate a number called the determinant of the matrix $A$.

If $A=\left[\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right]$,
then determinant of $A$ is written as
$|A|=\left|\begin{array}{cccc}a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n}\end{array}\right|$.

Minors

Let $A=\left[a_{i j}\right]_{3 \times 3}$ be a given square matrix of order $3$ . The minor of an arbitrary element $a_{i j}$ is the determinant obtained by deleting the $i^{\text {th }}$ row and $j^{\text {th }}$ column in which the element $a_{i j}$ stands. The minor of $a_{i j}$ is usually denoted by $M_{i j}$.

Cofactors

The cofactor is a signed minor. The cofactor of $a_{i j}$ is usually denoted by $A_{i j}$ and is defined as $A_{i j}=(-1)^{i+j} M_{i j}$.

For instance, consider the $3 \times 3$ matrix defined by $A=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$
Then the minors and cofactors of the elements $a_{11}, a_{12}, a_{13}$ are given as follows :
(i) Minor of $a_{11}$ is $\mathrm{M}_{11} \quad=\left|\begin{array}{ll}a_{22} & a_{23} \\ a_{32} & a_{33}\end{array}\right|=a_{22} a_{33}-a_{32} a_{23}$

Cofactor of $a_{11}$ is $A_{11}=(-1)^{1+1} M_{11}=\left|\begin{array}{ll}a_{22} & a_{23} \\ a_{32} & a_{33}\end{array}\right|=a_{22} a_{33}-a_{32} a_{23}$

(ii) Minor of $a_{12}$ is $M_{12} \quad=\left|\begin{array}{ll}a_{21} & a_{23} \\ a_{31} & a_{33}\end{array}\right|=a_{21} a_{33}-a_{31} a_{23}$

Cofactor $a_{12}$ is $A_{12} \quad=(-1)^{1+2}\left|\begin{array}{ll}a_{21} & a_{23} \\ a_{31} & a_{33}\end{array}\right|=-\left(a_{21} a_{33}-a_{31} a_{23}\right)$

(iii) Minor of $a_{13}$ is $M_{13} \quad=\left|\begin{array}{ll}a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right|=a_{21} a_{32}-a_{31} a_{22}$

Cofactor of $a_{13}$ is $A_{13}=(-1)^{1+3} M_{13}=\left|\begin{array}{ll}a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right|=a_{21} a_{32}-a_{31} a_{22}$.

Determinants Formula

The determinant formula for $2 \times 2$ matrices

$
\mathrm{A}=\left[\begin{array}{ll}
a_1 & a_2 \\
b_1 & b_2
\end{array}\right]
$

then $\operatorname{det} \mathrm{A}$ is :

$
|\mathrm{A}|=\left|\begin{array}{ll}
a_1 & a_2 \\
b_1 & b_2
\end{array}\right|=\mathrm{a}_1 \mathrm{b}_2-\mathrm{a}_2 \mathrm{b}_1
$

For a $3 \times 3$ matrix determinant can be calculated in the following way :

$
\text { let } \mathrm{A}=\left[\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right]
$

then we find $\operatorname{det} \mathrm{A}$ in following way

$
|A|=a_1\left(b_2 c_3-b_3 c_2\right)-a_2\left(b_1 c_3-c_1 b_3\right)+a_3\left(b_1 c_2-b_2 c_1\right)
$

This same process we follow to evaluate the determinant of the matrix of any order. Notice that we start the first term with the +ve sign, then the 2nd with the -ve sign and the 3rd again +ve sign; this sign sequence is followed for any order of the matrix.

How to Calculate the Determinant of Matrices?

To calculate the determinant of a matrix, first ensure it is square (same number of rows and columns). For a 2 × 2 matrix, multiply the diagonal elements and subtract the product of the off-diagonal elements. Let us understand in detail:

Calculating the Determinant of $2 \times 2$ Matrix

Let $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$ be a matrix of order 2. Then the determinant of $A$ is defined as

$
|A|=\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right|=a_{11} \quad a_{22}-a_{21} a_{12}
$
Example: Let $|A|=$ $\left|\begin{array}{cc}2 & 4 \\ -1 & 2\end{array}\right|=(2 \times 2)-(-1 \times 4)=4+4=8$.

Calculating the Determinant of $3 \times 3$ Matrix

The determinant of the matrix $
\mathrm{A}=\left[\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right]
$ is given by

$\qquad|\mathrm{A}|=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=a_1\left|\begin{array}{ll}b_2 & c_2 \\ b_3 & c_3\end{array}\right|-b_1\left|\begin{array}{ll}a_2 & c_2 \\ a_3 & c_3\end{array}\right|+c_1\left|\begin{array}{ll}a_2 & b_2 \\ a_3 & b_3\end{array}\right|$

Example: $|A|= \left|\begin{array}{ccc}0 & \sin \alpha & -\cos \alpha \\ -\sin \alpha & 0 & \sin \beta \\ \cos \alpha & -\sin \beta & 0\end{array}\right|$

$\begin{aligned} & =0\left|\begin{array}{cc}0 & \sin \beta \\ -\sin \beta & 0\end{array}\right|-\sin \alpha\left|\begin{array}{cc}-\sin \alpha & \sin \beta \\ \cos \alpha & 0\end{array}\right|-\cos \alpha\left|\begin{array}{cc}-\sin \alpha & 0 \\ \cos \alpha & -\sin \beta\end{array}\right| \\ & =0-\sin \alpha(0-\sin \beta \cos \alpha)-\cos \alpha(\sin \alpha \sin \beta-0) \\ & =\sin \alpha \sin \beta \cos \alpha-\cos \alpha \sin \alpha \sin \beta=0\end{aligned}$

Evaluation of determinants of matrices of order $3$ by the Sarrus rule,

Let $A=\left[a_{i j}\right]_{3 \times 3}=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$
Write the entries of Matrix $A$ as follows :

Then $|A|$ is computed as follows :

$
|A|=\left[a_{11} a_{22} a_{33}+a_{12} a_{23} a_{31}+a_{13} a_{21} a_{32}\right]-\left[a_{33} a_{21} a_{12}+a_{32} a_{23} a_{11}+a_{31} a_{22} a_{13}\right]
$

Notice that we start the first term with the +ve sign, then the 2nd with the -ve sign and the 3rd again +ve sign; this sign sequence is followed for any order of matrix.

Example: $A=\left[\begin{array}{ccc}3 & 4 & 1 \\ 0 & -1 & 2 \\ 5 & -2 & 6\end{array}\right]$

$|A| =[3(-1)(6)+4(2)(5)+1(0)(-2)]-[5(-1)(1)+(-2)(2) 3+6(0)(4)] $
$ =[-18+40+0]-[-5-12+0]$
$=22+17=39$

Multiplication of Determinants

There are two types of multiplication of determinants:

1) Multiplication of a determinant by a scalar quantity

2) Multiplication of a determinant by another determinant

Multiplication of a determinant by a scalar quantity

If $A$ is a square matrix and $k$ is a scalar quantity then, $|\mathrm{kA}|=\mathrm{k}^{\mathrm{n}}|\mathrm{A}|$, where $n$ is the order of $A$

Multiplication of a determinant by another determinant

Determinant multiplication is a binary operation that produces a determinant from two determinants. For determinant multiplication, the order of both determinants should be the same.

Multiplication of two determinants can be done by 4 methods, namely,

• row-by-row multiplication
• row-by-column multiplication
• column-by-row multiplication
• column-by-column multiplication

Multiplication of $2 \times 2$ Determinants

Let the two determinants of second order be

$\Delta_1=\left|\begin{array}{ll}a_1 & b_1 \\ a_2 & b_2\end{array}\right| \quad$ and $\quad \Delta_2=\left|\begin{array}{ll}\alpha_1 & \beta_1 \\ \alpha_2 & \beta_2\end{array}\right|$

We can multiply these by row-by-row or column-by-column or row-by-column or column-by-row

Row-by-row multiplication of these two determinants is given by

$\Delta_1 \times \Delta_2=\left|\begin{array}{ll}\left(a_1 \alpha_1+b_1 \alpha_2\right) & \left(a_1 \beta_1+b_1 \beta_2\right) \\ \left(a_2 \alpha_1+b_2 \alpha_2\right) & \left(a_2 \beta_1+b_2 \beta_2\right)\end{array}\right|$

Row-by-column multiplication of these two determinants is given by

$\Delta_1 \times \Delta_2=\left|\begin{array}{ll}\left(a_1 \alpha_1+b_1 \beta_1\right) & \left(a_1 \alpha_2+b_1 \beta_2\right) \\ \left(a_2 \alpha_1+b_2 \beta_1\right) & \left(a_2 \alpha_2+b_2 \beta_2\right)\end{array}\right|$

Column-by-row multiplication of these two determinants is given by

$\Delta_1 \times \Delta_2=\left|\begin{array}{ll}\left(a_1 \alpha_1+a_2 \alpha_2\right) & \left(b_1 \beta_1+b_2 \beta_2\right) \\ \left(a_1 \alpha_1+a_2 \alpha_2\right) & \left(b_1 \beta_1+b_2 \beta_2\right)\end{array}\right|$

Column-by-column multiplication of these two determinants is given by

$\Delta_1 \times \Delta_2=\left|\begin{array}{ll}\left(a_1 \alpha_1+a_2 \alpha_2\right) & \left(b_1 \beta_1+b_2 \beta_2\right) \\ \left(a_1 \alpha_1+a_2 \alpha_2\right) & \left(b_1 \beta_1+b_2 \beta_2\right)\end{array}\right|$

Multiplication of $3 \times 3$ Determinants

Let two determinants of third order be

$\begin{equation}
\begin{aligned}
&\Delta_1=\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right| \text { and } \Delta_2=\left|\begin{array}{ccc}
\alpha_1 & \beta_1 & \gamma_1 \\
\alpha_2 & \beta_2 & \gamma_2 \\
\alpha_3 & \beta_3 & \gamma_3
\end{array}\right|
\end{aligned}
\end{equation}$

We can multiply these row-by-row or column-by-column or row-by-column or column-by-row

Row-by-row multiplication of these two determinants is given by

$\begin{equation}
\begin{aligned}
&\Delta_1 \times \Delta_2=\left|\begin{array}{lll}
a_1 \alpha_1+b_1 \beta_1+c_1 \gamma_1 & a_1 \alpha_2+b_1 \beta_2+c_1 \gamma_2 & a_1 \alpha_3+b_1 \beta_3+c_1 \gamma_3 \\
a_2 \alpha_1+b_2 \beta_1+c_2 \gamma_1 & a_2 \alpha_2+b_2 \beta_2+c_2 \gamma_2 & a_2 \alpha_3+b_2 \beta_3+c_2 \gamma_3 \\
a_3 \alpha_1+b_3 \beta_1+c_3 \gamma_1 & a_3 \alpha_2+b_3 \beta_2+c_3 \gamma_2 & a_3 \alpha_3+b_3 \beta_3+c_3 \gamma_3
\end{array}\right|
\end{aligned}
\end{equation}$

Row-by-column multiplication of these two determinants is given by

$\Delta_1 \times \Delta_2=\left|\begin{array}{lll}a_1 \alpha_1+b_1 \alpha_2+c_1 \alpha_3 & a_1 \beta_1+b_1 \beta_2+c_1 \beta_3 & a_1 \gamma_1+b_1 \gamma_2+c_1 \gamma_3 \\ a_2 \alpha_1+b_2 \alpha_2+c_2 \alpha_3 & a_2 \beta_1+b_2 \beta_2+c_2 \beta_3 & a_2 \gamma_1+b_2 \gamma_2+c_2 \gamma_3 \\ a_3 \alpha_1+b_3 \alpha_2+c_3 \alpha_3 & a_3 \beta_1+b_3 \beta_2+c_3 \beta_3 & a_3 \gamma_1+b_3 \gamma_2+c_3 \gamma_3\end{array}\right|$

Column-by-row multiplication of these two determinants is given by

$\Delta_1 \times \Delta_2=\left|\begin{array}{lll}a_1 \cdot \alpha_1+a_2 \cdot \beta_1+a_3 \cdot \gamma_1 & b_1 \cdot \alpha_1+b_2 \cdot \beta_1+b_3 \cdot \gamma_1 & c_1 \cdot \alpha_1+c_2 \cdot \beta_1+c_3 \cdot \gamma_1 \\ a_1 \cdot \alpha_2+a_2 \cdot \beta_2+a_3 \cdot \gamma_2 & b_1 \cdot \alpha_2+b_2 \cdot \beta_2+b_3 \cdot \gamma_2 & c_1 \cdot \alpha_2+c_2 \cdot \beta_2+c_3 \cdot \gamma_2 \\ a_1 \cdot \alpha_3+a_2 \cdot \beta_3+a_3 \cdot \gamma_3 & b_1 \cdot \alpha_3+b_2 \cdot \beta_3+b_3 \cdot \gamma_3 & c_1 \cdot \alpha_3+c_2 \cdot \beta_3+c_3 \cdot \gamma_3\end{array}\right|$

Column-by-column multiplication of these two determinants is given by

$\Delta_1 \times \Delta_2=\left|\begin{array}{lll}a_1 \alpha_1+a_2 \alpha_2+a_3 \alpha_3 & b_1 \beta_1+b_2 \beta_2+b_3 \beta_3 & c_1 \gamma_1+c_2 \gamma_2+c_3 \gamma_3 \\ a_1 \alpha_1+a_2 \alpha_2+a_3 \alpha_3 & b_1 \beta_1+b_2 \beta_2+b_3 \beta_3 & c_1 \gamma_1+c_2 \gamma_2+c_3 \gamma_3 \\ a_1 \alpha_1+a_2 \alpha_2+a_3 \alpha_3 & b_1 \beta_1+b_2 \beta_2+b_3 \beta_3 & c_1 \gamma_1+c_2 \gamma_2+c_3 \gamma_3\end{array}\right|$

Properties of Determinants

The properties of determinants are given below:

If $A$ and $B$ are square matrices of the same order:

• The determinant of the transpose of matrix $A$ is equal to the determinant of matrix $A$. $\operatorname{det}\left(A^{\prime}\right)=\operatorname{det} A$
• The product of the determinant of matrices $A B$ is equal to the product of the determinants of individual matrices. $\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)$
• The determinant of the skew-symmetric matrix of odd order is zero. if $A$ is the skew-symmetric matrix of odd order then $|A|=0$.
• The determinant of the skew-symmetric matrix of even order is a perfect square. if $A$ is a skew-symmetric matrix of even order then $|A|$ is a perfect square.
• $|k A|=k^n|A|$, where $n$ is the order of matrix $A$ and $k$ is a constant.
• $\left|A^n\right|=|A|^n$, where $n$ belongs to $N$.
• $|A B|=|B A|$

Important Formulae related to Determinants

Determinants in Mathematics: Solved Previous Year Questions

Question 1:

Let $A=\left[\begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array}\right]$.
If $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A})))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}, \mathrm{m}, \mathrm{n} \in \mathrm{N}$, then $\mathrm{m}+\mathrm{n}$ is equal to

Solution:

$\begin{aligned} & |\operatorname{adj}(\operatorname{adj} 3 \mathrm{~A})|=|\operatorname{adj} 3 \mathrm{~A}|^2 \\ & =|3 \mathrm{~A}|^4=\left(3^3|\mathrm{~A}|\right) 4 \\ & =3^{12}|\mathrm{~A}|^4 \\ & \begin{aligned}|\mathrm{A}|=\left|\begin{array}{ccc}2 & 2+\mathrm{p} & 2+\mathrm{p}+\mathrm{q} \\ 4 & 6+2 \mathrm{p} & 8+3 \mathrm{p}+2 \mathrm{q} \\ 6 & 12+3 \mathrm{p} & 20+6 \mathrm{p}+3 \mathrm{q}\end{array}\right|=8 \\ \begin{aligned} \mathrm{C}_2=|\operatorname{adj}(\operatorname{adj} 3 \mathrm{~A})| & =3^{12}(8)^4 \\ & =2^{12} 3^{12} \\ & =\mathrm{m}+\mathrm{n}=24\end{aligned}\end{aligned}\end{aligned}$

Hence, the answer is 24.

Question 2:

Let $A=\left[\mathrm{a}_{\mathrm{ij}}\right]=\left[\begin{array}{cc}\log _5 128 & \log _4 5 \\ \log _5 8 & \log _4 25\end{array}\right]$. If $\mathrm{A}_{\mathrm{ij}}$ is the cofactor of $\mathrm{a}_{\mathrm{ij}}, \mathrm{C}_{\mathrm{ij}}=\sum_{\mathrm{k}=1}^2 \mathrm{a}_{\mathrm{ik}} \mathrm{A}_{\mathrm{jk}}, 1 \leq \mathrm{i}$, $\mathrm{j} \leq 2$, and $\mathrm{C}=\left[\mathrm{C}_{\mathrm{ij}}\right]$, then $8|\mathrm{C}|$ is equal to:

Solution:

We are given the matrix:
$
A = \begin{bmatrix}
\log_5 128 & \log_4 5 \\
\log_5 8 & \log_4 25
\end{bmatrix}
$

To compute the determinant of $ A $, use the original entries:

$
|A| = \log_5 128 \cdot \log_4 25 - \log_5 8 \cdot \log_4 5
$

Now, simplify each logarithms,

$
\log_5 128 = \frac{\log 128}{\log 5} = \frac{7 \log 2}{\log 5}, \quad
\log_5 8 = \frac{3 \log 2}{\log 5}
$

$
\log_4 5 = \frac{\log 5}{2 \log 2}, \quad
\log_4 25 = \frac{2 \log 5}{2 \log 2} = \frac{\log 5}{\log 2}
$

Now compute the determinant:
$
|A| = \left( \frac{7 \log 2}{\log 5} \cdot \frac{\log 5}{\log 2} \right) - \left( \frac{3 \log 2}{\log 5} \cdot \frac{\log 5}{2 \log 2} \right)
$

$
|A| = 7 - \frac{3}{2} = \frac{11}{2}
$

Now we compute the matrix $ C = [C_{ij}] $, where:
$
C_{ij} = \sum_{k=1}^2 a_{ik} A_{jk}
$

Because:
$
C_{11} = a_{11}A_{11} + a_{12}A_{12} = |A| = \frac{11}{2},\quad C_{12} = a_{11}A_{21} + a_{12}A_{22} = 0
$
$
C_{21} = a_{21}A_{11} + a_{22}A_{12} = 0,\quad C_{22} = a_{21}A_{21} + a_{22}A_{22} = |A| = \frac{11}{2}
$

So:
$
C = \begin{bmatrix}
\frac{11}{2} & 0 \\
0 & \frac{11}{2}
\end{bmatrix}
\Rightarrow |C| = \left( \frac{11}{2} \right)^2 = \frac{121}{4}
$

Finally:
$
8|C| = 8 \cdot \frac{121}{4} = 2 \cdot 121 = 242
$

Hence, the correct answer is 242.

Question 3:

Let M and m respectively be the maximum and the minimum values of

$
f(x)=\left|\begin{array}{ccc}
1+\sin ^2 x & \cos ^2 x & 4 \sin 4 x \\
\sin ^2 x & 1+\cos ^2 x & 4 \sin 4 x \\
\sin ^2 x & \cos ^2 x & 1+4 \sin 4 x
\end{array}\right|, x \in R
$

Then $\mathrm{M}^4-\mathrm{m}^4$ is equal to:

Solution:

We are given:

$
f(x)=\left|\begin{array}{ccc}
1+\sin^2 x & \cos^2 x & 4\sin 4x \\
\sin^2 x & 1+\cos^2 x & 4\sin 4x \\
\sin^2 x & \cos^2 x & 1+4\sin 4x
\end{array}\right|, \quad x \in \mathbb{R}
$

Apply row operations:

$
R_2 \rightarrow R_2 - R_1, \quad R_3 \rightarrow R_3 - R_1
$

$
f(x) = \left|\begin{array}{ccc}
1+\sin^2 x & \cos^2 x & 4\sin 4x \\
-1 & 1 & 0 \\
-1 & 0 & 1
\end{array}\right|
$

Now expand along the first row:

$
f(x) = (1+\sin^2 x) \cdot \left|\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right|
- \cos^2 x \cdot \left|\begin{array}{cc}
-1 & 0 \\
-1 & 1
\end{array}\right|
+ 4\sin 4x \cdot \left|\begin{array}{cc}
-1 & 1 \\
-1 & 0
\end{array}\right|
$

Calculate the 2×2 minors:

$
= (1+\sin^2 x)(1) - \cos^2 x ( -1 ) + 4\sin 4x (1)
$

$
= 1 + \sin^2 x + \cos^2 x + 4\sin 4x
$

Since $ \sin^2 x + \cos^2 x = 1 $, we get:

$
f(x) = 1 + 1 + 4\sin 4x = 2 + 4\sin 4x
$

Now find the maximum and minimum values:

- $ \max(\sin 4x) = 1 \Rightarrow f(x) = 2 + 4(1) = 6 $
- $ \min(\sin 4x) = -1 \Rightarrow f(x) = 2 + 4(-1) = -2 $

So,
$
M = 6, \quad m = -2
$

Then,

$
M^4 - m^4 = 6^4 - (-2)^4 = 1296 - 16 = 1280
$

Hence, the correct answer is 1280.

Question 4:

The system of equations

$
\begin{aligned}
& x+y+z=6 \\
& x+2 y+5 z=9 \\
& x+5 y+\lambda z=\mu
\end{aligned}
$

has no solution if:

Solution:

For the system to have no solutions, let us calculate the determinant,

$D=0$

$\left|\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & \lambda\end{array}\right|=0$

$1(2\lambda-25)-1(\lambda-5)+1(5-2)=0$

$\lambda-20+3=0$

$\lambda-17=0$

$\lambda=17$

Similarly, the determinant,

$D_3 \neq 0$

$\left|\begin{array}{lll}1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & \mu\end{array}\right| \neq 0$

$1(2\mu-45)-1(\mu-9)+6(5-2) \neq 0$

$\mu-36+18\neq0$

$\mu \neq 18$

Hence, the correct answer is "$\lambda=17, \mu \neq 18$".

Question 5:

If $\mathrm{A}, \mathrm{B}$ and $\left(\operatorname{adj}\left(\mathrm{A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)$ are non-singular matrices of same order, then the inverse of $\mathrm{A}\left(\operatorname{adj}\left(\mathrm{A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)^{-1} \mathrm{~B}$, is equal to:

Solution:

Given the expression:

$
\left[ A \left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right)^{-1} \cdot B \right]^{-1}
$

First, apply the inverse to the product inside the bracket:

$
= B^{-1} \cdot \left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right) \cdot A^{-1}
$

Distribute $ B^{-1} $ and $ A^{-1} $:

$
= B^{-1} \operatorname{adj}(A^{-1}) A^{-1} + B^{-1} \operatorname{adj}(B^{-1}) A^{-1}
$

Using the property of adjoint and determinant for inverse matrices:

$
\operatorname{adj}(M^{-1}) = |M^{-1}| \cdot M
$

Thus,

$
B^{-1} \operatorname{adj}(A^{-1}) A^{-1} = B^{-1} |A^{-1}| I
$

and

$
B^{-1} \operatorname{adj}(B^{-1}) A^{-1} = |B^{-1}| I A^{-1}
$

Since

$
|A^{-1}| = \frac{1}{|A|} \quad \text{and} \quad |B^{-1}| = \frac{1}{|B|}
$

we get:

$
= \frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}
$

Rewrite by expressing inverses in terms of adjoints and determinants:

$
= \frac{\operatorname{adj} B}{|B| |A|} + \frac{\operatorname{adj} A}{|A| |B|}
$

Combine the terms:

$
= \frac{1}{|A||B|} \left( \operatorname{adj} B + \operatorname{adj} A \right)
$

Hence, the correct answer is $\frac{1}{|\mathrm{AB}|}(\operatorname{adj}(\mathrm{B})+\operatorname{adj}(\mathrm{A}))$.

Determinants in Different Exams

The chapter on Determinants is an important part of Class 12 Mathematics and is frequently tested in board exams as well as competitive examinations. It focuses on evaluating determinants and understanding their properties and applications. The table below highlights the exam-wise focus areas, commonly asked topics, and effective preparation strategies to help students prepare efficiently.

List of topics related to determinants according to NCERT/JEE Mains

This section covers all the important topics from Determinants as prescribed in the NCERT syllabus and commonly asked in JEE Main. Studying these topics systematically will help you build strong fundamentals and improve problem-solving skills.

Important Books and Resources for Determinants

Find recommended books and study resources that explain determinants clearly and provide problems for thorough practice to excel in exams.

NCERT Resources for Determinants

Access NCERT notes, solutions, and exemplar problems for determinants that simplify the study and help students prepare effectively.

• NCERT Maths Notes for Class 12th Chapter 4 Determinants
• NCERT Maths Solutions for Class 12th Chapter 4 Determinants
• NCERT Maths Exemplar Solutions for Class 12th Chapter 4 Determinants

NCERT Subjectwise Resources

Explore organised NCERT resources chapter-wise and subject-wise for a systematic understanding.

Practice Questions based on Determinants

Practice a variety of questions on matrices to strengthen problem-solving skills and boost exam readiness. Regular practice helps build confidence and mastery of concepts.

Conclusion

Determinants form a crucial part of Class 12 Mathematics and provide a powerful concept for solving linear equations and analysing matrices. A clear understanding of formulas, properties, and applications helps in solving problems accurately and efficiently. With regular practice and proper revision, students can confidently handle determinant-based questions in board and competitive examinations.