Singular Matrix

Determinant of matrix

The determinant of a matrix A is a number that is calculated from the matrix. For a determinant to exist, matrix A must be a square matrix. The determinant of a matrix is denoted by det A or |A|.

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 \times \mathrm{b}_2-\mathrm{a}_2 \times \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 \cdot c_3-b_3 \cdot c_2\right)-a_2\left(b_1 \cdot 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 2nd with the -ve sign and 3rd again +ve sign, this sign sequence is followed for any order of matrix.

This whole process is row-dependent, the same process can be done using columns, which means we can select an element along a column delete their row and column compute the determinant of left out matrix, and then multiply it with the element that we select. And we will get the same result as we get while doing the whole process along the row.

Singular and non-singular matrix

A square matrix is called a singular matrix if its determinant is 0 otherwise it is called a non-singular matrix. Let's say A is a square matrix then it is singular if |A| = 0, otherwise, it will be non-singular if |A| ≠ 0.

Properties of Determinants

If A and B are square matrices of same order:

i) The determinant of the transpose of matrix A is equal to the determinant of matrix A. det (A’) = det A

ii) The product of the determinant of matrices AB is equal to the product of the determinant of individual matrices. det (AB) = det (A) det (B)

iii) 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.

iv) 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.

v) |kA| = kⁿ |A|, where n is the order of matrix A and k is a constant.

vii) |Aⁿ| = |A|ⁿ, where n belongs to N.

viii) |AB| = |BA|

Recommended Video Based on Singular and Non-Singular Matrix:

Solved Examples Based on Singular and Non-Singular Matrices

Example 1: Let $A_1, A_2, A_3$ be the three A.P. with the same common difference d and having their first terms as $\mathrm{A}, \mathrm{A}+1, \mathrm{~A}+2$, respectively. Let $a, b, c$ be the $7^{\text {th }}, 9^{\text {th }}, 17^{\text {th }}$ terms of $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3$, respectively such that $
\left|\begin{array}{ccc}
a & 7 & 1 \\
2 b & 17 & 1 \\
c & 17 & 1
\end{array}\right|+70=0
$

If $a=29$, then the sum of the first 20 terms of an AP
whose first term is $c-a-b$ and the common difference is $\frac{d}{12}$, is equal to [JEE MAINS 2023]

Solution

$\begin{aligned} & \left|\begin{array}{ccc}A+6 d & 7 & 1 \\ 21(A+1+8 d) & 17 & 1 \\ A+2+16 d & 17 & 1\end{array}\right|+70=0 \\ & A=-7, d=6 \\ & \therefore \mathrm{c}-\mathrm{a}-\mathrm{b}=20 \\ & \therefore 5_{20}=495\end{aligned}$

Hence, the answer is 495.

Example 2:

Let $\alpha$ be a root of the equation $(a-c) x^2+(b-a) x+(c-b)=0$ where $\mathrm{a}, \mathrm{b}$, and c are distinct real numbers such that the matrix $\left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{array}\right]$ is singular. Then, the value of $\frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$ is

[JEE MAINS 2023]
Solution: $(a-c) x^2+(b-a) x+(c-b)=0$
$(a \neq c)$ $\mathrm{x}=1$ is one root \& other root is $\frac{\mathrm{c}-\mathrm{b}}{\mathrm{a}-\mathrm{c}}$
now $\left|\begin{array}{ccc}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{array}\right|$ is singular

$
\Rightarrow\left|\begin{array}{ccc}
\alpha^2 & \alpha & 1 \\
1 & 1 & 1 \\
a & b & c
\end{array}\right|=0 \Rightarrow \alpha^2(c-b)-\alpha(c-a)+(b-a)=0
$
$\Rightarrow \alpha^2(c-b)+\alpha(a-c)+(b-a)=0
$

satisfied by $\alpha=1$ or $\alpha=\frac{b-a}{c-b}$
Now, if $\alpha=1$ then $\forall \mathrm{a} \neq \mathrm{b} \neq \mathrm{c}$

$
\begin{aligned}
& \sum \frac{(a-c)^2}{(b-a)(c-b)}=\frac{\sum(a-c)^3}{(a-b)(b-c)(c-a)} \\
& =\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)} \\
& =3
\end{aligned}
$

if $\left.A+B+C=0 \quad \Rightarrow \quad A^3+B^3+C^3=3 A B C\right]$

Hence, the answer is 3

Example 3: The probability that a randomly chosen $2 \times 2$ matrix with all the entries from the set of first 10 primes, is singular, is equal to :
[JEE MAINS 2022]
Solution: Let the matrix A be $\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
If A is singular then $|A|=0$

$
\begin{aligned}
& \Rightarrow \mathrm{ad}-\mathrm{bc}=0 \\
& \Rightarrow \mathrm{ad}=\mathrm{bc}
\end{aligned}
$

Case I $: \mathrm{a}=\mathrm{d}$
a can take 10 values.
d can take the value which equals a
Now for ad $=$ be to hold

$
\mathrm{bc}=\mathrm{a}^2 \Rightarrow \mathrm{b}=\mathrm{a}, \mathrm{c}=\mathrm{a}(\mathrm{as} \mathrm{a} \text { is prime })
$

$\therefore$ only one option for b and c
Total such matrices $=10 \times 1 \times 1 \times 1=10$

Case II: $:$ ㅇd
In this case, can take 10 values and $d$ can take 9 values.
Now for $\mathrm{ad}=\mathrm{bc}, \quad(\mathrm{b}=\mathrm{a}$ and $\mathrm{c}=\mathrm{d})$
or $\quad(b=d$ and $c=a)$

$
\therefore \quad 10 \times 9 \times 2=180
$

$
\text { favourable }=10+180=190
$$
Total matrices $=10^4$

$
\text { Probability }=\frac{190}{10^4}=\frac{19}{10^3}
$

Hence, the answer is $\frac{19}{10^3

The value of $\left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|_{\text {is : }}$
[MAINS 2021]
Solution: Put a $=0$, we get

$
\begin{aligned}
\Delta & =\left|\begin{array}{lll}
(a+1)(a+2) & a+2 & 1 \\
(a+2)(a+3) & a+3 & 1 \\
(a+3)(a+4) & a+4 & 1
\end{array}\right|=\left|\begin{array}{ccc}
2 & 2 & 1 \\
6 & 3 & 1 \\
12 & 4 & 1
\end{array}\right| \\
\Delta & =2(3-4)-2(6-12)+1(24-36)=-2
\end{aligned}
$

OR
Given matrix is

$
\begin{aligned}
\Delta & =\left|\begin{array}{lll}
(a+1)(a+2) & a+2 & 1 \\
(a+2)(a+3) & a+3 & 1 \\
(a+3)(a+4) & a+4 & 1
\end{array}\right| \\
\mathrm{R}_2 & \rightarrow \mathrm{R}_2-\mathrm{R}_1 \text { and } \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\
\Delta & =\left|\begin{array}{ccc}
(a+1)(a+2) & a+2 & 1 \\
(a+2)(a+3-a-1) & 1 & 0 \\
a^2+7 a+12-a^2-3 a-2 & 2 & 0
\end{array}\right| \\
& =\left|\begin{array}{ccc}
a^2+3 a+2 & a+2 & 1 \\
2(a+2) & 1 & 0 \\
4 a+10 & 2 & 0
\end{array}\right| \\
& =4(a+2)-4 a-10 \\
& =4 a+8-4 a-10=-2
\end{aligned}
$
Hence, the answer is -2

example 5: If the minimum and the maximum values of the function

$
\begin{aligned}
& f:\left[\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow R, \text { defined by } \\
& f(\theta)=\left|\begin{array}{ccc}
-\sin ^2 \theta & -1-\sin ^2 \theta & 1 \\
-\cos ^2 \theta & -1-\cos ^2 \theta & 1 \\
12 & 10 & -2
\end{array}\right|
\end{aligned}
$

are $m$ and $M$ respectively, then the ordered pair $(m, M)$ is equal to:
[JEE
[MAINS 2020]
Solution

$
\begin{aligned}
& \mathrm{C}_3 \rightarrow \mathrm{C}_3-\left(\mathrm{C}_1-\mathrm{C}_2\right) \\
& f(\theta)=\left|\begin{array}{ccc}
\sin ^2 \theta & -1-\sin ^2 \theta & 0 \\
-\cos ^2 \theta & -1-\cos ^2 \theta & 0 \\
12 & 10 & -4
\end{array}\right| \\
& =-4\left[\left(1+\cos ^2 \theta\right) \sin ^2 \theta-\cos ^2 \theta\left(1+\sin ^2 \theta\right)\right] \\
& =-4\left[\sin ^2 \theta+\cos ^2 \theta-\cos ^2 \theta-\cos ^2 \theta \sin ^2 \theta\right] \\
& f(\theta)=4 \cos 2 \theta \\
& \theta \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \\
& 2 \theta \in\left[\frac{\pi}{2}, \pi\right] \\
& (\theta) \in[-4,0] \\
& (\mathrm{m}, \mathrm{M})=(-4,0)
\end{aligned}
$

Hence, the answer is $(-4,0)$