# Determinant of a Matrix

Get started with the Determinant of a matrix with this beginner-friendly post, covering the basics of the Determinant of a Matrix, Singular Matrix, Non-Singular Matrix, Adjoint and Multiplicative Inverse of a Matrix.

## Determinant of a Matrix

Determinant of a denoted by |A| or det A.
Determinant of a Square Matrix
Determinant of A denoted by |A| or det A.
Let A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]
|A|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|
|A|=a d-c b
Example 1:
A=\left[\begin{array}{cc}5 & 6 \\ -4 & 1\end{array}\right]
Solution:
A=\left[\begin{array}{cc}5 & 6 \\ -4 & 1\end{array}\right]
|A|=\left|\begin{array}{ll}5 & 6 \\ -4 & 1\end{array}\right|
|A|=(5)(1)-(-4)(6)
|A|=5-(-24)
|A|=5+24
|A|=29
Example 2:
F=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]
|F|=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|
Expand by Row 1:
|F|=1\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|-0\left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right|+0\left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right|
|F|=1(1-0)-0+0
|F|=1(1)
|F|=1
Singular Matrix
If |A|=0 then A is Singular Matrix.
D=\left[\begin{array}{cc}-3 & 6 \\ 2 & -4\end{array}\right]
|D|=\left|\begin{array}{cc} -3 & 6 \\ 2 & -4 \end{array}\right| \\
|D|=12-12 \\
|D|=0
Thus D is a singular matrix.
Non-Singular Matrix
If |A| \neq 0 then \mathrm{A} is Non-Singular Matrix.
C=\left[\begin{array}{cc}3 a & -2 b \\ 2 a & b\end{array}\right]
Solution:
C=\left[\begin{array}{cc}3 a & -2 b \\ 2 a & b\end{array}\right]
|C|=\left|\begin{array}{cc}3 a & -2 b \\ 2 a & b\end{array}\right|
|C|=3 a b-(-4 a b)
|C|=3 a b+4 a b
|C|=7 a b \neq 0
Adjoint of Square Matrix
Let A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]
As change the places of a \ and \ d with each other and change the size of b \ and \ c . So
{Adj} A=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]
B=\left[\begin{array}{cc}-3 & -1 \\ 2 & 3\end{array}\right]
{Adj} \mathrm{B}=\left[\begin{array}{cc}3 & 1 \\ -2 & -3\end{array}\right]
D=\left[\begin{array}{cc}-3 & 6 \\ 2 & -4\end{array}\right]
{Adj} {D}=\left[\begin{array}{rr}-4 & -6 \\ -2 & -3\end{array}\right]

## MCQs

1. |A| is called as
O Determinant
O A symmetric
O None

Explanation:
This is the way to write Determinant

2. |A|=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|=
O a b+c d
O a b+a d
O abcd
O a d-b c

Explanation:
In this way, find the Determinant

3. \left|\begin{array}{cc}-2 & 2 \\ 3 & 5\end{array}\right|=
O 16
O -6
O -16
O None of these

Explanation:
(-2)(5)-(3)(2)
-10-6
-16

4. \left|\begin{array}{cc}4 & -2 \\ -2 & 1\end{array}\right|=
O 16
O 0
O 2
O 4

Explanation:
(4)(1)-(-2)(-2)
4-4
0

5. The determinant is a
O number
O Transpose
O Matrix

Explanation:

6. \left|\begin{array}{cc}4 & -2 \\ -2 & 1\end{array}\right|=
O Singular
O Non-singular
O None

Explanation:
(4)(1)-(-2)(-2)
4-4
0
As Determinant is 0. Thus, it is singular matrix.

7. |A|=0
O Singular
O Non-singular
O None

Explanation:
When the Determinant of matrix is zero, then the matrix is singular matrix.

8. |A| \neq 0
O Singular
O Non-singular
O None

Explanation:
When the Determinant of matrix is not equal zero, then the matrix is Non-singular matrix.

9. If A=\left[\begin{array}{ll}7 & 8 \\ 3 & 2\end{array}\right] \ then \ adj \ A=
O \left[\begin{array}{cc}2 & 8 \\ -3 & 7\end{array}\right]
O \left[\begin{array}{cc}2 & -8 \\ -3 & 7\end{array}\right]
O \left[\begin{array}{cc}7 & 8 \\ -3 & 2\end{array}\right]
O None

Answer: \left[\begin{array}{cc}2 & -8 \\ -3 & 7\end{array}\right]
Explanation:
In Adjoint, change the places of a and d with each other and change the signs of b and c.

10. A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \ the \ adjA \ =
O \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]
O \left[\begin{array}{ll}a & c \\ d & a\end{array}\right]
O a d-b c
O All of these

Answer: \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]
Explanation:
In Adjoint, change the places of a and d with each other and change the signs of b and c.

11. Multiplicative inverse of A is
O AB
O B
O A
O A^{-1}

Explanation:

12. A \cdot A^{-1}=A^{-1} \cdot A
O I
O A
O A^{-1}
O None

Explanation:
A matrix and its multiplicative inverse is equal to Multiplicative Identity “I”.

13. A^{-1}=\frac{1}{|A|} {adj} of
O B
O I
O A
O None

Explanation:
Definition of Multiplicative inverse

14. F^{-1}=\frac{1}{|F|} adj of
O A
O I
O F
O None

Explanation:
Definition of Multiplicative inverse

15. If A=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right] \ then \ A^{-1}=
O -\frac{1}{8}\left[\begin{array}{cc}-1 & -3 \\ 2 & 2 \end{array}\right]
O -\frac{1}{8}\left[\begin{array}{cc} -2 & -3 \\ -2 & 1 \end{array}\right]
O \frac{1}{8}\left[\begin{array}{cc} -2 & -3 \\ -2 & 1 \end{array}\right]
O None of these

Explanation:
A=\left[\begin{array}{cc}1 & 3 \\ 2 & -2\end{array}\right]
We have
A^{-1}=\frac{1}{|A|} \ adj of A
|A|=\left|\begin{array}{ll}1 & 3 \\ 2 & -2\end{array}\right|
|A|=(1)(-2)-(2)(-3)
|A|=-2-6
|A|=-8
Adj \ A=\left[\begin{array}{cc}-2 & -3 \\ -2 & 1\end{array}\right]
Now
A^{-1}=\frac{1}{-8} \left[\begin{array}{cc}-2 & -3 \\ -2 & 1\end{array}\right]

1. \left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] is
a. An identity matrix w.r.t multiplication
b. A column matrix
c. An identity matrix w.r.t addition
d. A row matrix

Explanation:
Zero (0) is called additive identity. Thus Zero or Null matrix is additive identity matrix.

2. The matrix \left[\begin{array}{cc}4 & 0 \\ 0 & -12\end{array}\right] is
a. A scalar matrix
b. \quad 2 \times 3 matrix
c. A diagonal matrix
d. None of these

Answer: A diagonal matrix
Explanation:
A square matrix on which all elements are zero except diagonal elements is known as diagonal matrix.

3. If A=\left[\begin{array}{cc}-1 & -2 \\ 3 & 1\end{array}\right] , then adj A is equal to
a. \quad\left[\begin{array}{cc}-1 & -2 \\ 3 & 1\end{array}\right]
b. \quad\left[\begin{array}{cc}1 & 2 \\ -3 & -1\end{array}\right]
c. \left[\begin{array}{cc}-1 & 2 \\ 3 & 1\end{array}\right]
d. \quad\left[\begin{array}{cc}1 & -2 \\ 3 & 1\end{array}\right]

Answer: \quad\left[\begin{array}{cc}-1 & -2 \\ 3 & 1\end{array}\right]
Explanation:
Let A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]
As change the places of a \ and \ d with each other and change the signs of b \ and \ c . So
Adj \ A=\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]

4. If A =\left[\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right] then A^{-1}=
a. \quad\left[\begin{array}{cc}4 & 3 \\ -3 & 2\end{array}\right]
b. \quad\left[\begin{array}{cc}4 & -3 \\ -3 & 2\end{array}\right]
c. \quad\left[\begin{array}{cc}-2 & 3 \\ 3 & -4\end{array}\right]
d. \quad\left[\begin{array}{cc}-4 & 3 \\ 3 & -2\end{array}\right]

Explanation:
A=\left[\begin{array}{ll} 2 & 3 \\ 3 & 4 \end{array}\right]

A^{-1}=\frac{1}{|A|} { Adj } A

|A|=\left|\begin{array}{ll}2 & 3 \\ 3 & 4\end{array}\right|
|A|=8-9
|A|=-1 \neq 0
{Adj} A=\left[\begin{array}{cc}4 & -3 \\ -3 & 2\end{array}\right]
Put the values in equation
A^{-1}=\frac{1}{-1}\left[\begin{array}{cc} 4 & -3 \\ -3 & 2 \end{array}\right]

A^{-1}=-\left[\begin{array}{cc} 4 & -3 \\ -3 & 2 \end{array}\right]

A^{-1}=\left[\begin{array}{cc} -4 & 3 \\ 3 & -2 \end{array}\right]

5. For what value of d is the 2 \times 2 matrix \left[\begin{array}{cc}1 & 1.5 \\ 2 & d\end{array}\right] not invertible?
a. -0.6
b. 0
c. 0.6
d. 3

Explanation:
Singular matrix is also called NOT invertible.
Thus |A|=0
\left|\begin{array}{ll}5 & 1.5 \\ 2 & d\end{array}\right|=0
5 \times d-2 \times 1.5=0
5d-3=0
5d=3
d=\frac{3}{5}
d=0.6

6. Suppose A and B are 2 \times 5 matrices, which of the following are the dimensions of the matrix A+B ?
a. 2 \times 5
b. 10 \times 10
c. \quad 7 \times 7
d. \quad 7 \times 1

Answer: 2 \times 5
Explanation:
For Addition of two matrices, the dimensions of the matrices must be same. Thus A+B have the dimensions 2 \times 5

7.Which of the following is the multiplicative inverse of \left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] is
a. \quad\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]
b. \quad\left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right]
c. \quad\left[\begin{array}{cc}-1 & 2 \\ 0 & 1\end{array}\right]
d. \quad\left[\begin{array}{ll}1 & -2 \\ 0 & -1\end{array}\right]

Answer: \quad\left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right]
Explanation:
Let \ A=\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right]

A^{-1}=\frac{1}{|A|} { Adj } A

Let \ |A|=\left|\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right|
|A|=1-0
|A|=1 \neq 0
{Adj} A=\left[\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right]
Put the values in equation
A^{-1}=\frac{1}{1}\left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right]

A^{-1}=\left[\begin{array}{cc} 1 & -2 \\ 0 & 1 \end{array}\right]

8. The determinant of \left[\begin{array}{cc}4 & -1 \\ -9 & 2\end{array}\right] is
a. 17
b. 1
c. -1