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By using properties of determinants, show that:
(i) 
(ii) 
(i) 
Applying R1 → R1 − R3 and R2 → R2 − R3, we have:
Applying R1 → R1 + R2, we have:
Expanding along C1, we have:
Hence, the given result is proved.
(ii) Let
.
Applying C1 → C1 − C3 and C2 → C2 − C3, we have:
Applying C1 → C1 + C2, we have:
Expanding along C1, we have:
Hence, the given result is proved.
Choose the correct answer.
Let A be a square matrix of order 3 × 3, then 
is equal to
A. 
B. 
C. 
D. 
Answer: C
A is a square matrix of order 3 × 3.
Hence, the correct answer is C.
By using properties of determinants, show that:
We have,
Here, the two rows R1 and R3 are identical.
∴Δ = 0.
Using the property of determinants and without expanding, prove that:
Here, the two rows R1 and R3 are identical.
Δ = 0.
Using the property of determinants and without expanding, prove that:
By applying C3 → C3 + C2, we have:
Here, two columns C1 and C3 are proportional.
Δ = 0.
Using the property of determinants and without expanding, prove that:
Applying R2 → R2 − R3, we have:
Applying R1 ↔R3 and R2 ↔R3, we have:
Applying R1 → R1 − R3, we have:
Applying R1 ↔R2 and R2 ↔R3, we have:
From (1), (2), and (3), we have:
Hence, the given result is proved.
By using properties of determinants, show that:
Applying R2 → R2 + R1 and R3 → R3 + R1, we have:
By using properties of determinants, show that:
(i) 
(ii) 
(i) Step 1
R1 = R1 + R2 + R3
Step 2
(5x+4) common in R1, bring that out, you are left with all 1s in R1
Applying C2 → C2 − C1, C3 → C3 − C1, we have:
Step 3
R2 = R2 - 2x(R1)
R3 = R3 - 2x(R1)
Now matrix is :
(5x+4)[(1,1,1),(0,4-x,0),(0,0,4-x)]
=(5x+4)(4-x)(4-x)
Expanding along C3, we have:
Hence, the given result is proved.
(ii) Here you need to subtract y(R1) from R2 and R3.
Applying R1 → R1 + R2 + R3, we have:
Applying C2 → C2 − C1 and C3 → C3 − C1, we have:
Expanding along C3, we have:
By using properties of determinants, show that:
(i) 
(ii) 
(i) 
Applying R1 → R1 + R2 + R3, we have:
Applying C2 → C2 − C1, C3 → C3 − C1, we have:
Expanding along C3, we have:
Hence, the given result is proved.
(ii) 
Applying C1 → C1 + C2 + C3, we have:
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
Expanding along R3, we have:
Hence, the given result is proved.
By using properties of determinants, show that:
Applying R1 → R1 + R2 + R3, we have:
Applying C2 → C2 − C1 and C3 → C3 − C1, we have:
Expanding along R1, we have:
Hence, the given result is proved.
By using properties of determinants, show that:
Applying R1 → R1 + bR3 and R2 → R2 − aR3, we have:
Expanding along R1, we have:
By using properties of determinants, show that:
Taking out common factors a, b, and c from R1, R2, and R3 respectively, we have:
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
Applying C1 → aC1, C2 → bC2, and C3 → cC3, we have:
Expanding along R3, we have:
Hence, the given result is proved.
Which of the following is correct?
A. Determinant is a square matrix.
B. Determinant is a number associated to a matrix.
C. Determinant is a number associated to a square matrix.
D. None of these
Answer: C
We know that to every square matrix, 
of order n. We can associate a number called the determinant of square matrix A, where 
element of A.
Thus, the determinant is a number associated to a square matrix.
Hence, the correct answer is C.
By using properties of determinants, show that:
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
Applying R3 → R3 + R2, we have:
Expanding along R3, we have:
Hence, the given result is proved.
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