Comprehension Based Questions: Matrices and Determinants - Free MCQ Practice

A diagonal matrix is a matrix in which all of the elements not on the diagonal of a square matrix are 0.

⇒ Sum of elements of  

Each element of set A is 3 × 3 symmetric matrix with five of its entries as 1 and four of its entries as 0, we can keep in diagonal either 2 zero and one 1 or no zero and three 1 so that the left over zeros and one’s are even in number.
Hence taking 2 zeros and one 1 in diagonal the possible cases are 
and taking 3 ones in diagonal the possible cases are 
∴ Total elements A can have = 9 + 3 = 12

The given system will have unique solution if | A |≠ 0 which is so for the matrices.

which are 6 in number.

For the given system to be inconsistent |A| = 0. The matrices for which | A | = 0 are

We find for A = (i)
By Cramer’s rule D₁ = 0 = D₂=D₃
∴ infinite many solution
For A  = (ii)
By Cramer ’s rule D₁ ≠ 0
⇒ no solution i.e. inconsistent.
Similarly we find the system as inconsistent in cases (iii), (v) and (vi).
Hence for four cases system is inconsistent.

If A is symmetrie then b = c
⇒ A = a² – b² = (a+b)(a –b)
Which is divisible by p if (a + b) is divisible by p or (a – b) is divisible by p.
Now (a + b) is divisible by p if (a, b) can take values (1, p – 1), (2, p – 2),(3, p – 3),...(p–1, 1)
∴ (p – 1)ways.
Also (a – b) is divisible by p only when a – b = 0 i.e. a = b, then (a, b) can take values (0,0) (1,1), (2,2)..... (pÝ, p–2)
∴ p ways.
If A is skew symmetric, then a = 0 and b = – c or b + c = 0 which gives |A| = 0 when b2
∴ b = 0, c = 0 But this possibility is already included when A is symmetric and (a, b) = (0, 0).
Again if A is both symmetric and skew symmetric, then clearly A is null matrix which case is already included.
Hence total number of ways = p + (p – 1) = 2p – 1

Trace A = a + a = 2a is not divisible by p ⇒ a is not divisible by p
⇒ a ≠ 0
But |A| is divisible by p
⇒ a² – bc is divisible by p
It will be so if on dividing a² by p suppose we get   then on dividing bc by p we should get  for some
integeral values of m, n and l. i.e. the remainder should be same in each case, so that 

For this to happen a can take any value from 1 to p–1, also if b takes any value from 1 to p–1 then c should take only that value corresponding to which the remainder is same.
∴ No. of ways = (p – 1) × (p – 1) × 1 = (p – 1)².

Total number of matrices = total number of ways a, b, c can be selected = p × p × p = p³.
Number of ways when det (A) is divisible by p and trace (A) ≠ 0 are equal to number of ways det (A) is divisible by p and trace (A) is not divisible by p = (p – 1)²

Also number of ways when det (A) is divisible by p and trace A = 0 are the ways when bc is multiple of p
⇒ b = 0 or c = 0 for b = 0, c can take values 0, 1, 2, ......., p – 1
For c = 0, b can take values 0, 1, 2, ............, p – 1
Here (b, c) = (0, 0) is coming twice.
∴ Total ways of selecting b and c = p + p – 1 = 2p – 1
∴ Total number of ways when det (A) is divisible by p = (p – 1)² + 2p – 1 = p² Hence the number of ways when det (A) is not divisible by p = p³ – p².

From equation (E), we get
a + 8b + 7c = 0
9a + 2b + 3c = 0
a + b + c  = 0

Therefore system has infinite many solutions.
Solving these, we get b = 6a and c = – 7a
Now (a, b, c) lies on 2x + y + z = 1
⇒ b = 6, c = – 7
∴  2a + 6a – 7a = 1 ⇒ a = 1
∴ 7a + b + c = 7 + 6 – 7 = 6 ⇒ b = 6, c = – 7

If a = 2 then b = 12, c = –14

If b = 6 then a = 1, c = – 7
∴ Equation becomes x² + 6x – 7 = 0 or (x + 7) (x – 1) = 0 whose roots are 1 and –7.
Let α = 1 and β = – 7

The given equations are
x – 2y + 3z = – 1
– x + y – 2z = k
x – 3y + 4z =1

∴ If k ≠ 3 , the system has no solutions.
Hence statement-1 is true and statement-2 is a correct explanation for statement - 1.