Question

Let A be a 6 × 5 matrix with entries in R and B be a 5 × 4 matrix with entries in R. Consider the following two statements.

P: For all such nonzero matrices A and B, there is a nonzero matrix Z such that AZB is the 6 × 4 zero matrix.

Q: For all such nonzero matrices A and B, there is a nonzero matrix Y such that BYA is the 5 × 5 zero matrix.

Which one of the following holds?

  1. Both P and Q are true
  2. P is true but Q is false
  3. P is false but Q is true
  4. Both P and Q are false

Answer: Both P and Q are true

Solution :-

Given that order of matrix A is 6×5 and order of matrix B is 5×4 and also in statement ‘P’ given that the order of AZB is 6×4. Then matrix multiplication AZB is possible only if order of Z is 5×5. Because A6×5Z5×5B5×4=(AZ)6×5B5×4=(AZB)6×4.
So, order of Z is 5×5.

Here, in general we can observe that the order of A is n×(n-1) and the order of B is (n-1)×(n-2) and order of AZB is n×(n-2) then matrix multiplication AZB is possible only when the order of Z is (n-1)×(n-1).

Now, if statement P is true for lower order matrix then it is also true for the higher order matrices.

So, we have to prove this statement first for lower order and then it can be proved in the same manner for higher order.

Now, Let A is 3×2 order and B is 2×1 order matrices then Z is 2×2 order and AZB is 3×1 order matrices according to above observation.

Now, let A=[a11a12a21a22a31a32]3×2,B=[b11b21]2×1andZ=[z11z12z21z22]2×2. Here, we are interested in finding the matrix Z2×2 which makes matrix (AZB)3×1 zero matrix.

So,AZB=[000]3×1[a11a12a21a22a31a32][z11z12z21z22][b11b21]=[000][a11z11+a12z21a11z12+a12z22a21z11+a22z21a21z12+a22z22a31z11+a32z21a31z12+a32z22][b11b21]=[000][(a11b11)z11+(a12b11)z21+(a11b21)z12+(a12b21)z22(a21b11)z11+(a22b11)z21+(a21b21)z12+(a22b21)z22(a31b11)z11+(a32b11)z21+(a31b21)z12+(a32b21)z22]=[000] So,(a11b11)z11+(a12b11)z21+(a11b21)z12+(a12b21)z22=0(1)(a21b11)z11+(a22b11)z21+(a21b21)z12+(a22b21)z22=0(2)(a31b11)z11+(a32b11)z21+(a31b21)z12+(a32b21)z22=0(3)

Now, solve this system of linear equation for z11,z21,z12,z22.
So, for finding solution of this system, the augmented matrix is

[a11b11a12b11a11b21a12b210a21b11a22b11a21b21a22b210a31b11a32b11a31b21a32b210]

Applying R1R1a11b11

[1a12a11b21b11a12b21a11b110a21b11a22b11a21b21a22b210a31b11a32b11a31b21a32b210]

Applying R2R2(a21b11)R1andR3R3(a31b11)R1

[1a12a11b21b11a12b21a11b1100a22b11a12a21b11a110a22b21a21a12b22a1100a32b11a12a31b11a110a32b21a12a31b21a110]

Applying R2R2(a22b11a12a21b11a11)

[1a12a11b21b11a12b21a11b110010b21b1100a32b11a12a31b11a110a32b21a12a31b21a110]

Applying R3R3(a32b11a12a31b11a11)R2

[1a12a11b21b11a12b21a11b110010b21b11000000]

So, z11+a12a11z21+b21b11z12+a12b21a11b11z22=0(4)

and z21+b21b11z22=0z21=b21b11z22.

Puting the value of z21 in eq(4), we get

z11a12b21a11b11z22+b21b11z12+a12b21a11b11z22=0z11=b21b11z12.

Hence, Z=[b21b11z12z12b21b11z22z22]

Therefore, for any matrix A=[a11a12a21a22a31a32]3×2 and B=[b11b21]2×1 we find matrix Z=[b21b11z12z12b21b11z22z22]2×2 which makes matrix AZB of order 3×1, zero matrix.

Let’s check whether it is true or not.

So,AZB=[a11a12a21a22a31a32][b21b11z12z12b21b11z22z22][b11b21]=[a11b21b11z12a12b21b11z22a11z12+a12z22a21b21b11z12a22b21b11z22a21z12+a22z22a31b21b11z12a32b21b11z22a31z12+a32z22][b11b21]=[a11b21z12a12b21z22+a11b21z12+a12b21z22a21b21z12a22b21z22+a21b21z12+a22b21z22a31b21z12a32b21z22+a21b21z12+a22b21z22]=[000]3×1

Therefore, this same process we can apply for the matrix A of order 6×5 and B of order 5×4 than we can find matrix Z of order 5×5 which makes AZB of order 6×4, zero matrix.

So, the statement ‘P’ is true.

In Short if A and B is matrix of any order and if matrix multiplication AZB or BZA is possible then there is matrix Z which makes AZB or BZA, zero matrix.

In same manner we can see that Statement ‘Q’ is also true.
Matrix multiplication BYA is possible only when order of Y is 4×6.
Here, B5×4Y4×6A6×5=(BY)5×6A6×5=(BYA)5×5 .
Hence, matrix multiplication is possible then we can find matrix Y which makes BYA of zero matrix.

So, the correct answer is option-1, Both P and Q are true