Q. 48 B5.0( 3 Votes )

# Find the matrix A

Answer : We know that the two matrices are eligible for their product only when the number of columns of first matrix is equal to the number of rows of the second matrix.

The matrix given on the RHS of the equation is a 2×3 matrix and the one given on the LHS of the equation is a 2×3 matrix.

Therefore, A has to be a 2×2 matrix.

Let, So the given question becomes, Now we will multiply the two matrices on LHS, we get [as cij = ai1b1j + ai2b2j + … + ainbnj] To satisfy the above equality condition, corresponding entries of the matrices should be equal, i.e.,

a + 4b = – 7, 2a + 5b = – 8, 3a + 6b = – 9

c + 4d = 2, 2c + 5d = 4, 3c + 6d = 6

Now, a + 4b = – 7 a = – 7 – 4b…………(i)

2a + 5b = – 8

2( – 7 – 4b) + 5b = – 8 (by substituting the value of a from eqn(i))

– 14 – 8b + 5b = – 8

3b = – 14 + 8

b = – 2

Hence substitute the value of b in eqn(i), we get

a = – 7 – 4b

a = – 7 – 4( – 2) = – 7 + 8 = 1

a = 1

Now, c + 4d = 2 c = 2 – 4d…………(ii)

2c + 5d = 4

2(2 – 4d) + 5d = 4 by substituting the value of a from eqn(ii))

4 – 8d + 5d = 4

3d = 0

d = 0

Hence substitute the value of d in eqn(ii), we get

c = 2 – 4d

c = 2 – 4(0)

c = 2

Now substituting these values in matrix A, we get is the matrix A.

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