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MHTCET 20 Apr 2026 Shift 1 Paper

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Question : 116 of 150

Marks: +1, -0

If

A=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right], B=\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right]

X A=B^{T}

A^{-1} Y=B

X Y=

[-1]
[1]
[-2]
[2]

Solution:

Concept:

Use matrix inverses and associativity to simplify the product

XY

.

Explanation:

Given

XA = B^T

and

A^{-1}Y = B

.

Multiply

XA = B^T

on the right by

A^{-1}

to get

X = B^T A^{-1}

.

Multiply

A^{-1}Y = B

on the left by

A

to get

Y = AB

.

Now,

XY = (B^T A^{-1})(AB)

.

Using associativity,

XY = B^T (A^{-1}A) B

.

Since

A^{-1}A = I

, we get

XY = B^T I B = B^T B

.

Here,

B^T = \begin{bmatrix}1 & 0 & 1\end{bmatrix}

and

B = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}

.

Therefore,

B^T B = 1^2 + 0^2 + 1^2 = 2

.

So,

XY = [2]

.

Answer:

Option D:

[2]

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