Let A=≤ft[{array}{cc}-3 & 2 \\ 1 & 4{array}] and if A²-2 A+I=≤ft[{array}{cc}18 & p \\ q & 11{array}], then ...

Correct Answer is : p=−2,q=−1p=-2, q=-1p=−2,q=−1

Correct Answer is : p=−2,q=−1p=-2, q=-1p=−2,q=−1

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MHTCET 18 Apr 2026 Shift 2 Paper

Question : 114 of 150

Marks: +1, -0

Let

A=\left[\begin{array}{cc}-3 & 2 \\ 1 & 4\end{array}\right]

A^{2}-2 A+I=\left[\begin{array}{cc}18 & p \\ q & 11\end{array}\right]

, then ...

Solution:

Concept:

For a given matrix

A

, compute

A^2

by matrix multiplication, then substitute into the expression

A^2 - 2A + I

and compare the corresponding entries of the matrices.

Explanation:

Given

A=\begin{bmatrix}-3 & 2\\ 1 & 4\end{bmatrix}

First, find

A^2 = A \cdot A

A^2=\begin{bmatrix}-3 & 2\\ 1 & 4\end{bmatrix}\begin{bmatrix}-3 & 2\\ 1 & 4\end{bmatrix}

=\begin{bmatrix}(-3)(-3)+(2)(1) & (-3)(2)+(2)(4)\\ (1)(-3)+(4)(1) & (1)(2)+(4)(4)\end{bmatrix}

=\begin{bmatrix}9+2 & -6+8\\ -3+4 & 2+16\end{bmatrix}=\begin{bmatrix}11 & 2\\ 1 & 18\end{bmatrix}

Next, compute

2A

2A=2\begin{bmatrix}-3 & 2\\ 1 & 4\end{bmatrix}=\begin{bmatrix}-6 & 4\\ 2 & 8\end{bmatrix}

The identity matrix is

I=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}

Now substitute into

A^2-2A+I

A^2-2A+I=\begin{bmatrix}11 & 2\\ 1 & 18\end{bmatrix}-\begin{bmatrix}-6 & 4\\ 2 & 8\end{bmatrix}+\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}

=\begin{bmatrix}11+6+1 & 2-4+0\\ 1-2+0 & 18-8+1\end{bmatrix}=\begin{bmatrix}18 & -2\\ -1 & 11\end{bmatrix}

Compare this with the given matrix

\begin{bmatrix}18 & p\\ q & 11\end{bmatrix}

Thus,

p=-2

and

q=-1

Answer:

p=-2

q=-1

, so the correct option is Option A.

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