Matrices & DeterminantsmediumFree

Matrices & Determinants — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

A

is a

2\times 2

matrix such that

A\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix}

and

A^2\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}

. The sum of the elements of

A

is:

-1

0

2

5

correct

Solution

Step 1: Obtain equations from the first condition Let

A = \begin{pmatrix}a & b\\ c & d\end{pmatrix}

. Then:

\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}a-b\\c-d\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix}

a - b = -1 \quad \text{and} \quad c - d = 2 \qquad \cdots (1)

Step 2: Use the second condition

A^2\begin{pmatrix}1\\-1\end{pmatrix} = A\!\left(A\begin{pmatrix}1\\-1\end{pmatrix}\right) = A\begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}

\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}-a+2b\\-c+2d\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}

-a + 2b = 1 \quad \text{and} \quad -c + 2d = 0 \qquad \cdots (2)

Step 3: Solve the system From

(1)

and

(2)

: adding

a-b=-1

and

-a+2b=1

gives

b = 0

, hence

a = -1

. Adding

c-d=2

and

-c+2d=0

gives

d = 2

, hence

c = 4

a + b + c + d = -1 + 0 + 4 + 2 = 5

Answer: (4)

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