Test Index

Matrices and Determinants Part 1

Section: Mathematics

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Question : 9 of 100

Marks: +1, -0

If

A=\begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix}

I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},

10A^{-1}

[8 Jan 2020 Shift 2]

$4I - A$
$A-6I$
$6I-A$
$A-4I$

Solution:

Given

A=\begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix}

and

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

.

|A| = (2)(4) - (2)(9) = 8 - 18 = -10

.

\operatorname{adj} A = \begin{pmatrix} 4 & -2 \\ -9 & 2 \end{pmatrix}

.

A^{-1} = \dfrac{\operatorname{adj} A}{|A|} = \dfrac{\begin{pmatrix} 4 & -2 \\ -9 & 2 \end{pmatrix}}{-10} = \begin{pmatrix} -\frac{2}{5} & \frac{1}{5} \\ \frac{9}{10} & -\frac{1}{5} \end{pmatrix}

.

Multiplying by

10

:

10A^{-1} = \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix}

.

Now compute the candidate

A-6I

:

A - 6I = \begin{pmatrix} 2-6 & 2 \\ 9 & 4-6 \end{pmatrix} = \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix}

.

Since this matches

10A^{-1}

exactly,

10A^{-1} = A - 6I

.

Therefore the correct option is

A-6I

.

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