Question

Find the inverse of the matrix [6, 7] [-6, 7] if it exists

Answer 1

For any square matrix A, if it is invertible, its inverse is given by

`A^(-1) = \frac1{\det(A)} \mathrm{adj}(A)`

where adj(A) is the so-called adjugate matrix, which is the transpose of the cofactor matrix of A. If you don't know what that is, that's not terribly important. What is important is that the inverse does not exist if the determinant, det(A), is zero.

We have

`A=\begin{bmatrix}6&7\\-6&7\end{bmatrix} \implies \det(A) = 6*7-7*(-6)=84 \\eq 0`

so the inverse does indeed exist. (And we eliminate D as an answer.)

From the given choices, it's quite clear that C must the correct answer, since we know det(A) = 84, and we can easily confirm that

`\begin{bmatrix}7&-7\\6&6\end{bmatrix} \begin{bmatrix}6&7\\-6&7\end{bmatrix} = \begin{bmatrix}84&0\\0&84\end{bmatrix}`

so that multiplying by 1/84 recovers the identity matrix.

Answer 2

We are given with a square matrix and we are asked to find it's inverse if it exists . So , let's recall some important points first :-

• Inverse can only be found of square matrices.

• If
  `{\bf A}` is a square matrix than , it's inverse is denoted by
  `{\bf A^(-1)}` , and is given by
  `{\bf{A^(-1)=(1)/(adj(A))det(A)}}` , where adj(A) is the adjoint of the matrix A and det(A) is the determinant of the matrix A

• Inverse of a matrix exist if A is non-singular

• Non-Singular means that det(A) ≠ 0

• Adjoint of a matrix is the matrix of transpose of cofactors

• Transpose of a matrix is founded by exchanging it's rows by columns and columns by rows and is denoted by
  `{\bf{A^(T)}}`

Now , in this question let's assume that
`{\sf A=\begin{bmatrix}6 & 7 \\ -6 & 7\end{bmatrix}}`

Now , Calculating det(A) :-

`{:\implies \quad \sf det(A)=\begin{vmatrix}6 & 7 \\ -6 & 7 \end{vmatrix}}`

`{:\implies \quad \sf det(A)=42-(-42)}`

`{:\implies \quad \sf det(A)=42+42}`

`{:\implies \quad \sf det(A)=84}`

As , det(A) ≠ 0 . So ,
`{\bf A^(-1)}` exists . Now , we need to find the matrix of cofactors first , but let's find cofactors first , so here ;

`{\blacktriangleright \sf C_(11)=7,\: C_(12)=6,\: C_(21)=-7,\: C_(22)=6}`

Now , let's assume that matrix of cofactors is C , so putting the cofactors as elements of the matrix , C will be ;

`{:\implies \quad \sf C=\begin{bmatrix}7 & 6 \\ -7 & 6\end{bmatrix}}`

Now , adj(A) will be found by interchanging it's rows by columns and vice versa.

`{:\implies \quad \sf adj(A)=\begin{bmatrix}7 & -7 \\ 6 & 6\end{bmatrix}}`

Now as
`{\sf A^(-1)}` is given by
`{\bf{A^(-1)=(1)/(adj(A))det(A)}}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{A^(-1)=(1)/(84)\begin{bmatrix}7 & -7 \\ 6 & 6\end{bmatrix}}}}`

Hence , Option C)
`{\sf (1)/(84)\begin{bmatrix}7 & -7 \\ 6 & 6\end{bmatrix}}` is correct :D