Find the inverse of the matrix \(\begin{bmatrix} 1& -3 \\ 4& 2 \end{b

1.	Find the inverse of the matrix \(\begin{bmatrix} 1& -3 \\ 4& 2 \end{bmatrix}\)1. \({1\over14}\begin{bmatrix} 2& -3 \\ 4& -1 \end{bmatrix}\)2. \({1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\)3. \({1\over14}\begin{bmatrix} 2& -3 \\ -4& 1 \end{bmatrix}\)4. \({1\over14}\begin{bmatrix} 2& 3 \\ 4& 1 \end{bmatrix}\)
Answer» Correct Answer - Option 2 : \({1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\) Concept: For any matrix A, If matrix A = [aij] A-1 = \(\rm \text{(adj A)}\over\|A\|\) The adj A matrix is the transpose of the matrix [Aij] is the cofactor of the element aij. Calculation: A = \(\begin{bmatrix} 1& -3 \\ 4& 2 \end{bmatrix}\) = [aij] \|A\| = \(\begin{vmatrix} 1& -3 \\ 4& 2 \end{vmatrix}\) ⇒ \|A\| = 2 × 1 - (-3) × 4 ⇒ \|A\| = 14 Now, finding cofactor of matrix a₁₁ = 1, cof(a11) = 2 a₁₂ = -3, cof(a12) = -4 a₂₁ = 4, cof(a21) = 3 a₂₂ = 2, cof(a22) = 1 Cof(A) = \(\begin{bmatrix} 2& -4 \\ 3& 1 \end{bmatrix}\) adj(A) = [cof(A)]^T adj(A) = \(\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\) A^-1 = \(\rm \text{(adj A)}\over\|A\|\) ⇒ A^-1 = \(\boldsymbol{{1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}}\)

Find the inverse of the matrix \(\begin{bmatrix} 1& -3 \\ 4& 2 \end{bmatrix}\)1. \({1\over14}\begin{bmatrix} 2& -3 \\ 4& -1 \end{bmatrix}\)2. \({1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\)3. \({1\over14}\begin{bmatrix} 2& -3 \\ -4& 1 \end{bmatrix}\)4. \({1\over14}\begin{bmatrix} 2& 3 \\ 4& 1 \end{bmatrix}\)

Answer» Correct Answer - Option 2 : \({1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}\)

Concept:

For any matrix A, If matrix A = [aij]

A-1 = \(\rm \text{(adj A)}\over|A|\)

The adj A matrix is the transpose of the matrix [Aij] is the cofactor of the element aij.

Calculation:

A = \(\begin{bmatrix} 1& -3 \\ 4& 2 \end{bmatrix}\) = [aij]

|A| = \(\begin{vmatrix} 1& -3 \\ 4& 2 \end{vmatrix}\)

⇒ |A| = 2 × 1 - (-3) × 4

⇒ |A| = 14

Now, finding cofactor of matrix

a₁₁ = 1, cof(a11) = 2

a₁₂ = -3, cof(a12) = -4

a₂₁ = 4, cof(a21) = 3

a₂₂ = 2, cof(a22) = 1

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

adj(A) = [cof(A)]^T

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

A^-1 = \(\rm \text{(adj A)}\over|A|\)

⇒ A^-1 = \(\boldsymbol{{1\over14}\begin{bmatrix} 2& 3 \\ -4& 1 \end{bmatrix}}\)