Exercise 507Linear algebra › Matrices and linear systemsL1Added 3 Nov 2026 Inverse of a 2×2 matrix #inverse#determinant#2x2 matrices StatementCompute the inverse ofA=(3512).A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}.A=(3152). HintFor A=(abcd)A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}A=(acbd), if ad−bc≠0ad-bc \neq 0ad−bc=0 thenA−1=1ad−bc(d−b−ca).A^{-1} = \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.A−1=ad−bc1(d−c−ba). Show solutionHide solutionSolutionDeterminant. detA=3⋅2−5⋅1=6−5=1\det A = 3\cdot2 - 5\cdot1 = 6 - 5 = 1detA=3⋅2−5⋅1=6−5=1.Since detA≠0\det A \neq 0detA=0, AAA is invertible andA−1=11(2−5−13)=(2−5−13).A^{-1} = \frac{1}{1}\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}.A−1=11(2−1−53)=(2−1−53).Verification.AA−1=(3512)(2−5−13)=(6−5−15+152−2−5+6)=(1001).AA^{-1} = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 6-5 & -15+15 \\ 2-2 & -5+6 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.AA−1=(3152)(2−1−53)=(6−52−2−15+15−5+6)=(1001).