Matrices
A square matrix \(A\) of order \(n\) is invertible if there is a matrix \(B\) of the same order with
\[AB = BA = I_n.\]
Then \(B\) is called the inverse of \(A\), written \(A^{-1}\). If an inverse exists it is unique: if \(B\) and \(C\) are both inverses, \(B = BI = B(AC) = (BA)C = IC = C\).
In the next chapter you will see that a square matrix is invertible exactly when its determinant is non-zero.
\[\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\]
Valid only when \(ad - bc \ne 0\). Swap the diagonal, negate the off-diagonal, divide by the determinant.
\[(AB)^{-1} = B^{-1}A^{-1}\]
Also \((A^{-1})^{-1} = A\) and \((A^T)^{-1} = (A^{-1})^T\).
Inverse needs AB = BA = I, which requires square matrices of the same order.
If \(A^2-3A+2I=O\) then \(A(3I-A)=2I\), so \(A^{-1}=\tfrac12(3I-A)\). Multiply the equation by \(A^{-1}\) whenever the constant term is non-zero.
Swapping the diagonal entries but forgetting to change the signs of the off-diagonal entries (or vice versa).
Correct: swap \(a\) and \(d\), change the signs of \(b\) and \(c\), then divide by \(ad - bc\). Verify with \(AA^{-1} = I\).
Find the inverse of \(A = \begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\) and verify.
Final Answer: \(A^{-1} = \begin{bmatrix} 5 & -2 \\ -7 & 3 \end{bmatrix}\)
A fresh set of questions from this topic. Change answers freely – solutions appear after you submit.
Start Practice\(B = A^{-1} \iff AB = BA = I\). Inverse is unique. \((AB)^{-1} = B^{-1}A^{-1}\). 2×2 inverse: swap diagonal, negate off-diagonal, divide by \(ad - bc\).
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