Matrices flash cards

1. 1.For a square matrix $A$ of order $n$, what is the relation between $adj(A)$, $A$, and $|A|$, and what is $|adj(A)|$ in terms of $|A|$?

   $A\cdot adj(A) = adj(A)\cdot A = |A|\, I_n$, and $|adj(A)| = |A|^{n-1}$.

   Hint: Think of the adjugate identity used to derive $A^{-1}$.

2. 2.How is the inverse of a square matrix $A$ defined in terms of $|A|$ and $adj(A)$, and what is the necessary condition for $A^{-1}$ to exist?

   $A^{-1} = \dfrac{1}{|A|}\,adj(A)$, which exists iff $|A|\neq 0$ (i.e., $A$ is non-singular).

   Hint: What happens when the determinant is zero?

3. 3.For a symmetric matrix $A$ ($A^T=A$) and a skew-symmetric matrix $B$ ($B^T=-B$), state one key property each about their diagonal elements and about $A^2, B^2$.

   Every diagonal element of a skew-symmetric matrix is $0$ (since $a_{ii} = -a_{ii} \Rightarrow a_{ii}=0$); diagonal elements of a symmetric matrix can be anything. Also, both $A^2$ and $B^2$ are symmetric: $(A^2)^T = (A^T)^2 = A^2$, and $(B^2)^T = (B^T)^2 = (-B)^2 = B^2$.

   Hint: Recall $a_{ii} = -a_{ii}$ for skew-symmetric.

4. 4.Any square matrix $A$ can be uniquely expressed as the sum of a symmetric and a skew-symmetric matrix. Write this decomposition.

   $A = \dfrac{1}{2}(A+A^T) + \dfrac{1}{2}(A-A^T)$, where the first term is symmetric and the second is skew-symmetric.

   Hint: Add and subtract $A^T$.

5. 5.For an orthogonal matrix $A$ (i.e., $A^TA = AA^T = I$), what are the possible values of $|A|$, and how is $A^{-1}$ related to $A^T$?

   $|A| = \pm 1$ and $A^{-1} = A^T$.

   Hint: Take determinant of $A^TA=I$.

6. 6.If $A$ is a $3\times3$ matrix with $|A|=4$, find $|adj(3A)|$.

   Use $adj(kA)=k^{n-1}adj(A)$ for $n=3$, so $adj(3A)=9\,adj(A)$. Then $|adj(3A)|=9^3|adj(A)|=729\cdot|A|^{2}=729\times16=\mathbf{11664}$.

   Hint: Scale the adjoint formula first, then the determinant scales by $9^3$.

7. 7.For an invertible $3\times3$ matrix $A$, prove $adj(A^{-1})=(adj\,A)^{-1}$ and find $|adj(A^{-1})|$ in terms of $|A|$.

   Since $A\,adjA=|A|I$, we get $adjA=|A|A^{-1}$. Replacing $A$ by $A^{-1}$: $adj(A^{-1})=|A^{-1}|(A^{-1})^{-1}=\dfrac{A}{|A|}=(adjA)^{-1}$. Also $|adj(A^{-1})|=|A^{-1}|^{n-1}=|A|^{-(n-1)}=\mathbf{|A|^{-2}}$ for $n=3$.

   Hint: Start from $A\,adjA=|A|I$ and substitute $A\to A^{-1}$.

8. 8.If $A,B$ are invertible $3\times3$ matrices, express $adj(AB)$ in terms of $adjA$ and $adjB$.

   Since $(AB)(adjB\,adjA)=A(B\,adjB)adjA=A(|B|I)adjA=|B|(A\,adjA)=|B||A|I=|AB|I$, uniqueness of the adjugate (for invertible matrices) gives $adj(AB)=\mathbf{adjB\cdot adjA}$ — the order reverses, just like $(AB)^{-1}=B^{-1}A^{-1}$.

   Hint: Multiply $AB$ by the candidate $adjB\,adjA$ and see what you get.

9. 9.For $A=\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}$, compute $A^{-1}$ using the adjugate method.

   Expanding, $|A|=2(4-1)-(-1)(-2-0)+0=6-2=4$. Computing all nine cofactors gives $adjA=\begin{bmatrix}3&2&1\\2&4&2\\1&2&3\end{bmatrix}$ (symmetric, matching $A$). Hence $A^{-1}=\dfrac{1}{4}adjA=\begin{bmatrix}3/4&1/2&1/4\\1/2&1&1/2\\1/4&1/2&3/4\end{bmatrix}$, verified by $A\cdot adjA=4I$.

   Hint: Compute the determinant, then all nine 2x2 cofactors carefully.

10. 10.For a $3\times3$ matrix $A$ with $|A|=2$, find $|adj(adj(adj\,A))|$.

    $|adjA|=|A|^{2}$. Then $|adj(adjA)|=|adjA|^{2}=|A|^{4}$. Applying the rule once more, $|adj(adj(adjA))|=|adj(adjA)|^{2}=|A|^{8}$. With $|A|=2$, this equals $2^8=\mathbf{256}$.

    Hint: Apply $|adjM|=|M|^{n-1}$ repeatedly, squaring the exponent each time.

Open the interactive deck for the other 75 cards, with self-grading so the ones you keep missing come back.