Determinant flash cards

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

   $|adj(A)| = |A|^{n-1}$, and also $adj(adj(A)) = |A|^{n-2}A$ (for $n \geq 2$, and provided $A$ is invertible).

   Hint: Think about the order of the matrix as the exponent.

2. 2.State the condition on the determinant of coefficient matrix $\Delta$ for a system of 3 linear equations in 3 unknowns (using Cramer's Rule) to have a unique solution.

   Unique solution exists iff $\Delta \neq 0$, given by $x=\dfrac{\Delta_1}{\Delta}, y=\dfrac{\Delta_2}{\Delta}, z=\dfrac{\Delta_3}{\Delta}$.

   Hint: Non-zero determinant means matrix is invertible.

3. 3.For a system $AX=B$ with $|A|=0$, what are the two possible outcomes based on $\Delta_1,\Delta_2,\Delta_3$?

   If $\Delta=0$ and all $\Delta_1=\Delta_2=\Delta_3=0$: infinite solutions or no solution (system may be consistent or inconsistent); if $\Delta=0$ and at least one $\Delta_i \neq 0$: no solution.

   Hint: Consider consistency of the equations.

4. 4.What is the area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ in determinant form, and what does zero area imply?

   $\text{Area}=\dfrac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{vmatrix}\right|$; zero value implies the three points are collinear.

   Hint: A degenerate triangle has no area (take the modulus since area cannot be negative).

5. 5.How is a skew-symmetric determinant of odd order evaluated, and what is the formula for $A^{-1}$ in terms of $adj(A)$?

   A skew-symmetric determinant of odd order equals $0$; also $A^{-1}=\dfrac{1}{|A|}adj(A)$, valid only when $|A|\neq 0$.

   Hint: Recall $A^T=-A$ and its effect on odd-order determinants.

6. 6.If $\Delta=\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}$ and $\Delta'$ is formed by replacing $R_2$ with $R_2+\lambda R_1+\mu R_3$, how does $\Delta'$ compare to $\Delta$?

   $\Delta'=\Delta$. By multilinearity in $R_2$, $\Delta' = \det(R_1,R_2,R_3)+\lambda\det(R_1,R_1,R_3)+\mu\det(R_1,R_3,R_3)$. The last two determinants have a repeated row, so they vanish, leaving $\Delta'=\Delta$.

   Hint: Split the determinant using linearity in $R_2$; repeated-row terms vanish.

7. 7.If $A$ is a $3\times3$ matrix with $|A|=5$, find $|2A|$ and $|-A|$.

   For an $n\times n$ matrix, $|kA|=k^n|A|$ since each of the $n$ rows is scaled by $k$. Here $n=3$: $|2A|=2^3\cdot5=40$ and $|-A|=(-1)^3\cdot5=-5$.

   Hint: Scaling every row of a $3\times3$ matrix by $k$ multiplies the determinant by $k^3$.

8. 8.If $R_1=(a+p,\;b+q,\;c+r)$ in a $3\times3$ determinant with other rows $(x,y,z)$ and $(l,m,n)$ unchanged, express the determinant as a sum of two determinants.

   By linearity of the determinant in a single row, $\begin{vmatrix}a+p&b+q&c+r\\x&y&z\\l&m&n\end{vmatrix}=\begin{vmatrix}a&b&c\\x&y&z\\l&m&n\end{vmatrix}+\begin{vmatrix}p&q&r\\x&y&z\\l&m&n\end{vmatrix}$, since each entry of $R_1$ splits and the other two rows stay fixed.

   Hint: A determinant is linear in each row taken separately, so split entry-wise.

9. 9.$\Delta=\begin{vmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix}=5$. Find $\begin{vmatrix}b_2&a_2&c_2\\b_1&a_1&c_1\\b_3&a_3&c_3\end{vmatrix}$.

   First swap columns 1 and 2 of $\Delta$ (rows become $(b_1,a_1,c_1),(b_2,a_2,c_2),(b_3,a_3,c_3)$), flipping the sign to $-5$. Then swap rows 1 and 2 of that result to get $(b_2,a_2,c_2),(b_1,a_1,c_1),(b_3,a_3,c_3)$, flipping the sign again to $+5$. The two sign changes cancel, so the answer is $\mathbf{5}$, not $-5$.

   Hint: Track two separate sign-flipping operations — they may cancel out.

10. 10.For a $3\times3$ matrix $A$, if $\text{adj}(A)=O$ (the zero matrix) but $A\neq O$, what can you conclude about $|A|$ and $\text{rank}(A)$?

    Since $A\cdot\text{adj}(A)=|A|I$, if $\text{adj}(A)=O$ then $|A|I=O$, forcing $|A|=0$. Moreover, $\text{adj}(A)=O$ for a $3\times3$ matrix happens precisely when every $2\times2$ minor of $A$ vanishes, i.e. $\text{rank}(A)\le1$. So $A$ is singular with rank at most 1.

    Hint: Use $A\,\text{adj}A=|A|I$ and think about when every $2\times2$ minor is zero.

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