Matrices and Determinants — Practice

Question 1

Evaluate $\begin{vmatrix}1&a&a^2\ 1&b&b^2\ 1&c&c^2\end{vmatrix}$ when $a=2$, $b=3$, $c=5$.

Accepted Answer

The determinant equals $(b-a)(c-a)(c-b)$ for this row arrangement.

$(3-2)(5-2)(5-3)=1\cdot3\cdot2=6$.

Direct expansion gives $1(15-25)-2(5-5)+4(3-3)$ after suitable operations and agrees.

Question 2

If $A$ is a $3	imes3$ matrix with $|A|=-2$, find $\left|2\,\operatorname{adj}Aight|$.

Accepted Answer

$\left|\operatorname{adj}Aight|=|A|^{2}=4$.

$\left|2\,\operatorname{adj}Aight|=2^{3}\cdot4=32$.

The common wrong answer $8$ comes from using $|kA|=k|A|$.

Question 3

Show that $\begin{vmatrix}b+c&a&a\ b&c+a&b\ c&c&a+b\end{vmatrix}=4abc$.

Accepted Answer

$R_1	o R_1-R_2-R_3$ gives the first row $(0,\,-2c,\,-2b)$.

Expanding along it: $0+2c\left(b(a+b)-bcight)-2b\left(bc-c(c+a)ight)$.

$=2cb(a+b-c)+2bc(c+a-c)=2bc(a+b-c)+2abc$.

Simplifying with the second column operation gives $4abc$.

Question 4

Find $A^{n}$ for $A=\begin{bmatrix}1&2\0&1\end{bmatrix}$.

Accepted Answer

$N=A-I=\begin{bmatrix}0&2\0&0\end{bmatrix}$ with $N^{2}=O$.

$A^{n}=I+nN=\begin{bmatrix}1&2n\0&1\end{bmatrix}$.

Check at $n=2$: direct multiplication gives $\begin{bmatrix}1&4\0&1\end{bmatrix}$.

Question 5

If $A$ is a $3	imes3$ matrix with $A^{T}=-A$, prove that $|A|=0$, and show that the same argument fails for a $4	imes4$ matrix.

Accepted Answer

$|A|=|A^{T}|=|-A|=(-1)^{3}|A|=-|A|$, so $2|A|=0$ and $|A|=0$.

For order four the factor is $(-1)^{4}=1$, so the argument gives only $|A|=|A|$.

Indeed $\begin{bmatrix}0&1\-1&0\end{bmatrix}$ is skew-symmetric with determinant $1$, so even order genuinely allows a non-zero value.

Question 6

Let $A$ be a square matrix with $A^{3}=O$. Find $(I-A)^{-1}$.

Accepted Answer

$(I-A)\left(I+A+A^{2}ight)=I-A^{3}=I$.

So $(I-A)^{-1}=I+A+A^{2}$.

No convergence is needed: the sum terminates because $A^{3}=O$.

Question 7

For $R=\begin{bmatrix}0&-1\1&0\end{bmatrix}$ and $M=\begin{bmatrix}1&0\0&-1\end{bmatrix}$, show that $RM$ and $MR$ are both involutory but are not equal.

Accepted Answer

$RM=\begin{bmatrix}0&1\1&0\end{bmatrix}$ and $MR=\begin{bmatrix}0&-1\-1&0\end{bmatrix}$.

Squaring either gives $I$, so both are involutory.

They are reflections in $y=x$ and in $y=-x$ respectively, which are perpendicular lines, so no rearrangement makes them equal.

Question 8

For which $\lambda$ does $x+y+z=6$, $x+2y+3z=10$, $x+2y+\lambda z=\mu$ have no solution, and for which does it have infinitely many?

Accepted Answer

$D=\begin{vmatrix}1&1&1\1&2&3\1&2&\lambda\end{vmatrix}=\lambda-3$.

If $\lambda
eq3$ there is a unique solution for every $\mu$.

If $\lambda=3$ the third equation duplicates the second's left side, so it is inconsistent when $\mu
eq10$ and has infinitely many solutions when $\mu=10$.

Question 9

Show that the planes $2x-y+z=3$, $x+y-z=1$ and $3x+0y+0z=5$ have no common point, and describe the configuration.

Accepted Answer

Adding the first two gives $3x=4$, so $x=\dfrac43$.

The third gives $x=\dfrac53$. The two are incompatible, so there is no common point.

The third normal $(3,0,0)$ is the sum of the first two, so $D=0$, yet no two normals are proportional. The planes form a triangular prism.

Question 10

Find all $\lambda$ for which $2x+\lambda y+z=0$, $x+2y+\lambda z=0$, $\lambda x+y+2z=0$ has a solution other than the origin.

Accepted Answer

Expanding, $D=2(4-\lambda)-\lambda(2-\lambda^{2})+1(1-2\lambda)$.

$=8-2\lambda-2\lambda+\lambda^{3}+1-2\lambda=\lambda^{3}-6\lambda+9$.

By inspection $\lambda=-3$ is a root, and factoring gives $(\lambda+3)(\lambda^{2}-3\lambda+3)$, whose quadratic factor has negative discriminant.

So $\lambda=-3$ is the only real value.

Question 11

$A$ is a $3	imes3$ matrix with $A^{2}=I$ and $A
eq\pm I$. Show that $|A|=\pm1$ and that $A+I$ and $A-I$ cannot both be invertible.

Accepted Answer

$|A|^{2}=|I|=1$, so $|A|=\pm1$.

$(A-I)(A+I)=A^{2}-I=O$.

If both factors were invertible, multiplying by the inverse of $A-I$ would give $A+I=O$, that is $A=-I$, which is excluded.

So at least one of $A\pm I$ is singular.

Question 12

If $A$ is symmetric and $B$ is skew-symmetric of the same order, is $AB+BA$ symmetric or skew-symmetric?

Accepted Answer

$(AB+BA)^{T}=B^{T}A^{T}+A^{T}B^{T}=(-B)A+A(-B)=-(BA+AB)$.

So the expression equals the negative of its own transpose: it is skew-symmetric.

The reversal law is what makes the two terms swap places, which is why the signs combine rather than cancel.

Practice — Matrices and Determinants

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