Solution. The two \(2\) by \(2\) determinants are immediate: \[2\cdot 4-1\cdot 3=5,\qquad (-2)(3)-5(-17)=-6+85=79.\] For the third, expand along the first row: \begin {align*} \begin {vmatrix}1&-1&1\\3&3&2\\-7&4&1\end {vmatrix} &=1\begin {vmatrix}3&2\\4&1\end {vmatrix} +1\begin {vmatrix}3&2\\-7&1\end {vmatrix} +1\begin {vmatrix}3&3\\-7&4\end {vmatrix}\\ &=(3-8)+(3+14)+(12+21)=-5+17+33=45. \end {align*}

For the fourth, choose the second column, which has two zeros — the whole point of being allowed to expand along any row or column: \[\begin {vmatrix}9&8&1\\-2&0&1\\3&0&-2\end {vmatrix} =-8\begin {vmatrix}-2&1\\3&-2\end {vmatrix}=-8(4-3)=-8.\]

Solution. Clear the first column using column operations, which by the determinant properties do not change the value: \[c_2\rightarrow c_2-c_1,\qquad c_3\rightarrow c_3-c_1\] gives \[\begin {vmatrix}1&0&0\\a&b-a&c-a\\a^2&b^2-a^2&c^2-a^2\end {vmatrix} =\begin {vmatrix}b-a&c-a\\b^2-a^2&c^2-a^2\end {vmatrix},\] expanding along the first row. Now factor \(b-a\) out of the first column and \(c-a\) out of the second, using \(b^2-a^2=(b-a)(b+a)\): \[=(b-a)(c-a)\begin {vmatrix}1&1\\b+a&c+a\end {vmatrix} =(b-a)(c-a)\big ((c+a)-(b+a)\big )=(b-a)(c-a)(c-b).\] This is the \(3\) by \(3\) case of the Vandermonde determinant. Notice what it says: the determinant vanishes exactly when two of \(a,b,c\) are equal, which is the algebraic form of the statement that three points with a repeated \(x\)-coordinate do not determine a unique parabola.

Solution. The coefficient determinant is \[\det A=\begin {vmatrix}2&-1&1\\1&1&3\\-3&4&-1\end {vmatrix}=-11\neq 0,\] so the system has a unique solution and Cramer’s rule applies. Replacing each column in turn by the right-hand side \((-1,0,5)^t\): \[\det A_1=\begin {vmatrix}-1&-1&1\\0&1&3\\5&4&-1\end {vmatrix}=-7,\qquad \det A_2=\begin {vmatrix}2&-1&1\\1&0&3\\-3&5&-1\end {vmatrix}=-17,\qquad \det A_3=\begin {vmatrix}2&-1&-1\\1&1&0\\-3&4&5\end {vmatrix}=8.\] Hence \[x=\frac {-7}{-11}=\frac {7}{11},\qquad y=\frac {-17}{-11}=\frac {17}{11},\qquad z=\frac {8}{-11}=-\frac {8}{11}.\] Checking in the second equation: \(\frac {7}{11}+\frac {17}{11}+3\left (-\frac {8}{11}\right ) =\frac {7+17-24}{11}=0\), as required.

Solution. The coefficient determinant factors as \[\det \begin {pmatrix}k&1&1\\1&k&1\\1&1&k\end {pmatrix}=(k-1)^2(k+2).\]

(a)

If \(k\neq 1\) and \(k\neq -2\) the determinant is non-zero, so there is a unique solution.

(c)

If \(k=1\) all three equations become \(x+y+z=1\): one equation in three unknowns, so infinitely many solutions.

(b)

If \(k=-2\) the three equations are \[-2x+y+z=1,\qquad x-2y+z=1,\qquad x+y-2z=1.\] Adding all three gives \(0=3\), which is impossible, so there is no solution.

Both roots of the determinant give a non-unique situation, but they differ, and the determinant alone cannot distinguish them — exactly the point made in the Section 3 problem on the parameter \(a\).

Solution. Expand along the third column, which has only one non-zero entry: \[\begin {vmatrix}\sin \alpha& \cos \alpha& 0\\-\cos \alpha& \sin \alpha& 0\\ \sin \alpha -\cos \alpha& \sin \alpha +\cos \alpha& 1\end {vmatrix} =1\cdot \begin {vmatrix}\sin \alpha& \cos \alpha \\-\cos \alpha& \sin \alpha \end {vmatrix} =\sin ^2\alpha +\cos ^2\alpha =1.\] The value is \(1\) for every \(\alpha \), so it certainly does not depend on \(\alpha \). The third row, which looks like the complicated part, contributes nothing at all — choosing the right column to expand along removes it before any trigonometry is needed.

Solution. Use the addition formula on the third column. For each row, \[\sin (\theta +\delta )=\sin \theta \cos \delta +\cos \theta \sin \delta ,\] so the third column is \[\cos \delta \cdot (\text {first column})+\sin \delta \cdot (\text {second column}).\] The third column is therefore a linear combination of the other two. Applying the column operation \[c_3\rightarrow c_3-\cos \delta \,c_1-\sin \delta \,c_2,\] which does not change the determinant, turns the third column into a column of zeros, and a determinant with a zero column is \(0\).

This is the general principle behind the question: a determinant vanishes as soon as one row or column is a linear combination of the others, whatever the entries happen to look like.

Solution. By Theorem 3.3.3 (or part (4) of the \(n\) by \(n\) determinant theorem) a square homogeneous system has a non-trivial solution exactly when the coefficient determinant is zero. Compute it by subtracting the first row from the second, which does not change the value: \[\begin {vmatrix}1&1&\alpha \\1&1&\beta \\\alpha& \beta& 1\end {vmatrix} =\begin {vmatrix}1&1&\alpha \\0&0&\beta -\alpha \\\alpha& \beta& 1\end {vmatrix}.\] Expanding along the second row, whose only non-zero entry is in position \((2,3)\) with sign \((-1)^{2+3}=-1\), \[=-(\beta -\alpha )\begin {vmatrix}1&1\\\alpha& \beta \end {vmatrix} =-(\beta -\alpha )(\beta -\alpha )=-(\beta -\alpha )^2.\] This is zero if and only if \(\beta =\alpha \), which is the assertion. Note that the first two equations differ only in the coefficient of \(z\); when \(\alpha =\beta \) they are the same equation, leaving two equations in three unknowns.

Solution. The proof of Theorem 4.2.10 established the identity \[A\,\text {adj}\,A=(\det A)\,I,\] and this holds for every square matrix, of any size. Take determinants of both sides. On the left, by the multiplicative property, \[\det \big (A\,\text {adj}\,A\big )=\det A\cdot \det (\text {adj}\,A).\] On the right, \((\det A)I\) is a scalar multiple of the \(n\) by \(n\) identity, and multiplying an \(n\) by \(n\) matrix by a scalar \(c\) multiplies its determinant by \(c^n\), so \[\det \big ((\det A)I\big )=(\det A)^n\det I=(\det A)^n.\] Equating, \[\det A\cdot \det (\text {adj}\,A)=(\det A)^n.\] Since \(A\) is non-singular, \(\det A\neq 0\) and we may divide by it: \[\det (\text {adj}\,A)=(\det A)^{n-1}.\] The step that is easy to get wrong is \(\det (cI)=c^n\) rather than \(c\): the scalar multiplies every one of the \(n\) rows.