M301 Linear Algebra · Week 8

Properties of Determinants, Cramer's Rule, Volume

Section 3.2: Properties of determinants · Section 3.3: Cramer's rule, volume, and linear transformations

By the end of this week you will be able to:

compute a determinant by row reduction and say how each row operation changes it;
decide whether a matrix is invertible from its determinant and use \(\det(AB)=(\det A)(\det B)\);
solve small linear systems with Cramer's rule and find an inverse with the adjoint formula;
compute areas of parallelograms and volumes of parallelepipeds with determinants;
explain why a linear transformation multiplies areas and volumes by \(|\det A|\).

Day 1: Properties of Determinants

Today we see how row operations change a determinant, and we use this to compute determinants quickly and to test whether a matrix is invertible.

Facts used today

For a \(2\times 2\) matrix, \(\det\begin{bmatrix} a & b\\ c & d\end{bmatrix} = ad-bc\).
A determinant can be computed by a cofactor expansion across any row or down any column.
If \(A\) is triangular, \(\det A\) is the product of the entries on the main diagonal.
The three elementary row operations are: replacement (add a multiple of one row to another row), interchange (swap two rows) and scaling (multiply a row by a nonzero number).

1.1 Row operations and the determinant

Cofactor expansion is easy only when a matrix has many zeros. For a general matrix we first create zeros with row operations. To do this safely, we must know how each row operation changes the determinant.

Theorem 3 (Row operations) Let \(A\) be a square matrix.

(a) If a multiple of one row of \(A\) is added to another row to produce \(B\), then \(\det B = \det A\).

(b) If two rows of \(A\) are interchanged to produce \(B\), then \(\det B = -\det A\).

Example 1 (the three rules on a small matrix)

Let \(A=\begin{bmatrix} 1 & 2\\ 3 & 4\end{bmatrix}\), so \(\det A = 4-6=-2\).

(a) Replacement \(R_2\to R_2-3R_1\): \(\det\begin{bmatrix} 1 & 2\\ 0 & -2\end{bmatrix} = -2\). The determinant does not change.

(b) Interchange \(R_1\leftrightarrow R_2\): \(\det\begin{bmatrix} 3 & 4\\ 1 & 2\end{bmatrix} = 6-4 = 2 = -(-2)\). The sign changes.

(c) Scaling \(R_1\to 5R_1\): \(\det\begin{bmatrix} 5 & 10\\ 3 & 4\end{bmatrix} = 20-30 = -10 = 5\cdot(-2)\). The determinant is multiplied by 5.

In practice we use Theorem 3 in the following way. We reduce \(A\) to an echelon form \(U\). An echelon form of a square matrix is upper triangular, so \(\det U\) is the product of its diagonal entries. Then we go back to \(\det A\) using the rules above.

Be careful with scaling

Rule (c) is usually used to factor out a common number from one row:

\[\det\begin{bmatrix} * & * & *\\ 5k & -2k & 3k\\ * & * & *\end{bmatrix} = k\cdot\det\begin{bmatrix} * & * & *\\ 5 & -2 & 3\\ * & * & *\end{bmatrix}.\]

If you divide a row by \(k\) to simplify it, the factor \(k\) must appear in front of the new determinant. Only one row is affected; the other rows stay as they are.

Example 2 (reduction to echelon form)

Compute \(\det A\), where \(A=\begin{bmatrix} 1 & -4 & 2\\ -2 & 8 & -9\\ -1 & 7 & 0\end{bmatrix}\).

Solution

We reduce \(A\) to echelon form. The replacements \(R_2\to R_2+2R_1\) and \(R_3\to R_3+R_1\) do not change the determinant:

\[\det A=\begin{vmatrix} 1 & -4 & 2\\ -2 & 8 & -9\\ -1 & 7 & 0\end{vmatrix} =\begin{vmatrix} 1 & -4 & 2\\ 0 & 0 & -5\\ 0 & 3 & 2\end{vmatrix}.\]

The \((2,2)\)-entry is 0, so we interchange rows 2 and 3. An interchange changes the sign:

\[\det A = -\begin{vmatrix} 1 & -4 & 2\\ 0 & 3 & 2\\ 0 & 0 & -5\end{vmatrix} = -(1)(3)(-5) = 15.\]

The last matrix is triangular, so its determinant is the product of the diagonal entries.

Example 3 (factoring out a row)

Compute \(\det A\), where \(A=\begin{bmatrix} 2 & -8 & 6 & 8\\ 3 & -9 & 5 & 10\\ -3 & 0 & 1 & -2\\ 1 & -4 & 0 & 6\end{bmatrix}\).

Solution

Every entry of row 1 is even. We factor out 2 from row 1 (rule (c)). This gives a 1 in the upper-left corner. Then we use replacements in column 1: \(R_2\to R_2-3R_1\), \(R_3\to R_3+3R_1\), \(R_4\to R_4-R_1\). These do not change the determinant:

\[\det A = 2\begin{vmatrix} 1 & -4 & 3 & 4\\ 3 & -9 & 5 & 10\\ -3 & 0 & 1 & -2\\ 1 & -4 & 0 & 6\end{vmatrix} = 2\begin{vmatrix} 1 & -4 & 3 & 4\\ 0 & 3 & -4 & -2\\ 0 & -12 & 10 & 10\\ 0 & 0 & -3 & 2\end{vmatrix}.\]

Next we use the 3 in column 2 as a pivot: \(R_3\to R_3+4R_2\). Then \(R_4\to R_4-\tfrac12 R_3\) creates the last zero:

\[\det A = 2\begin{vmatrix} 1 & -4 & 3 & 4\\ 0 & 3 & -4 & -2\\ 0 & 0 & -6 & 2\\ 0 & 0 & -3 & 2\end{vmatrix} = 2\begin{vmatrix} 1 & -4 & 3 & 4\\ 0 & 3 & -4 & -2\\ 0 & 0 & -6 & 2\\ 0 & 0 & 0 & 1\end{vmatrix} = 2\,(1)(3)(-6)(1) = -36.\]

1.2 The determinant and invertibility

Suppose we reduce a square matrix \(A\) to an echelon form \(U\) using only replacements and interchanges, and suppose we use \(r\) interchanges. By Theorem 3, \(\det A = (-1)^r\det U\). Since \(U\) is triangular, \(\det U = u_{11}u_{22}\cdots u_{nn}\).

If \(A\) is invertible, every diagonal entry of \(U\) is a pivot, so none of them is zero. If \(A\) is not invertible, \(U\) has fewer than \(n\) pivots, so at least \(u_{nn}=0\). See Figure 1. Therefore

\[\det A=\begin{cases} (-1)^r\cdot(\text{product of the pivots in } U) & \text{when } A \text{ is invertible},\\[2pt] 0 & \text{when } A \text{ is not invertible.}\end{cases}\qquad (1)\]

Figure 1. Typical echelon forms of a square matrix. A filled square is a pivot (nonzero), * is any number. Left: \(A\) is invertible and every diagonal entry of \(U\) is a pivot. Right: \(A\) is not invertible and a diagonal entry of \(U\) is 0.

Theorem 4 A square matrix \(A\) is invertible if and only if \(\det A \neq 0\).

Theorem 4 adds the statement "\(\det A\neq 0\)" to the Invertible Matrix Theorem. Two useful consequences:

If the columns of \(A\) are linearly dependent, then \(\det A = 0\).
If the rows of \(A\) are linearly dependent, then \(\det A = 0\).

Dependence is easy to see when two rows (or two columns) are equal, when one row is a multiple of another, or when a row (or column) is zero.

Example 4 (two equal rows)

Compute \(\det A\), where \(A=\begin{bmatrix} 3 & -1 & 2 & -5\\ 0 & 5 & -3 & -6\\ -6 & 7 & -7 & 4\\ -5 & -8 & 0 & 9\end{bmatrix}\).

Solution

Add 2 times row 1 to row 3. This is a replacement, so the determinant does not change:

\[\det A = \det\begin{bmatrix} 3 & -1 & 2 & -5\\ 0 & 5 & -3 & -6\\ 0 & 5 & -3 & -6\\ -5 & -8 & 0 & 9\end{bmatrix} = 0.\]

The new matrix has two equal rows. Its rows are linearly dependent, so by Theorem 4 its determinant is 0.

Example 5 (testing linear independence)

Are the vectors \(\mathbf v_1=\begin{bmatrix} 7\\ -4\\ -6\end{bmatrix}\), \(\mathbf v_2=\begin{bmatrix} -8\\ 5\\ 7\end{bmatrix}\), \(\mathbf v_3=\begin{bmatrix} 7\\ 0\\ -5\end{bmatrix}\) linearly independent?

Solution

Put the vectors into the columns of \(A=\begin{bmatrix} 7 & -8 & 7\\ -4 & 5 & 0\\ -6 & 7 & -5\end{bmatrix}\). The columns are independent exactly when \(A\) is invertible, that is, when \(\det A\neq 0\). Row 2 has a zero, so we expand across row 2. The signs in row 2 are \(-,+,-\):

\[\det A = -(-4)\begin{vmatrix} -8 & 7\\ 7 & -5\end{vmatrix} + 5\begin{vmatrix} 7 & 7\\ -6 & -5\end{vmatrix} - 0 = 4(40-49) + 5(-35+42) = -36+35 = -1.\]

Since \(\det A = -1 \neq 0\), the vectors are linearly independent.

Exercise 1.A (row operations and the determinant)

Self-check only. These exercises are not graded. Entries may be typed as integers, decimals (1.5 or 1,5) or simple fractions (3/2).

Suppose \(\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i\end{vmatrix} = 7\). Find each determinant without expanding.

(a) \(\begin{vmatrix} d & e & f\\ a & b & c\\ g & h & i\end{vmatrix} =\)

(b) \(\begin{vmatrix} a & b & c\\ 2d+a & 2e+b & 2f+c\\ g & h & i\end{vmatrix} =\)

(d) \(\begin{vmatrix} g & h & i\\ a & b & c\\ d & e & f\end{vmatrix} =\)

(e) \(\det\left(2\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\end{bmatrix}\right) =\)

(a) Rows 1 and 2 are interchanged, so the sign changes: \(-7\).

(b) First do \(R_2\to R_2-R_1\) (replacement, no change). This gives rows \(a,b,c\); \(2d,2e,2f\); \(g,h,i\). Then factor 2 out of row 2: \(2\cdot 7=14\).

(d) Interchange rows 1 and 2 (order \(a,g,d\)), then rows 2 and 3 (order \(a,d,g\)). Two interchanges give \((-1)^2\cdot 7 = 7\).

(e) Each of the 3 rows is multiplied by 2, so we factor out 2 three times: \(2^3\cdot 7 = 56\).

Exercise 1.B (a singular matrix with no zero entries)

Self-check only. Many answers are possible; the check computes \(\det A\) for your matrix.

Find a \(3\times 3\) matrix \(A\) such that

all nine entries of \(A\) are nonzero,
no row of \(A\) is a multiple of another row,
\(\det A = 0\).

\(A=\)

By Theorem 4, \(\det A = 0\) exactly when the rows are linearly dependent. Since no row may be a multiple of another, make row 3 the sum of rows 1 and 2. For example

\[A=\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 7\\ 5 & 7 & 10\end{bmatrix}.\]

The replacement \(R_3\to R_3-R_1-R_2\) produces a zero row, so \(\det A = 0\).

Exercise 1.C (a \(4\times 4\) determinant)

Self-check only.

Use row reduction to echelon form to compute \(\det A\), where \(A=\begin{bmatrix} 1 & 2 & -1 & 3\\ -2 & -4 & 5 & -4\\ 3 & 7 & -2 & 10\\ 0 & 1 & 4 & 6\end{bmatrix}\).

\(\det A =\)

Replacements \(R_2\to R_2+2R_1\) and \(R_3\to R_3-3R_1\) give

\[\det A=\begin{vmatrix} 1 & 2 & -1 & 3\\ 0 & 0 & 3 & 2\\ 0 & 1 & 1 & 1\\ 0 & 1 & 4 & 6\end{vmatrix}.\]

The \((2,2)\)-entry is 0, so interchange rows 2 and 3 (sign changes). Then \(R_4\to R_4-R_2\) and \(R_4\to R_4-R_3\):

\[\det A=-\begin{vmatrix} 1 & 2 & -1 & 3\\ 0 & 1 & 1 & 1\\ 0 & 0 & 3 & 2\\ 0 & 1 & 4 & 6\end{vmatrix} =-\begin{vmatrix} 1 & 2 & -1 & 3\\ 0 & 1 & 1 & 1\\ 0 & 0 & 3 & 2\\ 0 & 0 & 0 & 3\end{vmatrix} = -(1)(1)(3)(3) = -9.\]

1.3 The transpose and column operations

Theorem 5 If \(A\) is an \(n\times n\) matrix, then \(\det A^T = \det A\).

The reason: a cofactor expansion of \(\det A\) across row 1 is the same calculation as a cofactor expansion of \(\det A^T\) down column 1. (The full proof uses induction on \(n\).)

A row operation on \(A^T\) is a column operation on \(A\). So Theorem 3 also holds when the word row is replaced by column: adding a multiple of one column to another does not change \(\det A\), interchanging two columns changes the sign, and scaling a column by \(k\) multiplies \(\det A\) by \(k\).

Example 6 (dependent columns)

Let \(A=\begin{bmatrix} 2 & 4 & 1\\ 1 & 2 & 0\\ 3 & 6 & 5\end{bmatrix}\). Column 2 is 2 times column 1, so the columns are linearly dependent and \(\det A = 0\). We can also see it with a column operation: \(C_2\to C_2-2C_1\) gives a zero column and does not change the determinant.

1.4 Determinants and matrix products

Theorem 6 (Multiplicative property) If \(A\) and \(B\) are \(n\times n\) matrices, then \(\det(AB) = (\det A)(\det B)\).

Example 7 (checking Theorem 6)

Verify Theorem 6 for \(A=\begin{bmatrix} 6 & 1\\ 3 & 2\end{bmatrix}\) and \(B=\begin{bmatrix} 4 & 3\\ 1 & 2\end{bmatrix}\).

Solution

\[AB=\begin{bmatrix} 6 & 1\\ 3 & 2\end{bmatrix}\begin{bmatrix} 4 & 3\\ 1 & 2\end{bmatrix}=\begin{bmatrix} 25 & 20\\ 14 & 13\end{bmatrix},\qquad \det(AB) = 25\cdot 13 - 20\cdot 14 = 325-280 = 45.\]

Also \(\det A = 12-3 = 9\) and \(\det B = 8-3=5\), so \((\det A)(\det B) = 45 = \det(AB)\).

There is no rule for sums

In general \(\det(A+B)\neq \det A + \det B\). For example, with \(A=B=I_2\): \(\det(A+B) = \det(2I_2) = 4\), but \(\det A+\det B = 2\).

Theorems 3, 4 and 6 give several rules that we use often. Let \(A\) be \(n\times n\).

Consequences

(a) If \(A\) is invertible, \(\det(A^{-1}) = \dfrac{1}{\det A}\). Reason: \(\det A\cdot\det(A^{-1}) = \det(AA^{-1}) = \det I = 1\).

(b) \(\det(A^k) = (\det A)^k\) for every positive integer \(k\). Reason: apply Theorem 6 again and again.

Example 8 (using the rules)

Let \(A\) and \(B\) be \(4\times 4\) matrices with \(\det A = -2\) and \(\det B = 3\). Find

(a) \(\det(AB)\)

(b) \(\det(B^{-1}A)\)

(d) \(\det(A^TA)\)

(e) \(\det(A^3)\)

Solution

(a) \(\det(AB) = (\det A)(\det B) = (-2)(3) = -6\).

(b) \(\det(B^{-1}A) = \det(B^{-1})\det A = \tfrac13\cdot(-2) = -\tfrac23\).

(d) \(\det(A^TA) = \det(A^T)\det A = (\det A)^2 = 4\), by Theorems 5 and 6.

(e) \(\det(A^3) = (\det A)^3 = -8\).

Day 2: Cramer's Rule and a Formula for \(A^{-1}\)

Today we write the solution of \(A\mathbf x=\mathbf b\) and the inverse \(A^{-1}\) directly with determinants.

Facts used today

\(\det\begin{bmatrix} a & b\\ c & d\end{bmatrix} = ad-bc\).
\(A_{ij}\) is the matrix obtained from \(A\) by deleting row \(i\) and column \(j\). The \((i,j)\)-cofactor of \(A\) is \(C_{ij} = (-1)^{i+j}\det A_{ij}\). The signs \((-1)^{i+j}\) follow the pattern \(\begin{bmatrix} + & - & +\\ - & + & -\\ + & - & +\end{bmatrix}\).
Cofactor expansion across row \(i\): \(\det A = a_{i1}C_{i1} + a_{i2}C_{i2} + \cdots + a_{in}C_{in}\). The same works down any column.
If \(\det A\neq 0\), then \(A\) is invertible and \(A\mathbf x=\mathbf b\) has exactly one solution for each \(\mathbf b\).

2.1 Cramer's rule

Notation \(A_i(\mathbf b)\) For an \(n\times n\) matrix \(A\) and a vector \(\mathbf b\) in \(\mathbb R^n\), \(A_i(\mathbf b)\) is the matrix obtained from \(A\) by replacing column \(i\) by \(\mathbf b\): \[A_i(\mathbf b) = \begin{bmatrix}\mathbf a_1 & \cdots & \mathbf b & \cdots & \mathbf a_n\end{bmatrix}\quad(\mathbf b \text{ in column } i).\]

Example 1 (the matrices \(A_i(\mathbf b)\))

If \(A=\begin{bmatrix} 1 & 4\\ 2 & 3\end{bmatrix}\) and \(\mathbf b=\begin{bmatrix} 5\\ 6\end{bmatrix}\), then \(A_1(\mathbf b) = \begin{bmatrix} 5 & 4\\ 6 & 3\end{bmatrix}\) and \(A_2(\mathbf b) = \begin{bmatrix} 1 & 5\\ 2 & 6\end{bmatrix}\).

Theorem 7 (Cramer's rule) Let \(A\) be an invertible \(n\times n\) matrix. For any \(\mathbf b\) in \(\mathbb R^n\), the unique solution \(\mathbf x\) of \(A\mathbf x = \mathbf b\) has entries \[x_i = \frac{\det A_i(\mathbf b)}{\det A},\qquad i=1,2,\ldots,n.\]

Why it works. Let \(I_i(\mathbf x)\) be the identity matrix with column \(i\) replaced by \(\mathbf x\). If \(A\mathbf x=\mathbf b\), then

\[A\,I_i(\mathbf x) = \begin{bmatrix} A\mathbf e_1 & \cdots & A\mathbf x & \cdots & A\mathbf e_n\end{bmatrix} = \begin{bmatrix}\mathbf a_1 & \cdots & \mathbf b & \cdots & \mathbf a_n\end{bmatrix} = A_i(\mathbf b).\]

By the multiplicative property, \((\det A)\cdot\det I_i(\mathbf x) = \det A_i(\mathbf b)\). A cofactor expansion of \(I_i(\mathbf x)\) across row \(i\) shows \(\det I_i(\mathbf x) = x_i\). Since \(\det A\neq 0\), we can divide.

Example 2 (a \(2\times 2\) system)

Use Cramer's rule to solve

\[\begin{aligned} 3x_1 - 2x_2 &= 6\\ -5x_1 + 4x_2 &= 8\end{aligned}\]

Solution

Write the system as \(A\mathbf x=\mathbf b\) with

\[A=\begin{bmatrix} 3 & -2\\ -5 & 4\end{bmatrix},\quad A_1(\mathbf b)=\begin{bmatrix} 6 & -2\\ 8 & 4\end{bmatrix},\quad A_2(\mathbf b)=\begin{bmatrix} 3 & 6\\ -5 & 8\end{bmatrix}.\]

\(\det A = 12-10 = 2\neq 0\), so there is a unique solution. Then

\[x_1 = \frac{\det A_1(\mathbf b)}{\det A} = \frac{24+16}{2} = 20,\qquad x_2 = \frac{\det A_2(\mathbf b)}{\det A} = \frac{24+30}{2} = 27.\]

Check: \(3(20)-2(27) = 6\) and \(-5(20)+4(27) = 8\).

Example 3 (a \(3\times 3\) system)

Use Cramer's rule to solve

\[\begin{array}{rcrcrcr} 2x_1 &+& x_2 & & &=& 7\\ -3x_1 & & &+& x_3 &=& -8\\ & & x_2 &+& 2x_3 &=& -3\end{array}\]

Solution

Here \(A=\begin{bmatrix} 2 & 1 & 0\\ -3 & 0 & 1\\ 0 & 1 & 2\end{bmatrix}\) and \(\mathbf b=\begin{bmatrix} 7\\ -8\\ -3\end{bmatrix}\). Expanding across row 1:

\[\det A = 2\begin{vmatrix} 0 & 1\\ 1 & 2\end{vmatrix} - 1\begin{vmatrix} -3 & 1\\ 0 & 2\end{vmatrix} + 0 = 2(-1) - (-6) = 4.\]

Replace one column at a time by \(\mathbf b\) and expand across row 1 again:

\[\det A_1(\mathbf b) = \begin{vmatrix} 7 & 1 & 0\\ -8 & 0 & 1\\ -3 & 1 & 2\end{vmatrix} = 7(0-1) - 1(-16+3) + 0 = -7+13 = 6,\] \[\det A_2(\mathbf b) = \begin{vmatrix} 2 & 7 & 0\\ -3 & -8 & 1\\ 0 & -3 & 2\end{vmatrix} = 2(-16+3) - 7(-6-0) + 0 = -26+42 = 16,\] \[\det A_3(\mathbf b) = \begin{vmatrix} 2 & 1 & 7\\ -3 & 0 & -8\\ 0 & 1 & -3\end{vmatrix} = 2(0+8) - 1(9-0) + 7(-3-0) = 16-9-21 = -14.\]

So \(x_1 = \tfrac64 = \tfrac32\), \(x_2 = \tfrac{16}{4} = 4\), \(x_3 = \tfrac{-14}{4} = -\tfrac72\).

Check: \(2\cdot\tfrac32 + 4 = 7\), \(\;-3\cdot\tfrac32 - \tfrac72 = -8\), \(\;4 + 2\cdot(-\tfrac72) = -3\).

Example 4 (a system with a parameter)

Systems like the one below appear in engineering, for example after a Laplace transform. Here \(s\) is a parameter. Find the values of \(s\) for which the system has a unique solution, and describe the solution with Cramer's rule.

\[\begin{aligned} 3s\,x_1 - 2x_2 &= 4\\ -6x_1 + s\,x_2 &= 1\end{aligned}\]

Solution

\(A=\begin{bmatrix} 3s & -2\\ -6 & s\end{bmatrix}\), \(A_1(\mathbf b)=\begin{bmatrix} 4 & -2\\ 1 & s\end{bmatrix}\), \(A_2(\mathbf b)=\begin{bmatrix} 3s & 4\\ -6 & 1\end{bmatrix}\).

\(\det A = 3s^2-12 = 3(s+2)(s-2)\). The system has a unique solution exactly when \(\det A\neq 0\), that is, when \(s\neq \pm 2\). For such \(s\):

\[x_1 = \frac{4s+2}{3(s+2)(s-2)},\qquad x_2 = \frac{3s+24}{3(s+2)(s-2)} = \frac{s+8}{(s+2)(s-2)}.\]

When to use Cramer's rule

Cramer's rule needs \(\det A\neq 0\). If \(\det A = 0\), the formula gives no information: the system may have no solution or infinitely many. Use row reduction in that case.
For large \(n\) it is very slow: computing one \(n\times n\) determinant costs about as much as solving the whole system by row reduction. It is useful for \(2\times 2\) and \(3\times 3\) systems, for systems with parameters, and when you need only one unknown.

Exercise 2.A (Cramer's rule)

Self-check only. In part (c) the check substitutes your \(x_1, x_2\) into both equations.

Solve with Cramer's rule:

\[\begin{aligned} 4x_1 + x_2 &= 6\\ 5x_1 + 2x_2 &= 7\end{aligned}\]

(a) \(\det A =\)

(b) \(\det A_1(\mathbf b) =\) \(\det A_2(\mathbf b) =\)

\[\det A = \begin{vmatrix} 4 & 1\\ 5 & 2\end{vmatrix} = 3,\quad \det A_1(\mathbf b)=\begin{vmatrix} 6 & 1\\ 7 & 2\end{vmatrix} = 5,\quad \det A_2(\mathbf b)=\begin{vmatrix} 4 & 6\\ 5 & 7\end{vmatrix} = -2.\]

So \(x_1 = \tfrac53\) and \(x_2 = -\tfrac23\). Check: \(4\cdot\tfrac53 - \tfrac23 = 6\) and \(5\cdot\tfrac53 - \tfrac43 = 7\).

2.2 A formula for \(A^{-1}\): the adjoint

Column \(j\) of \(A^{-1}\) is the solution \(\mathbf x\) of \(A\mathbf x = \mathbf e_j\). Its \(i\)-th entry is the \((i,j)\)-entry of \(A^{-1}\). By Cramer's rule,

\[(i,j)\text{-entry of } A^{-1} = \frac{\det A_i(\mathbf e_j)}{\det A}.\]

A cofactor expansion down column \(i\) of \(A_i(\mathbf e_j)\) gives \(\det A_i(\mathbf e_j) = (-1)^{i+j}\det A_{ji} = C_{ji}\). Note the order of the subscripts: the \((i,j)\)-entry of \(A^{-1}\) uses the cofactor \(C_{ji}\).

Definition (adjoint) The adjoint (also called the classical adjoint) of \(A\), written \(\operatorname{adj} A\), is the transpose of the matrix of cofactors: \[\operatorname{adj} A = \begin{bmatrix} C_{11} & C_{21} & \cdots & C_{n1}\\ C_{12} & C_{22} & \cdots & C_{n2}\\ \vdots & \vdots & & \vdots\\ C_{1n} & C_{2n} & \cdots & C_{nn}\end{bmatrix}.\]

Example 5 (the \(2\times 2\) case)

Let \(A=\begin{bmatrix} a & b\\ c & d\end{bmatrix}\). The cofactors are \(C_{11} = d\), \(C_{12} = -c\), \(C_{21} = -b\), \(C_{22} = a\). Transposing the cofactor matrix gives

\[\operatorname{adj} A = \begin{bmatrix} d & -b\\ -c & a\end{bmatrix}.\]

With Theorem 8 below this gives the formula you already know: \(A^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix} d & -b\\ -c & a\end{bmatrix}\).

Theorem 8 (An inverse formula) Let \(A\) be an invertible \(n\times n\) matrix. Then \[A^{-1} = \frac{1}{\det A}\,\operatorname{adj} A.\]

Example 6 (a \(3\times 3\) inverse)

Find the inverse of \(A=\begin{bmatrix} 2 & 1 & 3\\ 1 & -1 & 1\\ 1 & 4 & -2\end{bmatrix}\).

Solution

The nine cofactors are

\[\begin{aligned} C_{11} &= +\begin{vmatrix} -1 & 1\\ 4 & -2\end{vmatrix} = -2, & C_{12} &= -\begin{vmatrix} 1 & 1\\ 1 & -2\end{vmatrix} = 3, & C_{13} &= +\begin{vmatrix} 1 & -1\\ 1 & 4\end{vmatrix} = 5,\\ C_{21} &= -\begin{vmatrix} 1 & 3\\ 4 & -2\end{vmatrix} = 14, & C_{22} &= +\begin{vmatrix} 2 & 3\\ 1 & -2\end{vmatrix} = -7, & C_{23} &= -\begin{vmatrix} 2 & 1\\ 1 & 4\end{vmatrix} = -7,\\ C_{31} &= +\begin{vmatrix} 1 & 3\\ -1 & 1\end{vmatrix} = 4, & C_{32} &= -\begin{vmatrix} 2 & 3\\ 1 & 1\end{vmatrix} = 1, & C_{33} &= +\begin{vmatrix} 2 & 1\\ 1 & -1\end{vmatrix} = -3. \end{aligned}\]

The adjoint is the transpose of the cofactor matrix (for instance, \(C_{12}\) goes to position \((2,1)\)):

\[\operatorname{adj} A = \begin{bmatrix} -2 & 14 & 4\\ 3 & -7 & 1\\ 5 & -7 & -3\end{bmatrix}.\]

We can get \(\det A\) and check our work at the same time:

\[(\operatorname{adj} A)\,A = \begin{bmatrix} -2 & 14 & 4\\ 3 & -7 & 1\\ 5 & -7 & -3\end{bmatrix}\begin{bmatrix} 2 & 1 & 3\\ 1 & -1 & 1\\ 1 & 4 & -2\end{bmatrix} = \begin{bmatrix} 14 & 0 & 0\\ 0 & 14 & 0\\ 0 & 0 & 14\end{bmatrix} = 14I.\]

So \(\det A = 14\) and

\[A^{-1} = \frac{1}{14}\begin{bmatrix} -2 & 14 & 4\\ 3 & -7 & 1\\ 5 & -7 & -3\end{bmatrix} = \begin{bmatrix} -1/7 & 1 & 2/7\\ 3/14 & -1/2 & 1/14\\ 5/14 & -1/2 & -3/14\end{bmatrix}.\]

A useful identity For every square matrix \(A\) (invertible or not), \[A\,(\operatorname{adj} A) = (\operatorname{adj} A)\,A = (\det A)\,I.\] The diagonal entries are cofactor expansions of \(\det A\). The off-diagonal entries are cofactor expansions of matrices with two equal rows (or columns), so they are 0.

Example 7 (one entry of the inverse)

For the matrix \(A\) of Example 6, find only the \((3,2)\)-entry of \(A^{-1}\).

Solution

The \((3,2)\)-entry of \(A^{-1}\) is \(C_{23}/\det A\). Delete row 2 and column 3 of \(A\):

\[C_{23} = (-1)^{2+3}\begin{vmatrix} 2 & 1\\ 1 & 4\end{vmatrix} = -7,\qquad \text{so the entry is } \frac{-7}{14} = -\frac12.\]

This agrees with the full inverse in Example 6. We needed one \(2\times 2\) determinant (and \(\det A\)), not the whole inverse.

Numerical note

Theorem 8 is mainly a theoretical tool: it shows how the entries of \(A^{-1}\) depend on the entries of \(A\). To compute an inverse of a large matrix, row reduce \(\begin{bmatrix} A & I\end{bmatrix}\) instead.

Exercise 2.B (the adjoint)

Self-check only. In part (b) the check verifies that \(A\,(\operatorname{adj} A) = (\det A)\,I\).

Let \(A=\begin{bmatrix} 2 & 0 & 1\\ 1 & 1 & 0\\ 0 & 3 & 1\end{bmatrix}\).

(a) \(\det A =\)

(b) Enter \(\operatorname{adj} A\).

\(\operatorname{adj} A=\)

(a) \(\det A = 2\begin{vmatrix} 1 & 0\\ 3 & 1\end{vmatrix} - 0 + 1\begin{vmatrix} 1 & 1\\ 0 & 3\end{vmatrix} = 2(1) + 3 = 5\).

(b) The cofactors are

\[\begin{aligned} C_{11} &= 1, & C_{12} &= -1, & C_{13} &= 3,\\ C_{21} &= -(0\cdot 1 - 1\cdot 3) = 3, & C_{22} &= 2, & C_{23} &= -(6-0) = -6,\\ C_{31} &= 0\cdot 0 - 1\cdot 1 = -1, & C_{32} &= -(0-1) = 1, & C_{33} &= 2. \end{aligned}\] \[\operatorname{adj} A = \begin{bmatrix} 1 & 3 & -1\\ -1 & 2 & 1\\ 3 & -6 & 2\end{bmatrix},\qquad A\,(\operatorname{adj} A) = 5I.\]

Day 3: Determinants as Area and Volume

Today we see that \(|\det A|\) is an area (in \(\mathbb R^2\)) or a volume (in \(\mathbb R^3\)), and that a linear transformation multiplies every area or volume by \(|\det A|\).

Facts used today

\(\det\begin{bmatrix} a & 0\\ 0 & d\end{bmatrix} = ad\) and \(\det\begin{bmatrix} a & 0 & 0\\ 0 & b & 0\\ 0 & 0 & c\end{bmatrix} = abc\).
Interchanging two columns changes only the sign of \(\det A\). Adding a multiple of one column to another column does not change \(\det A\). So neither operation changes \(|\det A|\).
\(\det(AB) = (\det A)(\det B)\).
Area of a parallelogram = base \(\times\) height. Volume of a parallelepiped = (area of the base) \(\times\) height.

3.1 Area and volume

Theorem 9 If \(A\) is a \(2\times 2\) matrix, the area of the parallelogram determined by the columns of \(A\) is \(|\det A|\). If \(A\) is a \(3\times 3\) matrix, the volume of the parallelepiped determined by the columns of \(A\) is \(|\det A|\).

The parallelogram determined by \(\mathbf a_1, \mathbf a_2\) has vertices \(\mathbf 0\), \(\mathbf a_1\), \(\mathbf a_2\), \(\mathbf a_1+\mathbf a_2\). The parallelepiped determined by \(\mathbf a_1,\mathbf a_2,\mathbf a_3\) is the "slanted box" with edges \(\mathbf a_1, \mathbf a_2, \mathbf a_3\) at the origin.

Figure 2 shows the parallelogram of a \(2\times 2\) matrix. It starts with \(A=I\): the columns \(\mathbf e_1,\mathbf e_2\) give the unit square, and \(\det I = 1\) is its area. Change the entries of \(A\) and compare the area of the parallelogram with \(\det A\).

Interactive figure: the parallelogram of a \(2\times 2\) matrix

\(A=\)

Figure 2. The parallelogram determined by the columns \(\mathbf a_1,\mathbf a_2\) of \(A\). The figure starts with \(A=I\): the columns \(\mathbf e_1,\mathbf e_2\) give the unit square, and its area is \(1=\det I\). Enter any matrix or press "Random matrix": the area of the parallelogram is \(|\det A|\). The dashed square is the unit square for comparison.

Why it is true. The idea has three steps.

Diagonal matrices. For \(A=\begin{bmatrix} a & 0\\ 0 & d\end{bmatrix}\) the parallelogram is a rectangle with sides \(|a|\) and \(|d|\), so its area is \(|ad| = |\det A|\). In \(\mathbb R^3\) a diagonal matrix gives a rectangular box with volume \(|abc|\).
Reduce to the diagonal case. Column interchanges and column replacements can turn any \(A\) into a diagonal matrix, and they do not change \(|\det A|\). A column interchange does not change the parallelogram at all.
A column replacement does not change the area. Replace \(\mathbf a_2\) by \(\mathbf a_2 + c\,\mathbf a_1\). The new vertex moves along the line through \(\mathbf a_2\) parallel to \(\mathbf a_1\). The base (from \(\mathbf 0\) to \(\mathbf a_1\)) and the height stay the same, so the area stays the same. You can check this in Figure 2: enter \(A=I\), then \(A=\begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\). The second column changes from \(\mathbf e_2\) to \(\mathbf e_2+2\mathbf e_1\), and the area stays 1.

In \(\mathbb R^3\) the argument is the same. Take the base in the plane \(\operatorname{Span}\{\mathbf a_1,\mathbf a_3\}\). The height is the distance from the tip of \(\mathbf a_2\) to this plane. If we replace \(\mathbf a_2\) by \(\mathbf a_2 + c\,\mathbf a_1 + d\,\mathbf a_3\), the tip moves inside the plane \(\mathbf a_2 + \operatorname{Span}\{\mathbf a_1,\mathbf a_3\}\), which is parallel to the base. The height does not change, so the volume does not change. Figure 3 shows the parallelepiped of a \(3\times 3\) matrix. It starts with \(A=I\), the unit cube with volume \(\det I = 1\). Change the entries and compare the volume with \(\det A\).

3D figure: the parallelepiped of a \(3\times 3\) matrix

You can drag the figure (or focus it and use the arrow keys) to look at it from another side.

\(A=\)

Figure 3. The parallelepiped determined by the columns \(\mathbf a_1,\mathbf a_2,\mathbf a_3\) of \(A\). The figure starts with \(A=I\): the unit cube, with volume \(1=\det I\). Enter any matrix or press "Random matrix": the volume of the parallelepiped is \(|\det A|\). The dashed cube is the unit cube for comparison.

Example 1 (a parallelogram not at the origin)

Find the area of the parallelogram with vertices \((-2,-2)\), \((0,3)\), \((4,-1)\) and \((6,4)\).

Solution

Theorem 9 needs a parallelogram with a vertex at the origin. Translating a region does not change its area, so we subtract \((-2,-2)\) from every vertex. The new vertices are \((0,0)\), \((2,5)\), \((6,1)\) and \((8,6)\). See Figure 4. Since \((2,5)+(6,1) = (8,6)\), the sides from the origin are \((2,5)\) and \((6,1)\). So the parallelogram is determined by the columns of

\[A=\begin{bmatrix} 2 & 6\\ 5 & 1\end{bmatrix},\qquad \det A = 2-30 = -28.\]

The area is \(|-28| = 28\).

Figure 4. Translating a parallelogram does not change its area. After the translation one vertex is at the origin and the sides from the origin are \((2,5)\) and \((6,1)\).

Example 2 (a parallelepiped)

Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at \((1,0,-2)\), \((1,2,4)\) and \((7,1,0)\).

Solution

The three adjacent vertices are the edge vectors. Put them into the columns of \(A\) and expand down column 1:

\[\det A = \begin{vmatrix} 1 & 1 & 7\\ 0 & 2 & 1\\ -2 & 4 & 0\end{vmatrix} = 1\begin{vmatrix} 2 & 1\\ 4 & 0\end{vmatrix} - 0 + (-2)\begin{vmatrix} 1 & 7\\ 2 & 1\end{vmatrix} = (0-4) - 2(1-14) = 22.\]

The volume is \(|22| = 22\).

Example 3 (area of a triangle)

Find the area of the triangle with vertices \((1,1)\), \((4,2)\) and \((2,5)\).

Solution

A triangle is half of a parallelogram. Translate by \((-1,-1)\): the vertices become \((0,0)\), \((3,1)\), \((1,4)\). The parallelogram on \((3,1)\) and \((1,4)\) has area \(\left|\det\begin{bmatrix} 3 & 1\\ 1 & 4\end{bmatrix}\right| = |12-1| = 11\). The triangle has area \(\tfrac{11}{2}\).

Exercise 3.A (area of a parallelogram)

Self-check only.

Find the area of the parallelogram with vertices \((-1,0)\), \((0,5)\), \((1,-4)\) and \((2,1)\).

Area \(=\)

Subtract \((-1,0)\) from each vertex: \((0,0)\), \((1,5)\), \((2,-4)\), \((3,1)\). Since \((1,5)+(2,-4) = (3,1)\), the sides from the origin are \((1,5)\) and \((2,-4)\).

\[\det\begin{bmatrix} 1 & 2\\ 5 & -4\end{bmatrix} = -4-10 = -14,\qquad \text{area} = 14.\]

3.2 Linear transformations

Let \(T\) be a linear transformation and \(S\) a set in its domain. \(T(S)\) is the set of all images \(T(\mathbf x)\) with \(\mathbf x\) in \(S\). How does the area (or volume) of \(T(S)\) compare with that of \(S\)?

Theorem 10 Let \(T:\mathbb R^2\to\mathbb R^2\) be the linear transformation determined by a \(2\times 2\) matrix \(A\). If \(S\) is a parallelogram in \(\mathbb R^2\), then \[\{\text{area of } T(S)\} = |\det A|\cdot\{\text{area of } S\}.\] If \(T\) is determined by a \(3\times 3\) matrix \(A\), and \(S\) is a parallelepiped in \(\mathbb R^3\), then \[\{\text{volume of } T(S)\} = |\det A|\cdot\{\text{volume of } S\}.\]

Why it is true (2D). Let \(S\) be the parallelogram determined by \(\mathbf b_1,\mathbf b_2\), so \(S = \{s_1\mathbf b_1 + s_2\mathbf b_2 : 0\le s_1\le 1,\ 0\le s_2\le 1\}\). Since \(T\) is linear,

\[T(s_1\mathbf b_1 + s_2\mathbf b_2) = s_1 A\mathbf b_1 + s_2 A\mathbf b_2.\]

So \(T(S)\) is the parallelogram determined by the columns of \(\begin{bmatrix} A\mathbf b_1 & A\mathbf b_2\end{bmatrix} = AB\), where \(B=\begin{bmatrix}\mathbf b_1 & \mathbf b_2\end{bmatrix}\). By Theorem 9 and Theorem 6,

\[\{\text{area of } T(S)\} = |\det(AB)| = |\det A|\cdot|\det B| = |\det A|\cdot\{\text{area of } S\}.\]

A parallelogram that does not start at the origin has the form \(\mathbf p + S\). Its image is \(T(\mathbf p) + T(S)\), a translate of \(T(S)\), so it has the same area.

The simplest case is the unit square. Its sides are \(\mathbf e_1, \mathbf e_2\), so \(T\) maps it onto the parallelogram determined by \(T(\mathbf e_1) = \mathbf a_1\) and \(T(\mathbf e_2) = \mathbf a_2\), the columns of \(A\). In Figure 5 the red unit square and the green unit disk show the case \(A=I\). Enter a matrix \(A\) to see their images under \(T\). In \(\mathbb R^3\) the same happens to the unit cube: \(T\) maps it onto the parallelepiped on the columns of \(A\), as in Figure 3.

Interactive figure: \(T\) maps the unit square and the unit disk

\(A=\)

Figure 5. With \(A=I\) the red square is the unit square (area 1) and the green disk is the unit disk (area \(\pi\)). Enter another matrix \(A\) or press "Random matrix": the shapes are replaced by their images under \(T(\mathbf x)=A\mathbf x\), the parallelogram on \(T(\mathbf e_1), T(\mathbf e_2)\) and an ellipse. Both areas are multiplied by \(|\det A|\). The dashed outlines show the unit square and the unit disk for comparison.

Other regions

Theorem 10 holds for any region with a finite area (or volume), not only for parallelograms. Cover the region by small squares. \(T\) multiplies the area of each small square by \(|\det A|\). As the squares get smaller, their total area approaches the area of the region, and the total area of their images approaches the area of \(T(S)\).

The sign of \(\det A\)

Area and volume use \(|\det A|\). The sign carries extra information: if \(\det A \lt 0\), then \(T\) reverses orientation, like a mirror. Enter \(A=\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}\) in Figure 5: \(\det A=-1\), the image of the square is the same square, but \(T(\mathbf e_1)\) and \(T(\mathbf e_2)\) have changed places. In this course we only use \(|\det A|\).

Example 4 (the area of an ellipse)

Let \(a\) and \(b\) be positive numbers. Find the area of the region \(E\) bounded by the ellipse \(\dfrac{x_1^2}{a^2} + \dfrac{x_2^2}{b^2} = 1\).

Solution

\(E\) is the image of the unit disk \(D\) under \(T\) with \(A=\begin{bmatrix} a & 0\\ 0 & b\end{bmatrix}\). Indeed, if \(\mathbf u = (u_1,u_2)\) and \(\mathbf x = A\mathbf u\), then \(u_1 = x_1/a\) and \(u_2 = x_2/b\). So \(\mathbf u\) is in \(D\) (that is, \(u_1^2+u_2^2\le 1\)) exactly when \(\mathbf x\) is in \(E\). By Theorem 10 (for general regions),

\[\{\text{area of } E\} = |\det A|\cdot\{\text{area of } D\} = ab\cdot\pi(1)^2 = \pi ab.\]

Enter \(A=\begin{bmatrix} 2 & 0\\ 0 & 1.5\end{bmatrix}\) in Figure 5 to see this with \(a=2\), \(b=1.5\).

Example 5 (applying \(T\) twice)

Let \(T(\mathbf x) = A\mathbf x\) with \(A=\begin{bmatrix} 2 & 1\\ 1 & 3\end{bmatrix}\), and let \(S\) be a triangle with area 4.

(a) Find the area of \(T(S)\).

(b) Find the area of \(T(T(S))\).

Solution

(a) \(\det A = 6-1 = 5\), so the area of \(T(S)\) is \(5\cdot 4 = 20\).

(b) Apply Theorem 10 again to the region \(T(S)\): the area is \(5\cdot 20 = 100\). Equivalently, \(T\circ T\) has matrix \(A^2\) and \(|\det A^2| = 25\), so the area is \(25\cdot 4 = 100\).

Example 6 (a volume in \(\mathbb R^3\))

Let \(T(\mathbf x) = A\mathbf x\) with \(A=\begin{bmatrix} 1 & 2 & 0\\ 0 & 1 & 0\\ 3 & 0 & 2\end{bmatrix}\). The image of the unit ball (volume \(\tfrac43\pi\)) is an ellipsoid. Find its volume.

Solution

Expand across row 2, which has only one nonzero entry: \(\det A = (+1)\cdot 1\cdot\begin{vmatrix} 1 & 0\\ 3 & 2\end{vmatrix} = 2\). The volume is \(2\cdot\tfrac43\pi = \tfrac83\pi\).

Exercise 3.B (area and volume under a transformation)

Self-check only. In (a) and (b) many answers are possible; the check computes the determinant of your matrix. In (c) you may type pi, for example 3pi or 3*pi.

(a) Enter a \(2\times 2\) matrix \(A\) with no zero entries such that \(T(\mathbf x)=A\mathbf x\) maps the unit square onto a parallelogram of area 6.

\(A=\)

(b) Enter a \(3\times 3\) matrix \(B\) that is not triangular (it has a nonzero entry above the diagonal and a nonzero entry below the diagonal) such that \(T(\mathbf x)=B\mathbf x\) doubles every volume.

\(B=\)

(c) Let \(A=\begin{bmatrix} 3 & 1\\ 0 & 2\end{bmatrix}\) and let \(D\) be the disk of radius 2 centered at the origin. The area of \(T(D)\) is

(a) We need \(|\det A| = 6\). One example: \(A=\begin{bmatrix} 2 & 1\\ 2 & 4\end{bmatrix}\), \(\det A = 8-2 = 6\). A matrix with \(\det A = -6\) is also correct.

(b) We need \(|\det B| = 2\). One example: \(B=\begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 1 & 0 & 2\end{bmatrix}\). Expand across row 2: \(\det B = 1\cdot\begin{vmatrix} 1 & 0\\ 1 & 2\end{vmatrix} = 2\).

(c) \(\det A = 6\) and the area of \(D\) is \(\pi\cdot 2^2 = 4\pi\). So the area of \(T(D)\) is \(6\cdot 4\pi = 24\pi\approx 75.4\).

Homework 8

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Question 115 points

Compute \(\det A\) by row reduction to echelon form, where

\[A=\begin{bmatrix} 1 & 2 & -1 & 3\\ 2 & 4 & 1 & 5\\ -1 & 1 & 2 & 0\\ 3 & 7 & 0 & 4\end{bmatrix}.\]

Enter \(A\) in the first grid. For each step, write the row operation(s) you used (for example R2 -> R2 - 2R1 or R2 <-> R3) and the new matrix.

Explain how each step changes the determinant and how you get \(\det A\) from your echelon form.

\(\det A=\)

Question 220 points

Let \(A\) and \(B\) be \(3\times 3\) matrices with \(\det A = 4\) and \(\det B = -3\). Find the following. Name the property you use in each step. Each part is worth 4 points.

(a) \(\det(2A^T)\)

(b) \(\det(A^{-1}B^2)\)

(d) Suppose \(\begin{vmatrix} a & b & c\\ d & e & f\\ g & h & i\end{vmatrix} = -5\). Find \(\begin{vmatrix} g & h & i\\ 2a-d & 2b-e & 2c-f\\ d & e & f\end{vmatrix}\).

(e) True or false? Justify your answer. "If \(M\) is a \(3\times 3\) matrix with \(\det M = 0\), then \(M\) has a zero row, or two of its rows are multiples of each other." If it is false, give a counterexample and show that its determinant is 0.

Question 320 points

Consider the system, where \(s\) is a parameter:

\[\begin{aligned} s\,x_1 + 2x_2 &= 3\\ 2x_1 + s\,x_2 &= 1\end{aligned}\]

(a) Find all values of \(s\) for which the system has a unique solution. For these values, use Cramer's rule to write \(x_1\) and \(x_2\) in terms of \(s\). (8 points)

(b) For each value of \(s\) excluded in (a), Cramer's rule does not apply. Use another method to decide whether the system has no solution or infinitely many solutions. (5 points)

(c) Use Cramer's rule to find only \(x_2\) for the system below. Do not solve for \(x_1\) or \(x_3\). (7 points) \[\begin{array}{rcrcrcr} 2x_1 &+& x_2 & & &=& 1\\ x_1 &+& 3x_2 &+& x_3 &=& 0\\ & & x_2 &+& 2x_3 &=& 2\end{array}\]

Question 420 points

Let \(A=\begin{bmatrix} 1 & 2 & 1\\ 0 & 1 & 3\\ 2 & 1 & 1\end{bmatrix}\).

(a) Compute all nine cofactors \(C_{ij}\) of \(A\) and write \(\operatorname{adj} A\). (8 points)

\(\operatorname{adj} A=\)

(b) Compute \(A\,(\operatorname{adj} A)\) and use the result to find \(\det A\). (4 points)

\(A^{-1}=\)

(d) Let \(M\) be a square matrix whose entries are all integers, with \(\det M = 1\) or \(\det M = -1\). Explain why all entries of \(M^{-1}\) are integers. (4 points)

Question 525 points

Each part is worth 5 points.

(a) Find the area of the triangle with vertices \((-2,1)\), \((3,2)\) and \((1,-3)\).

(b) Find the volume of the parallelepiped with one vertex at the origin and adjacent vertices at \((1,2,0)\), \((-1,1,3)\) and \((2,0,1)\).

(c) Now replace the vertex \((2,0,1)\) in (b) by \((2,0,h)\). For which value of \(h\) is the volume 0? What does this mean geometrically for the three edge vectors?

(d) Let \(T(\mathbf x) = A\mathbf x\) with \(A=\begin{bmatrix} 2 & -1\\ 1 & 3\end{bmatrix}\), and let \(S\) be the triangle in (a). Find the area of \(T(S)\).

(e) A linear transformation \(T:\mathbb R^3\to\mathbb R^3\), \(T(\mathbf x) = B\mathbf x\), maps the unit cube \(C\) onto a solid of volume 12. What are the possible values of \(\det B\)? What is the volume of \(T(T(C))\)? Explain.