Proof.

i.

If \(E\) is obtained from \(I_n\) by the elementary row operation \(r_i\longrightarrow kr_i\), then \(\det E=k\det I_n=k\). If \(E\) is obtained from \(I_n\) by the elementary row operation \(r_i\longleftrightarrow r_j\), then \(\det E=-\det I_n=-1\). If \(E\) is obtained from \(I_n\) by the elementary row operation \(r_i\longrightarrow r_i+kr_j\), then \(\det E=\det I_n=1\). Hence \(\det E\neq 0\).

ii.

If \(E\) is obtained from \(I_n\) by the elementary row operation \(r_i\longleftrightarrow r_j\) then \(E^t=E\), so that the \(\det E=\det E^t\). If \(E\) is obtained from \(I_n\) by the elementary row operation \(r_i\longrightarrow kr_i\), \(k\neq 0\) then \(E=E^t\) and \(\det E=\det E^t\). If \(E\) is obtained from \(I_n\) using the elementary row operation \(r_i\longrightarrow r_i+kr_j\) then \(E\) is lower triangular or upper triangular and so is \(E^t\). The diagonal elements of \(E\) and \(E^t\) are the same, and since \(\det E\) is the product of the diagonal elements it follows that \(\det E^t=\det E\).

□

Proof.

1.

By Theorem 2.6.3(1), \(EA\) is \(A\) with the corresponding row operation applied, and the effect of each operation on the determinant is known: multiplying a row by \(k\) multiplies the determinant by \(k\), interchanging two rows changes its sign, and adding a multiple of one row to another leaves it unchanged. Applying the same three facts to \(E=e(I)\) gives \(\det E=k\), \(-1\) and \(1\) respectively. In each case \(\det (EA)=\det E\cdot \det A\).

2.

Let \(R\) be the reduced echelon form of \(A\), so \(R=E_k\cdots E_1A\) with each \(E_t\) elementary. By (1) applied repeatedly, \[\det R=\det E_k\cdots \det E_1\cdot \det A,\] and every \(\det E_t\) is non-zero. So \(\det A=0\) if and only if \(\det R=0\).

Now \(R\) is a square reduced echelon matrix, so it is either \(I_n\) — in which case \(\det R=1\neq 0\) — or it has a row of zeros, in which case expanding along that row gives \(\det R=0\). By the Section 2 problem on reduced echelon forms, \(R=I_n\) exactly when \(A\) is non-singular. Combining, \(A\) is non-singular if and only if \(\det A\neq 0\).

3.

If \(A\) is singular then by (2) \(\det A=0\); also \(AB\) is then singular (if \(AB\) had an inverse \(C\) then \(BC\) would be a right inverse of \(A\), forcing \(A\) non-singular), so \(\det AB=0=\det A\det B\). If \(A\) is non-singular, write \(A=E_1E_2\cdots E_k\) as a product of elementary matrices — possible because its reduced echelon form is \(I_n\) — and apply (1) \(k\) times to both \(\det (AB)\) and \(\det A\).

4.

By Theorem 3.3.3(2) the system has a non-trivial solution if and only if \(R\neq I_n\), and by the argument in (2) that happens exactly when \(\det A=0\).

5.

Both \(\det A\) and \(\det A^t\) can be written as the same sum over permutations, each term being a product of \(n\) entries, one from each row and one from each column. Transposing exchanges the roles of the row and column indices, which replaces each permutation by its inverse; a permutation and its inverse have the same sign, so the terms — and hence the sums — agree.

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Solution.

1.

(i). \( \begin {vmatrix} -3&2&1\\2&1&-1\\-2&4&5\\ \end {vmatrix} =-33\neq 0\), so it is non-singular.

(ii).

(iii).

2.

(i). \(A= \begin {pmatrix} 2&-1&1&3\\1&2&-1&1\\-1&1&3&2\\4&-3&-1&-1\\ \end {pmatrix} \) is the coefficient matrix. Since \(\det A=-72\), there is a non-trivial solution. (ii). \(A= \begin {pmatrix} 1&1&1&-1\\1&1&-1&-1\\1&-1&-1&-1\\-1&-1&-1&-1\\ \end {pmatrix} \) is the coefficient matrix. Since \(\det A=8\neq 0\) there is no non-trivial solution.