4.1 2 by 2 Determinants

Definition 4.1.1. Let \( A= \begin {pmatrix} a&b\\c&d\\ \end {pmatrix} \) be any 2 by 2 matrix. The determinant of \(A\) is the number \(ad-bc\). This is denoted by \(\det A=ad-bc\) or \(|A|=ad-bc\).

Example 4.1.2. If \(A= \begin {pmatrix} 1&2\\3&4\\ \end {pmatrix} \) then the \(\det A=(1\times 4)-(2\times 3)=-2\).

Determinats of 2 by 2 matrices satisfy the following properties:

Theorem 4.1.3.

1.: If \(A\) is any 2 by 2 matrix, then \(\det A=\det A^t\).
2.: \(\det \begin {pmatrix} ka&kb\\c&d\\ \end {pmatrix} =k\det \begin {pmatrix} a&b\\c&d\\ \end {pmatrix} \) for any real number \(k\).
3.: If \(A= \begin {pmatrix} a&b\\c&d \end {pmatrix} \) and \(B= \begin {pmatrix} c&d\\a&b\\ \end {pmatrix} \), then \(\det B=-\det A\).
4.: \( \begin {pmatrix} a&a\\b&b\\ \end {pmatrix} =0\).
5.: \( \det \begin {pmatrix} a&ka\\b&kb\\ \end {pmatrix} =0\).

Proof.

1.: Let \( A= \begin {pmatrix} a&b\\c&d\\ \end {pmatrix} \). Then \(A^t= \begin {pmatrix} a&c\\b&d\\ \end {pmatrix} \). Therefore \(\det A=ad-bc\) and \(\det A^t=ad-bc\), so that \(\det A=\det A^t\).
2.: \(\det \begin {pmatrix} ka&kb\\c&d\\ \end {pmatrix} =kad-kbc=k(ad-bc)= k\det \begin {pmatrix} a&b\\c&d\\ \end {pmatrix} \).
3.: \(\det \begin {pmatrix} a&b\\c&d\\ \end {pmatrix} =ad-bc \) and \(\det \begin {pmatrix} c&d\\a&b\\ \end {pmatrix} bc-cd=-(ad-bc)=-\det \begin {pmatrix} a&b\\c&d\\ \end {pmatrix} \).
i.e. \(\det B=-\det A\).
4.: \(\det \begin {pmatrix} a&a\\b&b\\ \end {pmatrix} =ab-ab=0 \).
5.: \(\det \begin {pmatrix} a&ka\\b&kb\\ \end {pmatrix} =kab-kab=0\).

The properties of 2 by 2 determinants in Theorem 4.1.3 can be used to calculate the determinant of a 2 by 2 matrix in terms of simpler determinants. □

Example 4.1.4.

1.: Let \(A= \begin {pmatrix} -1&2\\3&4\\ \end {pmatrix} \) and \(B= \begin {pmatrix} -1&2\\12&16\\ \end {pmatrix} \). Then the \(\det A=-10\) and by Theorem 4.1.3 \(\det B=4\det A=-40\).
2.: If \( \begin {pmatrix} xyz&-x^2y\\yz^2&3z\\ \end {pmatrix} \) then by Theorem 4.1.3 \begin {align*} \det A &=\det \begin {pmatrix} xyz&-x^2y\\yz^2&3z\\ \end {pmatrix} =x\det \begin {pmatrix} yz&-xy\\yz^2&3z\\ \end {pmatrix} =xy\det \begin {pmatrix} z&-x\\yz^2&3z\\ \end {pmatrix}\\\\ &=xyz\det \begin {pmatrix} z&-x\\yz&3\\ \end {pmatrix} =xyz^2\det \begin {pmatrix} 1&-x\\y&3\\ \end {pmatrix}\\\\ &=xyz^2(3+xy)\\ \end {align*}

If two or more 2 by 2 determinants have a common row (or column), then their sum can be expressed as a single determinant.
For example \[\det \begin {pmatrix} a_1&x_1\\a_2&x_2\\ \end {pmatrix} +\det \begin {pmatrix} a_1&y_1\\a_2&y_2\\ \end {pmatrix} +\det \begin {pmatrix} a_1&z_1\\a_2&z_2\\ \end {pmatrix} =\det \begin {pmatrix} a_1&x_1+y_1+z_1\\a_2&x_2+y_2+z_2\\ \end {pmatrix} \] You can verify this identity by expanding both sides. Here the common column is \( \begin {pmatrix} a_1\\a_2 \end {pmatrix} \). This property implies that the value of a 2 by 2 determinant is unaltered if you add to one row (or column) a multiple of the other row (or column).

From the discussion above, the properties of 2 by 2 determinants can be stated in terms of elementary row (or column) operations.

i.: If \(B\) is the matrix obtained from \(A\) by applying the elementary row operation \(r_i\longrightarrow kr_i\), then \(\det B=k\det A\).
ii.: If \(B\) is the matrix obtained from \(A\) by applying the elementary row operation \(r_i\longleftrightarrow r_j\), then \(\det B=-\det A\).
iii.: If \(B\) is the matrix obtained from \(A\) by applying the elementary row operation \(r_i\longrightarrow r_i+kr_j\), then \(\det A=\det B\).

Example 4.1.5. \begin {align*} \det \begin {pmatrix} 14&-11\\3&-17\\ \end {pmatrix} &=\det \begin {pmatrix} -1&74\\3&-17\\ \end {pmatrix} =\det \begin {pmatrix} -1&74\\0&205\\ \end {pmatrix}\\\\ &=205\det \begin {pmatrix} -1&74\\0&1\\ \end {pmatrix} =205\det \begin {pmatrix} -1&0\\0&1\\ \end {pmatrix} =-205\det \begin {pmatrix} 1&0\\0&1\\ \end {pmatrix}\\\\ &=-205\\ \end {align*}

Theorem 4.1.6. If \(A\) and \(B\) are any 2 by 2 matrices then \(\det AB=\det A\det B\).

Proof. Let \(A= \begin {pmatrix} a_1&a_2\\a_3&a_4\\ \end {pmatrix} \) and \(B= \begin {pmatrix} b_1&b_2\\b_3&b_4\\ \end {pmatrix} \). Then \(\det AB= \begin {bmatrix} \begin {pmatrix} a_1&a_2\\a_3&a_4\\ \end {pmatrix} & \begin {pmatrix} b_1&b_2\\b_3&b_4\\ \end {pmatrix}\\ \end {bmatrix} \), thus \begin {align*} \det \begin {pmatrix} a_1b_1+a_2b_3&a_1b_2+a_2b_4\\a_3b_1+a_4b_3&a_3b_2+a_4b_4\\ \end {pmatrix} &=(a_1b_1+a_2b_2)(a_3b_2+a_4b_4)-(a_1b_2+a_2b_4)(a_3b_1+a_4b_3)\\ &=a_1a_4b_1b_4-a_1a_4b_2b_3+a_2a_3b_2b_3-a_2a_3b_1b_4\\ &=a_1a_4(b_1b_4-b_2b_3)-a_2a_3(b_1b_4-b_2b_3)\\ &=(a_1a_4-a_2a_3)(b_1b_4-b_2b_3)\\ &=\det A\det B\\ \end {align*} □

Example 4.1.7. Given that \(A= \begin {pmatrix} -1&-2\\3&2\\ \end {pmatrix} \) and \(B= \begin {pmatrix} 1&2\\3&4\\ \end {pmatrix} \) verify that \(\det AB=\det A\det B\).

Proof. \(\det AB=\det \begin {bmatrix} \begin {pmatrix} -1&-2\\3&2\\ \end {pmatrix} & \begin {pmatrix} 1&2\\3&4\\ \end {pmatrix} \end {bmatrix} =\det \begin {pmatrix} -5&-6\\9&14\\ \end {pmatrix} =-16 \).

\(\det A=\det \begin {pmatrix} -1&-2\\3&2\\ \end {pmatrix} =8. \) and \(\det B=\det \begin {pmatrix} 1&2\\3&4\\ \end {pmatrix} =-2 \).

Therefore, \(\det A\det B=-16=\det AB\). □

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