<p>For any square matrix \(\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), where&nbsp;\(a\), \(b\), \(c\) and \(d\) are real numbers, the <strong>determinant</strong> of \(\mathbf{A}\) is denoted by</p>
<p style="text-align: center">\(|\mathbf{A}| = ad - bc .\)</p>
<p>If&nbsp;\(|\mathbf{A}| \neq 0\), the <strong>inverse matrix</strong> of \(\mathbf{A}\) denoted by</p>
<p style="text-align: center">\[\mathbf{A}^{-1} = \frac{1}{|\mathbf{A}|} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix},\]</p>
<p>such that&nbsp;\(\mathbf{A}\mathbf{A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}\), where&nbsp;\(\mathbf{I}\) is the identity matrix.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the inverse of: \(\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 5 & -3 \end{pmatrix}\).</div></li><li><div class="list-item-paragraph">Hence, solve the system of equations<br></div><div class="list-item-paragraph" style="text-align: center">\( \;\,\, 2x + y = 5\)<br>\(5x - 3y = 7\)</div></li></ol>

For any square matrix \(\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), where \(a\), \(b\), \(c\) and \(d\) are real numbers, the **determinant** of \(\mathbf{A}\) is denoted by

If \(|\mathbf{A}| \neq 0\), the **inverse matrix** of \(\mathbf{A}\) denoted by

\[\mathbf{A}^{-1} = \frac{1}{|\mathbf{A}|} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix},\]

such that \(\mathbf{A}\mathbf{A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}\), where \(\mathbf{I}\) is the identity matrix.

Find the inverse of: \(\mathbf{A} = \begin{pmatrix} 2 & 1 \ 5 & -3 \end{pmatrix}\).

Find the inverse of: \(\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 5 & -3 \end{pmatrix}\).

Hence, solve the system of equations\n\[\begin{aligned} 2x + y &= 5 \ 5x - 3y &= 7 \end{aligned}\]

IP3U8 HOT Q1

IP3U8 HOT Q1

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