Notes On Properties of Determinants - II - CBSE Class 12 Maths
Property 4: If all the elements of a row (or a column) of a square matrix are multiplied by a number, k, then the value of the determinant of the matrix obtained is k times the determinant of the given matrix. Verification: Let A = $\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right|$ |A| = a1(b2c3 - b3c2) - a2(b1c3 - b3c1) + a3(b1c2 - b2c1) |A| = a1b2c3 - a1b3c2 - a2b1c3 + a2b3c1 + a3b1c2 - a3b2c1 Let the elements of row 1 be multiplied by a constant, k. Then B = $\left[\begin{array}{ccc}{\text{ka}}_{\text{1}}& {\text{ka}}_{\text{2}}& {\text{ka}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$ Expanding along the first row, we get |B| = ka1(b2c3 - b3c2) - ka2(b1c3 - b3c1) + ka3(b1c2 - b2c1) ⇒|B| = k|A| i.e. the determinant of matrix B is k times the determinant of matrix A. Note: If the elements of a row (or a column) of a square matrix are k times the elements of another row (or column), then the value of its determinant is zero. If M and N are two square matrices such that M = kN, where k is a real number, then |M| = k|N|, where n is the order of the matrices M and N. Property 5: If the elements of a row (or a column) of the determinant of a matrix are expressed as the sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants. The elements in the first column of a determinant are expressed as the sum of two terms. $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+l}}_{\text{1}}\text{}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{l}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$     Consider $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+l}}_{\text{1}}\text{}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{}$. Expanding the determinant along the first column, we get =(a1 + l1)(b2c3 - b3c2) - (a2 + l2)(b1c3 - b3c1) + (a3 + l3)(b1c2 - b2c1) =a1(b2c3 - b3c2) + l1(b2c3 - b3c2) - a2(b1c3 - b3c1) - l2(b1c3 - b3c1) + a3(b1c2 - b2c1) + l3(b1c2 - b2c1) $\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{l}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$  Note: Similar results can be obtained by considering the other rows and columns. Property 6: If the elements of a row (or column) of a square matrix are added with k times (where k is a constant) the corresponding elements of another row (or column), then the value of the determinant of the given matrix does not change. Or The value of a determinant remains unchanged if we apply the operation, Ri → Ri + kRj or Ci → Ci + kCj. Let A = $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right]$      Applying operation C1 → C1 + kC2, we have            $\left[\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+kb}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right]$ = (B)say |B| = $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+kb}}_{\text{1}}\text{}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{}$ $\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{kb}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$   [By property 5] The second determinant, each element of the first column is multiplied by k. $\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+k}\left|\begin{array}{ccc}{\text{b}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{b}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{b}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$   [By property 4] Here the second determinant whose first and second columns are identical. = |A| + k(0)  [By property 3] = |A| Hence, $\text{=}$$\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$ Note: Similar results can be obtained by considering the other rows and columns.

#### Summary

Property 4: If all the elements of a row (or a column) of a square matrix are multiplied by a number, k, then the value of the determinant of the matrix obtained is k times the determinant of the given matrix. Verification: Let A = $\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{a}}_{\text{2}}& {\text{a}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right|$ |A| = a1(b2c3 - b3c2) - a2(b1c3 - b3c1) + a3(b1c2 - b2c1) |A| = a1b2c3 - a1b3c2 - a2b1c3 + a2b3c1 + a3b1c2 - a3b2c1 Let the elements of row 1 be multiplied by a constant, k. Then B = $\left[\begin{array}{ccc}{\text{ka}}_{\text{1}}& {\text{ka}}_{\text{2}}& {\text{ka}}_{\text{3}}\\ {\text{b}}_{\text{1}}& {\text{b}}_{\text{2}}& {\text{b}}_{\text{3}}\\ {\text{c}}_{\text{1}}& {\text{c}}_{\text{2}}& {\text{c}}_{\text{3}}\end{array}\right]$ Expanding along the first row, we get |B| = ka1(b2c3 - b3c2) - ka2(b1c3 - b3c1) + ka3(b1c2 - b2c1) ⇒|B| = k|A| i.e. the determinant of matrix B is k times the determinant of matrix A. Note: If the elements of a row (or a column) of a square matrix are k times the elements of another row (or column), then the value of its determinant is zero. If M and N are two square matrices such that M = kN, where k is a real number, then |M| = k|N|, where n is the order of the matrices M and N. Property 5: If the elements of a row (or a column) of the determinant of a matrix are expressed as the sum of two or more terms, then the determinant can be expressed as the sum of two or more determinants. The elements in the first column of a determinant are expressed as the sum of two terms. $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+l}}_{\text{1}}\text{}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{l}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$     Consider $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+l}}_{\text{1}}\text{}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{}$. Expanding the determinant along the first column, we get =(a1 + l1)(b2c3 - b3c2) - (a2 + l2)(b1c3 - b3c1) + (a3 + l3)(b1c2 - b2c1) =a1(b2c3 - b3c2) + l1(b2c3 - b3c2) - a2(b1c3 - b3c1) - l2(b1c3 - b3c1) + a3(b1c2 - b2c1) + l3(b1c2 - b2c1) $\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{l}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{l}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{l}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$  Note: Similar results can be obtained by considering the other rows and columns. Property 6: If the elements of a row (or column) of a square matrix are added with k times (where k is a constant) the corresponding elements of another row (or column), then the value of the determinant of the given matrix does not change. Or The value of a determinant remains unchanged if we apply the operation, Ri → Ri + kRj or Ci → Ci + kCj. Let A = $\left[\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right]$      Applying operation C1 → C1 + kC2, we have            $\left[\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+kb}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right]$ = (B)say |B| = $\left|\begin{array}{ccc}{{\text{a}}_{\text{1}}\text{+kb}}_{\text{1}}\text{}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {{\text{a}}_{\text{2}}\text{+kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {{\text{a}}_{\text{3}}\text{+kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{}$ $\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+}\left|\begin{array}{ccc}{\text{kb}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{kb}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{kb}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$   [By property 5] The second determinant, each element of the first column is multiplied by k. $\text{=}\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|\text{+k}\left|\begin{array}{ccc}{\text{b}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{b}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{b}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$   [By property 4] Here the second determinant whose first and second columns are identical. = |A| + k(0)  [By property 3] = |A| Hence, $\text{=}$$\left|\begin{array}{ccc}{\text{a}}_{\text{1}}& {\text{b}}_{\text{1}}& {\text{c}}_{\text{1}}\\ {\text{a}}_{\text{2}}& {\text{b}}_{\text{2}}& {\text{c}}_{\text{2}}\\ {\text{a}}_{\text{3}}& {\text{b}}_{\text{3}}& {\text{c}}_{\text{3}}\end{array}\right|$ Note: Similar results can be obtained by considering the other rows and columns.

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