Mathematics 11 (Punjab Board)
Exercise 4.1 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 4: Matrices and Determinants.
Question 1: If \(A = [a_{ij}]_{m \times n}\), then show that (i) \(I_m A = A\) and (ii) \(A I_n = A\)
Is sawal mein humein dikhana hai ke identity matrix se multiply karne par matrix change nahi hoti. Identity matrix ka kaam numbers mein 1 jaisa hai. Hum matrix multiplication ki definition use karenge aur dekhenge ke har entry waisi hi rehti hai.
(i) \(I_m A = A\)
Let \(A = [a_{ij}]_{m \times n}\) and let \(I_m = [\delta_{ik}]\) be the identity matrix of order \(m\), where \(\delta_{ik} = 1\) if \(i = k\) and \(0\) otherwise.
By the definition of matrix multiplication, the \((i,j)\)-entry of \(I_m A\) is:
\[
(I_m A)_{ij} = \sum_{k=1}^{m} \delta_{ik} \, a_{kj} = \delta_{i1}a_{1j} + \delta_{i2}a_{2j} + \cdots + \delta_{ii}a_{ij} + \cdots + \delta_{im}a_{mj}
\]
Since \(\delta_{ik} = 0\) for every \(k \neq i\) and \(\delta_{ii} = 1\), all terms vanish except the \(i\)-th term:
\[
(I_m A)_{ij} = 1 \cdot a_{ij} = a_{ij}
\]
Thus every entry of \(I_m A\) equals the corresponding entry of \(A\), so \(I_m A = A\).
(ii) \(A I_n = A\)
Similarly, with \(I_n = [\delta_{kj}]\), the \((i,j)\)-entry of \(A I_n\) is:
\[
(A I_n)_{ij} = \sum_{k=1}^{n} a_{ik} \, \delta_{kj} = a_{ij}
\]
Hence \(A I_n = A\). Proved.
Answer: \(I_m A = A\) and \(A I_n = A\) — the identity matrix is the multiplicative identity for matrices.
Question 2: If \(A = \begin{bmatrix} 0 & -1 & 2 \\ 3 & 2 & 1 \\ -1 & 0 & 4 \end{bmatrix}\), \(B = \begin{bmatrix} 2 & 1 & -1 \\ 1 & 2 & 4 \\ -1 & 2 & 1 \end{bmatrix}\) and \(C = \begin{bmatrix} 1 & 0 & -2 \\ -1 & 5 & 0 \\ 3 & 4 & -1 \end{bmatrix}\), find (i) \(A - B\), (ii) \(B - C\), (iii) \((A - B) - C\), (iv) \(A - (B - C)\)
Matrix subtraction mein corresponding entries ko minus karte hain. Is sawal mein hum pehle \(A - B\) aur \(B - C\) nikalenge, phir un results se parts (iii) aur (iv) solve karenge. Note karen ke \((A-B)-C\) aur \(A-(B-C)\) barabar nahi hote — subtraction associative nahi hoti.
(i) \(A - B\)
\[
A - B = \begin{bmatrix} 0-2 & -1-1 & 2-(-1) \\ 3-1 & 2-2 & 1-4 \\ -1-(-1) & 0-2 & 4-1 \end{bmatrix} = \begin{bmatrix} -2 & -2 & 3 \\ 2 & 0 & -3 \\ 0 & -2 & 3 \end{bmatrix}
\]
(ii) \(B - C\)
\[
B - C = \begin{bmatrix} 2-1 & 1-0 & -1-(-2) \\ 1-(-1) & 2-5 & 4-0 \\ -1-3 & 2-4 & 1-(-1) \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 2 & -3 & 4 \\ -4 & -2 & 2 \end{bmatrix}
\]
(iii) \((A - B) - C\)
\[
(A-B) - C = \begin{bmatrix} -2-1 & -2-0 & 3-(-2) \\ 2-(-1) & 0-5 & -3-0 \\ 0-3 & -2-4 & 3-(-1) \end{bmatrix} = \begin{bmatrix} -3 & -2 & 5 \\ 3 & -5 & -3 \\ -3 & -6 & 4 \end{bmatrix}
\]
(iv) \(A - (B - C)\)
\[
A - (B-C) = \begin{bmatrix} 0-1 & -1-1 & 2-1 \\ 3-2 & 2-(-3) & 1-4 \\ -1-(-4) & 0-(-2) & 4-2 \end{bmatrix} = \begin{bmatrix} -1 & -2 & 1 \\ 1 & 5 & -3 \\ 3 & 2 & 2 \end{bmatrix}
\]
Common mistake: Parts (iii) aur (iv) ke answers mukhtalif hain — is se sabit hota hai ke matrix subtraction associative nahi hoti, yani \((A-B)-C \neq A-(B-C)\) aam tor par.
Answer: (i) \(\begin{bmatrix} -2 & -2 & 3 \\ 2 & 0 & -3 \\ 0 & -2 & 3 \end{bmatrix}\), (ii) \(\begin{bmatrix} 1 & 1 & 1 \\ 2 & -3 & 4 \\ -4 & -2 & 2 \end{bmatrix}\), (iii) \(\begin{bmatrix} -3 & -2 & 5 \\ 3 & -5 & -3 \\ -3 & -6 & 4 \end{bmatrix}\), (iv) \(\begin{bmatrix} -1 & -2 & 1 \\ 1 & 5 & -3 \\ 3 & 2 & 2 \end{bmatrix}\)
Question 3: If \(A = \begin{bmatrix} i & 2i \\ -1 & -i \end{bmatrix}\), \(B = \begin{bmatrix} -i & 2i \\ 2i & -i \end{bmatrix}\) and \(C = \begin{bmatrix} 2i & -1 \\ -i & 1 \end{bmatrix}\), show that (i) \((AB)C = A(BC)\) and (ii) \(A(B + C) = AB + AC\)
Yahan complex entries wali matrices hain. Part (i) mein associative law of multiplication aur part (ii) mein distributive law verify karna hai. \(i^2 = -1\) yaad rakhen aur har product ko ehtiyat se calculate karen.
(i) \((AB)C = A(BC)\) — associative law
First compute \(AB\):
\[
AB = \begin{bmatrix} i(-i)+2i(2i) & i(2i)+2i(-i) \\ (-1)(-i)+(-i)(2i) & (-1)(2i)+(-i)(-i) \end{bmatrix} = \begin{bmatrix} -3 & 0 \\ 2+i & -1-2i \end{bmatrix}
\]
Now \((AB)C\):
\[
(AB)C = \begin{bmatrix} -3 & 0 \\ 2+i & -1-2i \end{bmatrix}\begin{bmatrix} 2i & -1 \\ -i & 1 \end{bmatrix} = \begin{bmatrix} -6i & 3 \\ -4+5i & -3-3i \end{bmatrix}
\]
Next compute \(BC\):
\[
BC = \begin{bmatrix} -i(2i)+2i(-i) & -i(-1)+2i(1) \\ 2i(2i)+(-i)(-i) & 2i(-1)+(-i)(1) \end{bmatrix} = \begin{bmatrix} 4 & 3i \\ -5 & -3i \end{bmatrix}
\]
Now \(A(BC)\):
\[
A(BC) = \begin{bmatrix} i & 2i \\ -1 & -i \end{bmatrix}\begin{bmatrix} 4 & 3i \\ -5 & -3i \end{bmatrix} = \begin{bmatrix} -6i & 3 \\ -4+5i & -3-3i \end{bmatrix}
\]
Since both sides equal \(\begin{bmatrix} -6i & 3 \\ -4+5i & -3-3i \end{bmatrix}\), \((AB)C = A(BC)\) is verified.
(ii) \(A(B + C) = AB + AC\) — distributive law
\[
B + C = \begin{bmatrix} -i+2i & 2i-1 \\ 2i-i & -i+1 \end{bmatrix} = \begin{bmatrix} i & -1+2i \\ i & 1-i \end{bmatrix}
\]
\[
A(B+C) = \begin{bmatrix} i & 2i \\ -1 & -i \end{bmatrix}\begin{bmatrix} i & -1+2i \\ i & 1-i \end{bmatrix} = \begin{bmatrix} -3 & i \\ 1-i & -3i \end{bmatrix}
\]
\[
AC = \begin{bmatrix} i & 2i \\ -1 & -i \end{bmatrix}\begin{bmatrix} 2i & -1 \\ -i & 1 \end{bmatrix} = \begin{bmatrix} 0 & i \\ -1-2i & 1-i \end{bmatrix}
\]
\[
AB + AC = \begin{bmatrix} -3+0 & 0+i \\ (2+i)+(-1-2i) & (-1-2i)+(1-i) \end{bmatrix} = \begin{bmatrix} -3 & i \\ 1-i & -3i \end{bmatrix}
\]
Both sides are equal, so \(A(B+C) = AB + AC\) is verified.
Answer: Both the associative law \((AB)C = A(BC)\) and the distributive law \(A(B+C) = AB + AC\) hold for the given matrices.
Question 4: If \(A\) and \(B\) are square matrices of the same order, explain why in general (i) \((A+B)^2 \neq A^2 + 2AB + B^2\), (ii) \((A-B)^2 \neq A^2 - 2AB + B^2\), (iii) \((A+B)(A-B) \neq A^2 - B^2\)
Numbers ke liye \((a+b)^2 = a^2 + 2ab + b^2\) hota hai, lekin matrices ke liye ye formula seedha apply nahi hota. Wajah sirf aik hai: matrix multiplication commutative nahi hoti, yani \(AB \neq BA\). Hum har expression ko expand kar ke dekhenge.
(i)
\[
(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2
\]
Since matrix multiplication is not commutative in general, \(AB \neq BA\), so \(AB + BA \neq 2AB\). Hence \((A+B)^2 \neq A^2 + 2AB + B^2\).
(ii)
\[
(A-B)^2 = (A-B)(A-B) = A^2 - AB - BA + B^2
\]
Again \(AB \neq BA\), so \(-AB - BA \neq -2AB\). Hence \((A-B)^2 \neq A^2 - 2AB + B^2\).
(iii)
\[
(A+B)(A-B) = A^2 - AB + BA - B^2
\]
Since \(AB \neq BA\), the terms \(-AB + BA\) do not cancel. Hence \((A+B)(A-B) \neq A^2 - B^2\).
Common mistake: Numbers wale formulas matrices par lagana board mein aam ghalti hai. Ye formulas sirf tab durust hote hain jab \(AB = BA\) ho.
Answer: All three differ from the number-formulas because \(AB \neq BA\) in general; the correct expansions are \((A+B)^2 = A^2 + AB + BA + B^2\), \((A-B)^2 = A^2 - AB - BA + B^2\), \((A+B)(A-B) = A^2 - AB + BA - B^2\).
Question 5: If \(A = \begin{bmatrix} -1 & 2 & 3 \\ 1 & 0 & 2 \\ 3 & 5 & 3 \end{bmatrix}\), find \(A + A^t\), \(A - A^t\), \(AA^t\), \(A^tA\) and \((A^t)^t\)
Pehle \(A\) ka transpose nikalenge (rows aur columns interchange kar ke). Phir addition, subtraction aur multiplication ke rules apply karenge. Yaad rakhen ke \(A + A^t\) hamesha symmetric aur \(A - A^t\) hamesha skew-symmetric hoti hai.
Transpose of \(A\):
\[
A^t = \begin{bmatrix} -1 & 1 & 3 \\ 2 & 0 & 5 \\ 3 & 2 & 3 \end{bmatrix}
\]
\(A + A^t\):
\[
A + A^t = \begin{bmatrix} -2 & 3 & 6 \\ 3 & 0 & 7 \\ 6 & 7 & 6 \end{bmatrix}
\]
\(A - A^t\):
\[
A - A^t = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & -3 \\ 0 & 3 & 0 \end{bmatrix}
\]
\(AA^t\):
\[
AA^t = \begin{bmatrix} -1 & 2 & 3 \\ 1 & 0 & 2 \\ 3 & 5 & 3 \end{bmatrix}\begin{bmatrix} -1 & 1 & 3 \\ 2 & 0 & 5 \\ 3 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 14 & 5 & 16 \\ 5 & 5 & 9 \\ 16 & 9 & 43 \end{bmatrix}
\]
\(A^tA\):
\[
A^tA = \begin{bmatrix} -1 & 1 & 3 \\ 2 & 0 & 5 \\ 3 & 2 & 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 3 \\ 1 & 0 & 2 \\ 3 & 5 & 3 \end{bmatrix} = \begin{bmatrix} 11 & 13 & 8 \\ 13 & 29 & 21 \\ 8 & 21 & 22 \end{bmatrix}
\]
\((A^t)^t = A = \begin{bmatrix} -1 & 2 & 3 \\ 1 & 0 & 2 \\ 3 & 5 & 3 \end{bmatrix}\)
Common mistake: \(AA^t\) aur \(A^tA\) barabar nahi hote — dono ko alag alag calculate karen. (Note: the printed key's middle entries of \(A^tA\) contain a printing error; the values above are the verified ones.)
Answer: \(A+A^t = \begin{bmatrix} -2 & 3 & 6 \\ 3 & 0 & 7 \\ 6 & 7 & 6 \end{bmatrix}\), \(A-A^t = \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & -3 \\ 0 & 3 & 0 \end{bmatrix}\), \(AA^t = \begin{bmatrix} 14 & 5 & 16 \\ 5 & 5 & 9 \\ 16 & 9 & 43 \end{bmatrix}\), \(A^tA = \begin{bmatrix} 11 & 13 & 8 \\ 13 & 29 & 21 \\ 8 & 21 & 22 \end{bmatrix}\), \((A^t)^t = A\).
Question 6: Solve the matrix equation \(A^2 - 5A + 4I - X = 0\) if \(A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}\)
Equation ko rearrange karne par \(X = A^2 - 5A + 4I\) milta hai. Is liye humein pehle \(A^2\) (matlab \(A \times A\)), phir \(5A\) aur \(4I\) nikalna hoga, aur aakhir mein in sab ko jama/tafreeq kar ke \(X\) hasil karna hoga.
From \(A^2 - 5A + 4I - X = 0\), we get \(X = A^2 - 5A + 4I\).
Compute \(A^2 = A \cdot A\):
\[
A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}
\]
Compute \(5A\) and \(4I\):
\[
5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}, \qquad 4I = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{bmatrix}
\]
Now \(X = A^2 - 5A + 4I\):
\[
X = \begin{bmatrix} 5-10+4 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+4 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+4 \end{bmatrix} = \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix}
\]
Answer: \(X = \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix}\)
Question 7: If \(A\) and \(B\) are two matrices such that \(AB = B\) and \(BA = A\), show that \(A^2 + B^2 = A + B\)
Diye gaye relations mein \(A\) ko \(BA\) se aur \(B\) ko \(AB\) se replace karen ge, phir associative law use kar ke simplify karenge. Ye sawal associative property ki samajh ko test karta hai.
Given: \(AB = B\) and \(BA = A\).
Consider \(A^2\). Since \(A = BA\):
\[
A^2 = A \cdot A = A(BA)
\]
By the associative law of matrix multiplication, \(A(BA) = (AB)A\):
\[
A^2 = (AB)A
\]
But \(AB = B\), so \(A^2 = BA\). And since \(BA = A\):
\[
A^2 = A
\]
Similarly for \(B^2\). Since \(B = AB\):
\[
B^2 = B \cdot B = B(AB) = (BA)B = AB = B
\]
Adding the two results:
\[
A^2 + B^2 = A + B
\]
Hence proved.
Answer: From the given conditions, \(A^2 = A\) and \(B^2 = B\), so \(A^2 + B^2 = A + B\). Proved.
